Past Exam Questions, Solutions, and Explanations on Circle Theorems

$ (i) \\[3ex] \angle ABC = 90^\circ...\angle \:\:in\:\:a\:\:semicircle \\[3ex] \angle ABC = \angle DBA + \angle DBC ...diagram \\[3ex] 90 = \angle DBA + 46 \\[3ex] \angle DBA + 46 = 90 \\[3ex] \angle DBA = 90 - 46 \\[3ex] \angle DBA = 44^\circ \\[3ex] (ii) \\[3ex] \angle DAC = \angle DBC = 46^\circ ... \angle s \:\:in\:\:the\:\:same\:\:segment\;\;are\;\;equal \\[3ex] (iii) \\[3ex] \angle BCE = \angle A ... exterior\;\; \angle \:\:of\:\:a\:\:cyclic\:\:quadrilateral = interior\;\;opposite\;\;\angle \\[3ex] \angle A = \angle DAC + \angle CAB ...diagram \\[3ex] \angle A = 46 + 28 \\[3ex] \angle A = 74 \\[3ex] \therefore \angle BCE = 74^\circ $

$ (a) \\[3ex] \angle AOB = 2 * \angle ACB ...\angle\;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] \angle AOB = 2(40) \\[3ex] \angle AOB = 80^\circ \\[3ex] (b) \\[3ex] \triangle ABO \;\;is\;\;an\;\;isosceles\;\;\triangle \\[3ex] because\;\; |OA| = |OB| \\[3ex] ... same\;\;radius\;\;of\;\;the\;\;circle = same\;\;side\;\;of\;\;\triangle \\[3ex] and \\[3ex] because\;\;\angle OAB = \angle OBA ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle \\[3ex] (c) \\[3ex] \angle ABO = \angle BAO = p ..base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle \\[3ex] \\[3ex] \angle ABO + \angle BAO + \angle AOB = 180^\circ ... sum\;\;of\;\;\angle s\;\;in\;\;a\;\;\triangle \\[3ex] p + p + 80 = 180 \\[3ex] 2p = 180 - 80 \\[3ex] 2p = 100 \\[3ex] p = \dfrac{100}{2} \\[5ex] p = 50 \\[3ex] \therefore \angle ABO = \angle BAO = 50^\circ $

$ Obtuse\;\angle UOW = 2(x) = 2x ...\angle \:\:at\:\:center\:\:is\:\:twice\:\:\angle\:\:at\:\:circumference \\[3ex] 8x + 2x = 360 ...\angle s\:\:at\:\:a\:\:point \\[3ex] 10x = 360 \\[3ex] x = \dfrac{360}{10} \\[5ex] x = 36 \\[3ex] Reflex\;8x = 2 * \angle UVW ...\angle \:\:at\:\:center\:\:is\:\:twice\:\:\angle\:\:at\:\:circumference \\[3ex] 8x = 8 * 36 = 288 \\[3ex] \implies \\[3ex] 288 = 2 * \angle UVW \\[3ex] \angle UVW = \dfrac{288}{2} \\[5ex] \angle UVW = 144^\circ \\[3ex] OR \\[3ex] \angle UVW + \angle UYW = 180^\circ ...interior\:\:opposite\;\;\angle s\:\:of\:\:a\:\:cyclic\:\:quadrilateral\;\;are\;\;supplementary \\[3ex] \angle UYW = x = 36 \\[3ex] \angle UVW + 36 = 180 \\[3ex] \angle UVW = 180 - 36 \\[3ex] \angle UVW = 144^\circ $

$ \angle O = 235 + y = 360 ...\angle s \:\:at\:\:a\:\:point \\[3ex] y = 360 - 235 \\[3ex] y = 125 \\[3ex] y = 2x ...\angle \:\:at\:\:center\:\:is\:\:twice\:\:\angle\:\:at\:\:circumference \\[3ex] \implies \\[3ex] 125 = 2x \\[3ex] 2x = 125 \\[3ex] x = \dfrac{125}{2} \\[5ex] x = 62\dfrac{1}{2}^\circ $

Construction: Join the chord from point P to point R
$ \angle OPR = \angle ORP = y ...base\:\:\angle s \:\:of\:\:isosceles\:\: \triangle OPR \\[3ex] \underline{\triangle OPR} \\[3ex] \angle OPR + \angle ORP + \angle POR = 180^\circ ... sum\:\:of\:\:\angle s\:\:of\:\;\triangle OPR \\[3ex] y + y + 108 = 180 \\[3ex] 2y = 180 - 108 \\[3ex] 2y = 72 \\[3ex] y = \dfrac{72}{2} \\[5ex] y = 36 \\[3ex] \therefore \angle OPR = \angle ORP = 36^\circ \\[3ex] \angle POR = 2 * \angle PKR ...\angle \:\:at\:\:center\:\:is\:\:twice\:\:\angle\:\:at\:\:circumference \\[3ex] 108 = 2 * \angle PKR \\[3ex] \angle PKR = \dfrac{108}{2} \\[5ex] \angle PKR = 54 \\[3ex] \underline{\triangle PKR} \\[3ex] \angle KPR + \angle KRP + \angle PKR = 180^\circ ... sum\:\:of\:\:\angle s\:\:of\:\;\triangle PKR \\[3ex] \angle KPR = \angle OPK + \angle OPR...diagram \\[3ex] \angle KPR = \angle OPK + 36 \\[3ex] \angle KRP = 20 + \angle ORP...diagram \\[3ex] \angle KRP = 20 + 36 \\[3ex] \angle KRP = 56^\circ \\[3ex] \implies \\[3ex] \angle OPK + 36 + 56 + 54 = 180 \\[3ex] \angle OPK + 146 = 180 \\[3ex] \angle OPK = 180 - 146 \\[3ex] \angle OPK = 34^\circ $

WY is a diameter ... True
This is because it is a chord that passes through the centre of the circle.

WX is a radius. ... False
It does not pass through the centre of the circle.

The shaded section is a sector. ... False
It is a segment. It is the minor segment.
A sector will also include the area enclosed by both radii: CW and CX
In other words, the minor sector will be the area of ▵WCX and the area of minor segment WX

Arc XY is part of the circumference. ... True
Arcs of a circle are part of the circumference of the circle. $

$ (i) \\[3ex] \angle OJH = 90^\circ ... radius\;OJ \perp tangent\;JH\:\:at\:\:point\:\:of\:\:contact\;J \\[3ex] (ii) \\[3ex] \angle OGH = 90^\circ ... radius\;OG \perp tangent\;GH\:\:at\:\:point\:\:of\:\:contact\;G \\[3ex] \underline{Quadrilateral\:\:JOGH} \\[3ex] \angle J + \angle O + \angle G + \angle H = 360 ...sum\:\:of\:\:interior\:\:\angle s\:\:of\:\:a\:\:Quadrilateral \\[3ex] 90 + \angle O + 90 + 48 = 360 \\[3ex] 228 + \angle O = 360 \\[3ex] \angle O = 360 - 228 \\[3ex] \angle O = 132 \\[3ex] \angle O = \angle JOG = 132^\circ \\[3ex] (iii) \\[3ex] Obtuse\;\;\angle JOG = 2 * \angle JKG ...\angle \:\:at\:\:center\:\:is\:\:twice\:\:\angle\:\:at\:\:circumference \\[3ex] 132 = 2 * \angle JKG \\[3ex] 2 * \angle JKG = 132 \\[3ex] \angle JKG = \dfrac{132}{2} \\[5ex] \angle JKG = 66^\circ \\[3ex] (iv) \\[3ex] Reflex\;\;\angle JOG + Obtuse\;\;\angle JOG = 360^\circ ...\angle s\:\:at\:\:a\:\:point \\[3ex] Reflex\;\;\angle JOG + 132 = 360 \\[3ex] Reflex\;\;\angle JOG = 360 - 132 \\[3ex] Reflex\;\;\angle JOG = 228 \\[3ex] Reflex\;\;\angle JOG = 2 * \angle JLG ...\angle \:\:at\:\:center\:\:is\:\:twice\:\:\angle\:\:at\:\:circumference \\[3ex] 228 = 2 * \angle JLG \\[3ex] 2 * \angle JLG = 228 \\[3ex] \angle JLG = \dfrac{228}{2} \\[5ex] \angle JLG = 114^\circ \\[3ex] OR \\[3ex] \underline{Cyclic\;\;Quadrilateral\:\:KJLG} \\[3ex] \angle K + \angle L = 180^\circ ...sum\:\:of\:\:interior\:\:opposite\:\:\angle s\:\:of\:\:a\:\:Cyclic\:\:Quadrilateral \\[3ex] 66 + \angle L = 180 \\[3ex] \angle L = 180 - 66 \\[3ex] \angle L = 114 \\[3ex] \angle L = \angle JLG = 114^\circ $

$ Obtuse\;\;\angle POR = 2(65) ...\angle \:\:at\:\:center\:\:is\:\:twice\:\:\angle\:\:at\:\:circumference \\[3ex] Obtuse\;\;\angle POR = 130^\circ \\[3ex] x + Obtuse\;\;\angle POR = 360^\circ ...\angle s\:\:at\:\:a\:\:point \\[3ex] x + 130 = 360 \\[3ex] x = 360 - 130 \\[3ex] x = 230^\circ $

$ Reflex\:\:P\hat{O}S = 252^\circ \\[3ex] \angle SQR = 79^\circ \\[3ex] \angle QRS = x \\[3ex] Reflex\:\:P\hat{O}S + Obtuse\:\:P\hat{O}S = 360^\circ ...\angle s\:\:at\:\:a\:\:point \\[3ex] 252 + Obtuse\:\:P\hat{O}S = 360 \\[3ex] Obtuse\:\:P\hat{O}S = 360 - 252 \\[3ex] Obtuse\:\:P\hat{O}S = 108^\circ \\[3ex] Obtuse\:\:P\hat{O}S = 2 * \angle PQS ...\angle \:\:at\:\:center\:\:is\:\:twice\:\:\angle\:\:at\:\:circumference \\[3ex] 108 = 2 * \angle PQS \\[3ex] \angle PQS = \dfrac{108}{2} \\[5ex] \angle PQS = 54^\circ \\[3ex] $ Construction: Join the chord from point P to point S
$ \underline{\triangle PQS} \\[3ex] \angle QPS = \angle QSP = y...base\:\:\angle s\:\:of\:\:isosceles\:\:\triangle PQS \\[3ex] \angle QPS + \angle QSP + \angle PQS = 180^\circ ... sum\:\:of\:\:\angle s\:\:of\:\;\triangle PQS \\[3ex] \implies \\[3ex] y + y + 54 = 180 \\[3ex] 2y = 180 - 54 \\[3ex] 2y = 126 \\[3ex] y = \dfrac{126}{2} \\[5ex] y = 63 \\[3ex] \therefore \angle QPS = \angle QSP = 63^\circ \\[3ex] \angle QSR = \angle QPS = 63^\circ ...\angle\;\;between\;\;tangent\;SR\;\;and\;\;chord\;SQ = \angle\;\;in\;\;alternate\;\;segment \\[3ex] \underline{\triangle QRS} \\[3ex] \angle QSR + \angle SQR + \angle QRS = 180^\circ ... sum\:\:of\:\:\angle s\:\:of\:\;\triangle QRS \\[3ex] \implies \\[3ex] 63 + 79 + \angle QRS = 180 \\[3ex] 142 + \angle QRS = 180 \\[3ex] \angle QRS = 180 - 142 \\[3ex] \angle QRS = 38^\circ $

(i) Line OP is a radius

(ii) Line QR is a chord

(iii) Lines OP and QR are parallel because of the similar single arrows

$ \underline{\triangle PSQ} \\[3ex] \angle PSQ = 50^\circ ...\angle \:\:between\:\:tangent\;PT\:\:and\:\:chord\;PQ = \angle\:\:in\:\:alternate\:\:segment \\[3ex] \angle PSQ = \angle PQS = 50 ...base\:\: \angle s\:\:of\:\:isosceles\:\: \triangle PSQ \\[3ex] \angle PSQ + \angle PQS + \angle SPQ = 180 ...sum\:\:of\:\:\angle s\:\:of\:\;\triangle PSQ \\[3ex] 50 + 50 + \angle SPQ = 180 \\[3ex] 100 + \angle SPQ = 180 \\[3ex] \angle SPQ = 180 - 100 \\[3ex] \angle SPQ = 80 \\[3ex] \underline{Cyclic\:\:Quadrilateral\:\:SPQR} \\[3ex] \angle P + \angle R = 180 ...sum\:\:of\:\:interior\:\:opposite\:\:\angle s\:\:of\:\:a\:\:cyclic\:\:Quad \\[3ex] 80 + \angle R = 180 \\[3ex] \angle R = 180 - 80 \\[3ex] \angle R = \angle QRS = 100^\circ $

$ \angle OCA = \angle OAC = 48^\circ ...base\:\: \angle s\:\:of\:\:isosceles\:\: \triangle OAC \\[3ex] \angle ACB = 90^\circ ...\angle \:\:in\:\:a\:\:semicircle \\[3ex] \underline{\triangle ABC} \\[3ex] \angle CAB + \angle ABC + \angle ACB = 180^\circ ...sum\:\:of\:\:\angle s\:\:of\:\;\triangle ABC \\[3ex] \angle CAB = \angle OAC = 48^\circ ...diagram \\[3ex] \implies \\[3ex] 48 + \angle ABC + 90 = 180 \\[3ex] 138 + \angle ABC = 180 \\[3ex] \angle ABC = 180 - 138 \\[3ex] \angle ABC = 42^\circ $

$ (i) \\[3ex] H\hat{K}L = \angle HJL = 20^\circ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] (ii) \\[3ex] \underline{\triangle JOK} \\[3ex] \angle OKJ = \angle OJK = 50^\circ ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle JOK \\[3ex] \angle OKJ + \angle OJK + \angle JOK = 180^\circ ...sum\:\:of\:\:\angle s\:\:of\:\;\triangle JOK \\[3ex] 50 + 50 + \angle JOK = 180 \\[3ex] 100 + \angle JOK = 180 \\[3ex] \angle JOK = 180 - 100 \\[3ex] \angle JOK = 80^\circ \\[3ex] (iii) \\[3ex] \angle HJK = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] \underline{\triangle HJK} \\[3ex] \angle JHK + \angle HJK + \angle JKH = 180^\circ ...sum\:\:of\:\:\angle s\:\:of\:\;\triangle HJK \\[3ex] \angle JKH + 90 + 50 = 180 \\[3ex] \angle JHK + 140 = 180 \\[3ex] \angle JHK = 180 - 140 \\[3ex] \angle JHK = 40^\circ \\[5ex] OR \\[3ex] Another\;\;Approach\;\;to\;\;solving\;\;the\;\;question\;\;(longer\;\;approach) \\[3ex] (i) \\[3ex] \angle HJK = 90^\circ ...\angle \:\:in\:\:a\:\:semicircle \\[3ex] \angle HJK = \angle HJL + \angle LJK...diagram \\[3ex] 90 = 20 + \angle LJK \\[3ex] \angle LJK = 90 - 20 \\[3ex] \angle LJK = 70 \\[3ex] \angle LHK = \angle LJK ...\angle s \:\:on\:\:a\:\:straight\:\:line \\[3ex] \rightarrow \angle LHK = 70 \\[3ex] \angle HLK = 90^\circ ...\angle \:\:in\:\:a\:\:semicircle \\[3ex] \underline{\triangle HKL} \\[3ex] \angle LHK + \angle HLK + \angle HKL = 180 ...sum\:\:of\:\:\angle s\:\:of\:\:a\:\:\triangle \\[3ex] 70 + 90 + \angle HKL = 180 \\[3ex] 160 + \angle HKL = 180 \\[3ex] \angle HKL = 180 - 160 \\[3ex] \angle HKL = 20^\circ \\[3ex] (ii) \\[3ex] \angle OKJ = \angle OJK ...base\:\: \angle s\:\:of\:\:isosceles\:\: \triangle \\[3ex] \rightarrow \angle OJK = 50 \\[3ex] \underline{\triangle JOK} \\[3ex] \angle OKJ + \angle OJK + \angle JOK = 180 ...sum\:\:of\:\:\angle s\:\:of\:\:a\:\:\triangle \\[3ex] 50 + 50 + \angle JOK = 180 \\[3ex] 100 + \angle JOK = 180 \\[3ex] \angle JOK = 180 - 100 \\[3ex] \angle JOK = 80^\circ \\[3ex] (iii) \\[3ex] \underline{\triangle JHK} \\[3ex] \angle HJK + \angle JKH + \angle JHK = 180 ...sum\:\:of\:\:\angle s\:\:of\:\:a\:\:\triangle \\[3ex] 90 + 50 + \angle JHK = 180 \\[3ex] 140 + \angle JHK = 180 \\[3ex] \angle JHK = 180 - 140 \\[3ex] \angle JHK = 40^\circ $

$ \angle TQP = \angle QTR + \angle QRT \\[3ex] ...exterior\:\: \angle \:\:of\:\:a\:\: \triangle = sum\;\;of\;\;two\;\;interior\;\;opposite\;\;\angle s \\[3ex] \angle TQP = 31 + 58 \\[3ex] \angle TQP = 89^\circ \\[3ex] \angle TSP = \angle TQP ... \angle s \:\:in\:\:the\:\:same\:\:segment\;\;are\;\;equal \\[3ex] \therefore \angle TSP = 89^\circ $

$ x = y + 17...eqn.(1) ...\angle \:\:between\:\:tangent\:PZT\;\:and\:\:chord\;XZ = \angle\:\:in\:\:the\;\;alternate\:\:segment \\[3ex] 2x - 43 = y...eqn.(2) ...\angle \:\:between\:\:tangent\:PZT\:\;and\:\:chord\;YZ = \angle\:\:in\:\:the\;\;alternate\:\:segment \\[3ex] Substitute\:\:eqn.(1)\:\:into\:\:eqn.(2) \\[3ex] 2(y + 17) - 43 = y \\[3ex] 2y + 34 - 43 = y \\[3ex] 2y - y = 43 - 34 \\[3ex] y = 9^\circ $

$ \angle ORP = 90^\circ \\[3ex] ... radius\;OR \perp tangent\;RP\:\:at\:\:point\:\:of\:\:contact\;R \\[3ex] \angle ORP = \angle ORQ + \angle QRP...diagram \\[3ex] 90 = \angle ORQ + 34 \\[3ex] \angle ORQ = 90 - 34 \\[3ex] \angle ORQ = 56^\circ \\[3ex] \underline{\triangle ROQ} \\[3ex] \angle ORQ = \angle OQR = 56^\circ ...base\:\: \angle s\:\:of\:\:isosceles\:\: \triangle ROQ \\[3ex] \angle ORG + \angle OQR + \angle ROQ = 180 ...sum\:\:of\:\:\angle s\:\:of\:\;\triangle ORG \\[3ex] \implies \\[3ex] 56 + 56 + x = 180 \\[3ex] 112 + x = 180 \\[3ex] x = 180 - 112 \\[3ex] x = 68^\circ $

We can solve (i) in at least two ways

$ (i) \\[3ex] \underline{First\:\:Method} \\[3ex] \angle AOD = 2 * \angle ACD ...\angle \:\:at\:\:center\:\:is\:\:twice\:\:\angle\:\:at\:\:circumference \\[3ex] 114 = 2 * \angle ACD \\[3ex] \angle ACD = \dfrac{114}{2} \\[5ex] \angle ACD = 57^\circ \\[3ex] \underline{Second\:\:Method} \\[3ex] \angle ODG = 90 ... radius\;OD \perp tangent\;EDG\:\:at\:\:point\:\:of\:\:contact\;D \\[3ex] \angle ODA = \angle OAD = y ...base\:\: \angle s\:\:of\:\:isosceles\:\: \triangle \\[3ex] \underline{\triangle OAD} \\[3ex] \angle OAD + \angle ODA + \angle AOD = 180^\circ ...sum\:\:of\:\:\angle s\:\:of\:\:\triangle OAD \\[3ex] y + y + 114 = 180 \\[3ex] 2y + 114 = 180 \\[3ex] 2y = 180 - 114 \\[3ex] 2y = 66 \\[3ex] y = \dfrac{66}{2} \\[5ex] y = 33 \\[3ex] \angle ADE + \angle ODA + \angle ODG = 180^\circ ...\angle s\:\:on\:\:a\:\:straight\:\:line \\[3ex] \angle ADE + 33 + 90 = 180 \\[3ex] \angle ADE + 123 = 180 \\[3ex] \angle ADE = 180 - 123 \\[3ex] \angle ADE = 57 \\[3ex] \angle ADE = \angle ACD = 57^\circ ...\angle \:\:between\:\:tangent\;EDG\:\:and\:\:chord\;AD = \angle\:\:in\:\:the\;\;alternate\:\:segment \\[3ex] (ii) \\[3ex] \angle EAD = \angle ACD = 57^\circ ...\angle \:\:between\:\:tangent\;EAF\:\:and\:\:chord\;AD = \angle\:\:in\:\:the\;\;alternate\:\:segment \\[3ex] \underline{\triangle AED} \\[3ex] \angle ADE + \angle EAD + \angle AED = 180^\circ ...sum\:\:of\:\:\angle s\:\:of\:\;\triangle AED \\[3ex] 57 + 57 + \angle AED = 180 \\[3ex] 114 + \angle AED = 180 \\[3ex] \angle AED = 180 - 114 \\[3ex] \angle AED = 66^\circ \\[3ex] (iii) \\[3ex] \angle OAD = \angle OAC + \angle CAD ...diagram \\[3ex] \angle CDG = \angle CAD = 18^\circ ...\angle \:\:between\:\:tangent\;EDG\:\:and\:\:chord\;CD = \angle\:\:in\:\:the\;\;alternate\:\:segment \\[3ex] \implies \\[3ex] 33 = \angle OAC + 18 \\[3ex] \angle OAC = 33 - 18 \\[3ex] \angle OAC = 15^\circ \\[3ex] (iv) \\[3ex] \underline{Cyclic\;\;Quadrilateral\:\:ABCD} \\[3ex] \angle B + \angle D = 180^\circ ...sum\:\:of\:\:the\;\;two\;\;interior\:\:opposite\:\:\angle s\:\:of\:\:a\:\:cyclic\:\:quadrilateral \\[3ex] \angle D = \angle ODA + \angle ODC ...diagram \\[3ex] \angle ODG = \angle ODC + \angle CDG ...diagram \\[3ex] 90 = \angle ODC + 18 \\[3ex] \angle ODC = 90 - 18 \\[3ex] \angle ODC = 72 \\[3ex] So:\;\; \angle D = 33 + 72 \\[3ex] \angle D = 105^\circ \\[3ex] \implies \\[3ex] \angle B + 105 = 180 \\[3ex] \angle B = 180 - 105 \\[3ex] \angle B = 75^\circ \\[3ex] A\hat{B}C = \angle B = 75^\circ $

$ \angle MQP = 90 ...\angle \:\:in\:\:a\:\:semicircle \\[3ex] \angle MPQ = \angle MNQ = 42^\circ ...\angle s\:\:in\:\:same\:\:segment\;\;are\;\;equal \\[3ex] \underline{\triangle QMP} \\[3ex] \angle QMP + \angle MPQ + \angle MQP = 180^\circ ...sum\:\:of\:\:\angle s\:\:of\:\:\triangle QMP \\[3ex] \angle QMP + 42 + 90 = 180 \\[3ex] \angle QMP + 132 = 180 \\[3ex] \angle QMP = 180 - 132 \\[3ex] \angle QMP = 48^\circ $

$ (i) \\[3ex] \angle OQR = \angle ORQ = 32 ...base\:\: \angle s\:\:of\:\:isosceles\:\: \triangle \\[3ex] \underline{\triangle OQR} \\[3ex] \angle OQR + \angle ORQ + \angle QOR = 180^\circ ...sum\:\:of\:\:\angle s\:\:of\:\:\triangle OQR \\[3ex] 32 + 32 + \angle QOR = 180 \\[3ex] 64 + \angle QOR = 180 \\[3ex] \angle QOR = 180 - 64 \\[3ex] \angle QOR = 116 \\[3ex] \angle QOR = 2 * \angle QPR ...\angle \:\:at\:\:center\:\:is\:\:twice\:\:\angle\:\:at\:\:circumference \\[3ex] 116 = 2 * \angle QPR \\[3ex] \angle QPR = \dfrac{116}{2} \\[5ex] \angle QPR = 58^\circ \\[3ex] $ $ (ii) \\[3ex] \underline{Cyclic\;\;Quadrilateral\:\:QTPR} \\[3ex] \angle P + \angle Q = 180^\circ ...sum\:\:of\:\:the\;\;two\;\;interior\:\:opposite\:\:\angle s\:\:of\:\:a\:\:cyclic\:\:quadrilateral \\[3ex] \angle P = \angle TPQ + \angle QPR ...diagram \\[3ex] \angle P = 15 + 58 \\[3ex] \angle P = 73^\circ \\[3ex] \angle Q = \angle TQO + \angle OQR ...diagram \\[3ex] \angle Q = \angle TQO + 32 \\[3ex] \implies \\[3ex] 73 + \angle TQO + 32 = 180 \\[3ex] \angle TQO + 105 = 180 \\[3ex] \angle TQO = 180 - 105 \\[3ex] \angle TQO = 75^\circ $

$ \angle YXZ = 90^\circ ... \angle\;\;in\;\;a\;\;semicircle \\[3ex] \implies \\[3ex] \triangle XYZ \;\;is\;\;a\;\;right-angled\;\;\triangle $

$ a) \\[3ex] \angle SVW = \angle VZW = 51^\circ ...\angle \:\:between\:\:tangent\;SVT\:\:and\:\:chord\;VW = \angle\:\:in\;\;the\:\:alternate\:\:segment \\[3ex] b) \\[3ex] \angle TVZ = \angle VWZ = 78^\circ ...\angle \:\:between\:\:tangent\;SVT\:\:and\:\:chord\;VZ = \angle\:\:in\;\;the\:\:alternate\:\:segment \\[3ex] \underline{Cyclic\:\:Quadrilateral\:\:WXYZ} \\[3ex] \angle VWZ = \angle XYZ = 78^\circ ...exterior\:\: \angle \:\:of\:\:a\:\:cyclic\:\:quadrilateral = interior\:\:opposite\:\: \angle $

Intercepted arc = angle at center (central angle)

$ \angle AOB = \overset{\huge\frown}{AB} = 112^\circ \\[3ex] \underline{\triangle AOB} \\[3ex] \angle ABO = \angle BAO = p ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle AOB \\[3ex] \angle ABO + \angle BAO + \angle AOB = 180^\circ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle OAB \\[3ex] \implies \\[3ex] p + p + 112 = 180 \\[3ex] 2p = 180 - 112 \\[3ex] 2p = 68 \\[3ex] p = \dfrac{68}{2} \\[5ex] p = 34 \\[3ex] \therefore \angle ABO = \angle BAO = 34^\circ $

$ (23.1.1) \\[3ex] \underline{First\;\;angle\;\;equal\;\;to\;\;x} \\[3ex] \angle JOH = 2 * \angle JGH ...\angle \;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] 2x = 2 * \angle JGH \\[3ex] 2 * \angle JGH = 2x \\[3ex] \angle JGH = \dfrac{2x}{2} \\[5ex] \angle JGH = x \\[3ex] \underline{Second\;\;angle\;\;equal\;\;to\;\;x} \\[3ex] Parallel\;\;Lines:\;\;GJ\;\;and\;\;HK \\[3ex] \angle JGH = \angle GHK = x...alternate\;\;interior\;\; \angle s\;\;are\;\;equal \\[3ex] \underline{Third\;\;angle\;\;equal\;\;to\;\;x} \\[3ex] \angle GHK = \angle GJH = x ...\angle\;\;between\;\;tangent\;KH\;\;and\;\;chord\;GH = \angle\;\;in\;\;alternate\;\;segment \\[3ex] (23.1.2) \\[3ex] \underline{\triangle HOJ} \\[3ex] \angle OHJ + \angle HOJ + \angle HJO = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle HOJ \\[3ex] \hat{H_3} + 2x + \hat{H_1} = 180 ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle HOJ \\[3ex] \hat{H_3} = \hat{H_1} ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle HOJ \\[3ex] \hat{H_3} + 2x + \hat{H_3} = 180 \\[3ex] 2\hat{H_3} = 180 - 2x \\[3ex] \hat{H_3} = \dfrac{180 - 2x}{2} \\[5ex] \hat{H_3} = \dfrac{2(90 - x)}{2} \\[5ex] \hat{H_3} = 90 - x ...eqn.(1) \\[3ex] \underline{\triangle HGJ} \\[3ex] \angle GHJ + \angle JGH + \angle GJH = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle HGJ \\[3ex] (\hat{H_2} + \hat{H_3}) + x + x = 180 \\[3ex] \hat{H_2} + \hat{H_3} + 2x = 180 \\[3ex] \hat{H_2} = 180 - 2x - \hat{H_3} \\[3ex] From\;\;eqn.(1),\;\;substitute\;\;(90 - x)\;\;for\;\;\hat{H_3} \\[3ex] \hat{H_2} = 180 - 2x - (90 - x) \\[3ex] \hat{H_2} = 180 - 2x - 90 + x \\[3ex] \hat{H_2} = 90 - x...eqn.(2) \\[3ex] Comparing\;\;eqn.(1)\;\;and\;\;eqn.(2)\implies \\[3ex] \hat{H_2} = \hat{H_3} $

$ (i) \\[3ex] \angle BOA = 2 * \angle ACB \\[3ex] ...\angle \:\:at\:\:center\:\:is\:\:twice\:\:\angle\:\:at\:\:circumference \\[3ex] 130 = 2 * \angle ACB \\[3ex] \angle ACB = \dfrac{130}{2} \\[5ex] \angle ACB = 65^\circ \\[3ex] (ii) \\[3ex] \angle CBD = \angle CAD = 30^\circ ...\angle s\:\:in\:\:same\:\:segment\;\;are\;\;equal \\[3ex] (iii) \\[3ex] \angle ADB = \angle ACB = 65^\circ ...\angle s\:\:in\:\:same\:\:segment\;\;are\;\;equal \\[3ex] \underline{\triangle AED} \\[3ex] \angle DAE + \angle AED + \angle EDA = 180 ...sum\:\:of\:\:\angle s\:\:of\:\:\triangle AED \\[3ex] \angle EDA = \angle ADB = 65 ...diagram \\[3ex] 30 + \angle AED + 65 = 180 \\[3ex] \angle AED + 95 = 180 \\[3ex] \angle AED = 180 - 95 \\[3ex] \angle AED = 85^\circ $

$ \angle OTR = 90^\circ ...radius\;OT \perp tangent\;RT \;\;at\;\;point\;\;of\;\;contact\;T \\[3ex] \angle OSR = 90^\circ ...radius\;OS \perp tangent\;RS \;\;at\;\;point\;\;of\;\;contact\;S \\[3ex] \angle TOS = 2 * 68 ...\angle \;\;at\;\;centre = 2 * \angle \;\;at\;\;circumference \\[3ex] \angle TOS = 136^\circ \\[3ex] \underline{Quadrilateral\;\;TROS} \\[3ex] \angle T + \angle R + \angle S + \angle O = 360^\circ ...sum\;\;of\;\;the\;\;interior\;\;\angle s\;\;of\;\;a\;\;Quadrilateral \\[3ex] 90 + x + 90 + 136 = 360 \\[3ex] x + 316 = 360 \\[3ex] x = 360 - 316 \\[3ex] x = 44^\circ $

$ \angle ABC = 90^\circ...\angle \;\;in\;\;a\;\;semicircle $

$ (i.) \\[3ex] \angle ABC = 90^\circ ...\angle \;\;in\;\;a\;\;semicircle \\[3ex] (ii.) \\[3ex] \underline{\triangle ABC} \\[3ex] 2p + 90 + 3p = 180 \\[3ex] 5p + 90 = 180 \\[3ex] 5p = 180 - 90 \\[3ex] 5p = 90 \\[3ex] p = \dfrac{90}{5} \\[5ex] p = 18 \\[3ex] (a.) \\[3ex] \angle BAC = 2p \\[3ex] \angle BAC = 2(18) \\[3ex] \angle BAC = 36^\circ \\[3ex] (b.) \\[3ex] q = 2p...\angle s\;\;in\;\;the\;\;same\;\;segment \\[3ex] q = 36^\circ \\[3ex] (iii.) \\[3ex] \underline{\triangle DOC} \\[3ex] r = q ... base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle DOC \\[3ex] r = 36^\circ $

$ Center:\;\;O \\[3ex] \angle \;\;at\;\;center = 90^\circ \\[3ex] Let: \\[3ex] \angle MCN = \psi \\[3ex] \angle CMN = \theta \\[3ex] \angle CNM = \phi \\[3ex] \angle OMN = x \\[3ex] \angle ONM = y \\[3ex] 90 = 2 * \psi ...\angle \;\;at\;\;center = 2 * \angle\;\;at\;\;circumference \\[3ex] 2\psi = 90 \\[3ex] \psi = \dfrac{90}{2} \\[5ex] \psi = 45^\circ \\[3ex] Construction:\;\;Draw\;\;a\;\;chord\;\;to\;\;join\;\;M\;\;to\;\;N \\[3ex] \underline{\triangle OMN} \\[3ex] x = y ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle OMN \\[3ex] Because\;\;of\;\;same\;\;radii...OM = ON \\[3ex] \implies \\[3ex] 90 + x + y = 180 ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle OMN \\[3ex] 90 + x + x = 180 \\[3ex] 90 + 2x = 180 \\[3ex] 2x = 180 - 90 \\[3ex] 2x = 90 \\[3ex] x = \dfrac{90}{2} \\[5ex] x = 45 \\[3ex] x = y = 45^\circ \\[3ex] \underline{\triangle CMN} \\[3ex] \psi + \theta + x + y + \phi = 180 ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle CMN \\[3ex] 45 + \theta + 45 + 45 + \phi = 180 \\[3ex] 135 + \theta + \phi = 180 \\[3ex] \theta + \phi = 180 - 135 \\[3ex] \theta + \phi = 45^\circ $

$ \angle TAC = 30 ...\angle \:\:between\:\:tangent:\:TK\;\;and\:\:chord:\;TC = \angle\:\:in\:\:alternate\:\:segment \\[3ex] \angle ATC = 90 ...\angle \:\:in\:\:a\:\:semicircle \\[3ex] \underline{\triangle TKC} \\[3ex] \angle TCK = 90 + 30 ...exterior\:\: \angle \:\:of\:\:a\:\: \triangle \\[3ex] \angle TCK = 120 \\[3ex] \angle TKC + \angle TCK + \angle CTK = 180 ...sum\:\:of\:\:\angle s\:\:of\:\:a\:\:\triangle \\[3ex] \angle TKC + 120 + 30 = 180 \\[3ex] \angle TKC + 150 = 180 \\[3ex] \angle TKC = 180 - 150 \\[3ex] \angle TKC = 30^\circ $

$ \angle STP = x ...corresponding\;\;\angle s\;\;are\;\;equal \\[3ex] \underline{Cyclic\;\;Quadrilateral\;\;PQRS} \\[3ex] \angle P + \angle R = 180^\circ \\[3ex] ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quad\;\;are\;\;supplementary \\[3ex] \angle P + 110 = 180 \\[3ex] \angle P = 180 - 110 \\[3ex] \angle P = 70^\circ \\[3ex] \underline{\triangle PST} \\[3ex] \angle PST = \angle STP = x...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle PST \\[3ex] \angle PST + \angle STP + \angle SPT = 180^\circ ...sum\:\:of\:\:\angle s\:\:of\:\;\triangle PST \\[3ex] \angle SPT = \angle P = 70^\circ ...diagram \\[3ex] \implies \\[3ex] x + x + 70 = 180 \\[3ex] 2x + 70 = 110 \\[3ex] 2x = 180 - 70 \\[3ex] 2x = 110 \\[3ex] x = \dfrac{110}{2} \\[5ex] x = 55^\circ $

$ \angle MOP + 196 = 360^\circ ...sum\;\;of\;\;\angle s\;\;at\;\;a\;\;point \\[3ex] \angle MOP = 360 - 196 \\[3ex] \angle MOP = 164^\circ \\[3ex] \angle MOP = 2 * \angle MNP ...\angle \;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] 164 = 2 * \angle MNP \\[3ex] 2 * \angle MNP = 164 \\[3ex] \angle MNP = \dfrac{164}{2} \\[5ex] \angle MNP = 82^\circ \\[3ex] \underline{\triangle MOP} \\[3ex] \angle OMP = \angle OPM = p...base\;\;angle s\;\;of\;\;isosceles\;\;\triangle MOP \\[3ex] p + 164 + p = 180^\circ ...sum\;\;of\;\;of\;\;\angle s\;\;of\;\;\triangle MOP \\[3ex] 2p + 164 = 180 \\[3ex] 2p = 180 - 164 \\[3ex] 2p = 16 \\[3ex] p = \dfrac{16}{2} \\[5ex] p = 8^\circ \\[3ex] \underline{\triangle MNP} \\[3ex] \angle PMN + \angle MNP + \angle NPM = 180^\circ ...sum\;\;of\;\;of\;\;\angle s\;\;of\;\;\triangle MNP \\[3ex] (52 + p) + 82 + (m + p) = 180 \\[3ex] 52 + p + 82 + m + p = 180 \\[3ex] 52 + 8 + 82 + m + 8 = 180 \\[3ex] m + 150 = 180 \\[3ex] m = 180 - 150 \\[3ex] m = 30^\circ $

If QS is the diameter of the circle, then $\angle QPS = 90^\circ...\angle\;\;in\;\;a\;\;semicircle$
If QS is not the diameter of the circle, then $\angle QPS \ne 90^\circ$
So, let us calculate $\angle QPS$

$ \angle QPR = 27^\circ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] \underline{\triangle XPS} \\[3ex] \angle XPS = \angle XSP = p...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle PXS \\[3ex] p + p + 50 = 180...sum\;\;of\;\;\angle s\;\;in\;\;\triangle PXS \\[3ex] 2p + 50 = 180 \\[3ex] 2p = 180 - 50 \\[3ex] 2p = 130 \\[3ex] p = \dfrac{130}{2} \\[5ex] p = 65^\circ \\[3ex] \angle QPS = \angle QPR + \angle XPS...diagram \\[3ex] \angle QPS = 27 + 65 \\[3ex] \angle QPS = 92^\circ \\[3ex] 92^\circ \ne 90^\circ \\[3ex] $ Therefore, QS is not a diameter of the circle.

$ (i) \\[3ex] \angle MOL = 2 * \angle MNL ...\angle \;\;at\;\;center = 2 * \angle \;\;at\;\;circumference \\[3ex] 110 = 2 * \angle MNL \\[3ex] 2 * \angle MNL = 110 \\[3ex] \angle MNL = \dfrac{110}{2} \\[5ex] \angle MNL = 55^\circ \\[3ex] (ii) \\[3ex] \underline{\triangle MOL} \\[3ex] \angle LMO + \angle MLO + \angle MOL = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle MOL \\[3ex] \angle LMO = \angle MLO = p ...base \angle s\;\;of\;\;isosceles\;\;\triangle MOL \\[3ex] \implies \\[3ex] p + p + 110 = 180 \\[3ex] 2p + 110 = 180 \\[3ex] 2p = 180 - 110 \\[3ex] 2p = 70 \\[3ex] p = \dfrac{70}{2} \\[5ex] p = 35 \\[3ex] \therefore \angle LMO = 35^\circ $

$ \angle AOB = 2 * \angle ACB ...\angle \;\;at\;\;center = 2 * \angle \;\;at\;\;circumference \\[3ex] \angle AOB = 2 * 74 \\[3ex] \angle AOB = 148^\circ \\[3ex] \underline{\triangle AOB} \\[3ex] \angle OBA = \angle OAB = x ...base \angle s\;\;of\;\;isosceles\;\;\triangle AOB \\[3ex] \angle OBA + \angle OAB + \angle AOB = ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle AOB \\[3ex] x + x + 148 = 180 \\[3ex] 2x + 148 = 180 \\[3ex] 2x = 180 - 148 \\[3ex] 2x = 32 \\[3ex] x = \dfrac{32}{2} \\[5ex] x = 16^\circ $

Construction: Draw the chord from A to D

$ (i) \\[3ex] \underline{\triangle AOD} \\[3ex] \angle OAD + \angle ODA + \angle AOD = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle AOD \\[3ex] Let: \\[3ex] \angle OAD = ODA = p... base\;\; \angle s\;\;of\;\;isosceles\;\;\triangle AOD \\[3ex] \implies \\[3ex] p + p + 130 = 180 \\[3ex] 2p + 130 = 180 \\[3ex] 2p = 180 - 130 \\[3ex] 2p = 50 \\[3ex] p = \dfrac{50}{2} \\[5ex] p = 25 \\[3ex] \underline{\triangle ABD} \\[3ex] Let\;\; \angle ODB = k \\[3ex] \angle BAD + \angle ADB + \angle ABD = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle ABD \\[3ex] \angle BAD = 26 + p = 26 + 25 = 51^\circ ...diagram \\[3ex] \angle ADB = k + 25 ...diagram \\[3ex] \angle AOD = 2 * \angle ABD ...\angle \;\;at\;\;center = 2 * \angle \;\;at\;\;circumference \\[3ex] 130 = 2 * \angle ABD \\[3ex] 2 * \angle ABD = 130 \\[3ex] \angle ABD = \dfrac{130}{2} \\[5ex] \angle ABD = 65^\circ \\[3ex] \implies \\[3ex] 51 + (k + 25) + 65 = 180 \\[3ex] 51 + k + 25 + 65 = 180 \\[3ex] 141 + k = 180 \\[3ex] k = 180 - 141 \\[3ex] k = 39 \\[3ex] \therefore \angle ODB = 39^\circ \\[3ex] (ii) \\[3ex] $ Construction: Join O to B to complete the triangle BOD

$ \underline{\triangle BOD} \\[3ex] \angle ODB = \angle OBD = k = 39 ... base \angle s\;\;of\;\;isosceles\;\;\triangle BOD \\[3ex] \angle ODB + \angle OBD + \angle BOD = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle BOD \\[3ex] 39 + 39 + \angle BOD = 180 \\[3ex] 78 + \angle BOD = 180 \\[3ex] \angle BOD = 180 - 78 \\[3ex] \angle BOD = 102^\circ $

$ (a.) \\[3ex] (i.) \\[3ex] Straight\;\;line\;AC \;\;is\;\;the\;\;diameter \\[3ex] (ii.) \\[3ex] Straight\;\;line\;AB \;\;is\;\;a\;\;chord \\[3ex] (b.) \\[3ex] \angle ABC = 90^\circ...\angle\;\;in\;\;a\;\;semicircle\;\;is\;\;a\;\;right\;\;\angle $

$ \angle ACB = 56^\circ...\angle \;\;between\;\;tangent\;DAE\;\;and\;\;chord\;AB = \angle\;\;in\;\;alternate\;\;segment \\[3ex] \angle AOB = 2 * \angle ACB ...\angle \;\;at\;\;center = 2 * \angle \;\;at\;\;circumference \\[3ex] \angle AOB = 2 * 56 \\[3ex] \angle AOB = 112^\circ \\[3ex] \underline{\triangle AOB} \\[3ex] \angle BAO = \angle ABO = k ... base \angle s\;\;of\;\;isosceles\;\;\triangle AOB \\[3ex] \angle BAO + \angle ABO + \angle AOB = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle AOB \\[3ex] k + k + 112 = 180 \\[3ex] 2k + 112 = 180 \\[3ex] 2k = 180 - 112 \\[3ex] 2k = 68 \\[3ex] k = \dfrac{68}{2} \\[5ex] k = 34 \\[3ex] \therefore \angle BAO = \angle ABO = 34^\circ \\[3ex] \underline{\triangle ACB} \\[3ex] \angle ABC = \angle ABO + \angle CBO ...diagram \\[3ex] \angle ABC = 34 + 35 \\[3ex] \angle ABC = 69^\circ \\[3ex] \angle BAC = \angle BAO + \angle CAO ...diagram \\[3ex] \angle BAC = 34 + \angle CAO \\[3ex] \angle ACB + \angle ABC + \angle BAC = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle ACB \\[3ex] 56 + 69 + (34 + \angle CAO) = 180 \\[3ex] 125 + 34 + \angle CAO = 180 \\[3ex] 159 + \angle CAO = 180 \\[3ex] \angle CAO = 180 - 159 \\[3ex] \angle CAO = 21^\circ $

$ (i) \\[3ex] \underline{Smaller\;\;Circle:\;\;Centre:\;C} \\[3ex] \angle GCH = 2 * \angle GFH ...\angle\;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] 88 = 2 * \angle GFH \\[3ex] 2 * \angle GFH = 88 \\[3ex] \angle GFH = \dfrac{88}{2} \\[5ex] \angle GFH = 44^\circ \\[3ex] (ii) \\[3ex] \underline{Bigger\;\;Circle} \\[3ex] Cyclic\;\;Quadrilateral\;DGHE \\[3ex] \angle D + \angle H = 180^\circ \\[3ex] ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quad\;\;are\;\;supplementary \\[3ex] \angle D = \angle GDE \\[3ex] \angle H = \angle GHE = 126^\circ \\[3ex] \implies \\[3ex] \angle D + 126 = 180 \\[3ex] \angle D = 180 - 126 \\[3ex] \angle D = 54 \\[3ex] \angle GDE = 54^\circ \\[3ex] (iii) \\[3ex] \underline{Both\;\;Circles} \\[3ex] \triangle DFE \\[3ex] \angle DEF + \angle EDF + \angle DFE = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle DFE \\[3ex] \angle DEF = ? \\[3ex] \angle EDF = \angle GDE = 54^\circ \\[3ex] \angle DFE = \angle GFH = 44^\circ \\[3ex] \implies \\[3ex] \angle DEF + 54 + 44 = 180 \\[3ex] \angle DEF + 98 = 180 \\[3ex] \angle DEF = 180 - 98 \\[3ex] \angle DEF = 82^\circ $

$ (a) \\[3ex] \triangle CBT\;\;is\;\;right-angled\;\;because \\[3ex] \angle CBT = 90^\circ ... radius\;CB \perp tangent\;BT \;\;at\;\;point\;\;of\;\;contact\; B \\[5ex] (b) \\[3ex] At\;\;least\;\;three\;\;ways\;\;to\;\;show\;\;Congruency \\[3ex] Side:\;\; CA = CB ...radius \\[3ex] Angle:\;\; \angle CAT = \angle CBT = 90^\circ ... radius \perp tangent \;\;at\;\;point\;\;of\;\;contact \\[3ex] Side:\;\; AT = BT ...two\;\;tangents:\;\;AT\;\;and\;\;BT\;\;drawn\;\;from\;\;the\;\;same\;\;external\;\;point:\;T\;\;are\;\;equal\;\;in\;\;length \\[3ex] \implies \\[3ex] \triangle CAT \cong \triangle CBT...Side-Angle-Side \\[3ex] OR \\[3ex] Angle:\;\; \angle CAT = \angle CBT = 90^\circ ... radius \perp tangent \;\;at\;\;point\;\;of\;\;contact \\[3ex] Side:\;\; AT = BT ...two\;\;tangents:\;\;AT\;\;and\;\;BT\;\;drawn\;\;from\;\;the\;\;same\;\;external\;\;point:\;T\;\;are\;\;equal\;\;in\;\;length \\[3ex] Angle:\;\; \angle ATC = \angle BTC ... \\[3ex] $ For two tangents: AT and BT drawn from the same external point: T, the line joining the external point: T, and the centre of the circle: C; bisects the angle between the tangents.

$ \implies \\[3ex] \triangle CAT \cong \triangle CBT...Angle-Side-Angle \\[3ex] OR \\[3ex] Side:\;\; CA = CB ...radius \\[3ex] Side:\;\; AT = BT ...two\;\;tangents:\;\;AT\;\;and\;\;BT\;\;drawn\;\;from\;\;the\;\;same\;\;external\;\;point:\;T\;\;are\;\;equal\;\;in\;\;length \\[3ex] Side:\;\; CT = CT ... common\;\;side\;\;of\;\;the\;\;two\;\;\triangle s \\[3ex] \implies \\[3ex] \triangle CAT \cong \triangle CBT...Side-Side-Side \\[5ex] (c) \\[3ex] Because\;\;the\;\;polygon\;ACBT\;\;is\;\;a\;\;quadrilateral \\[3ex] where \\[3ex] \angle CAT = \angle CBT = 90^\circ ... radius \perp tangent \;\;at\;\;point\;\;of\;\;contact \\[3ex] and \\[3ex] \angle CAT \;\;and\;\; \angle CBT \;\;are\;\;opposite\;\;angles \\[3ex] \implies \\[3ex] Quadrilateral\;\;ACBT\;\;is\;\;a\;\;cyclic\;\;quadrilateral \\[3ex] ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quadrilateral\;\;are\;\;supplementary\;\;(90^\circ + 90^\circ = 180^\circ) $

$ \angle ABC + 62 = 180^\circ ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] \angle ABC = 180 - 62 \\[3ex] \angle ABC = 118^\circ \\[3ex] Reflex\;\; \angle AOC = 2 * \angle ABC \\[3ex] ...\angle \;\;at\;\;center = 2 * \angle \;\;at\;\;circumference \\[3ex] Reflex\;\; \angle AOC = 2 * 118 \\[3ex] Reflex\;\; \angle AOC = 236^\circ \\[3ex] (i.) \\[3ex] interior\;\;\angle AOC + Reflex\;\;\angle AOC = 360^\circ...\angle s\;\;at\;\;a\;\;point \\[3ex] interior\;\;\angle AOC + 236 = 360 \\[3ex] interior\;\;\angle AOC = 360 - 236 \\[3ex] interior\;\;\angle AOC = 124^\circ \\[3ex] (ii.) \\[3ex] \underline{\triangle ABC} \\[3ex] \angle BAC + \angle ABC + \angle ACB = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle ABC \\[3ex] \angle BAC + 118 + 39 = 180 \\[3ex] \angle BAC + 157 = 180 \\[3ex] \angle BAC = 180 - 157 \\[3ex] \angle BAC = 23^\circ $

$ (a.) \\[3ex] \underline{\triangle DOC} \\[3ex] \angle ODC = \angle OCD = 57^\circ ... base\;\; \angle s\;\;of\;\;isosceles\;\;\triangle DOC \\[3ex] \angle ODC + \angle OCD + \angle DOC = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle DOC \\[3ex] 57 + 57 + x = 180 \\[3ex] 114 + x = 180 \\[3ex] x = 180 - 114 \\[3ex] x = 66^\circ \\[3ex] (b.) \\[3ex] Let\;\;\angle ADO = p \\[3ex] \underline{Cyclic\;\;Quadrilateral\;ABCD} \\[3ex] \angle B = 82^\circ \\[3ex] \angle D = \angle ADO + \angle ODC ...diagram \\[3ex] \angle D = p + 57 \\[3ex] \angle D + \angle B = 180 ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quadrilateral\;\;are\;\;supplementary \\[3ex] \implies \\[3ex] (p + 57) + 82 = 180 \\[3ex] p + 57 + 82 = 180 \\[3ex] p + 139 = 180 \\[3ex] p = 180 - 139 \\[3ex] p = 41^\circ \\[3ex] \angle ODF = 90^\circ ...radius\;OD \perp tangent\;EDF\;\;at\;\;point\;\;of\;\;contact\;D \\[3ex] y + \angle ADO + \angle ODF = 180^\circ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] y + p + 90 = 180 \\[3ex] y + 41 + 90 = 180 \\[3ex] y + 131 = 180 \\[3ex] y = 180 - 131 \\[3ex] y = 49^\circ $

$ (i) \\[3ex] \underline{\triangle GOF} \\[3ex] \angle OGF = \angle OFG = p...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle GOF \\[3ex] \angle OGF + \angle OFG + \angle GOF = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle GOF \\[3ex] p + p + 118 = 180 \\[3ex] 2p = 180 - 118 \\[3ex] 2p = 62 \\[3ex] p = \dfrac{62}{2} \\[5ex] p = 31 \\[3ex] \therefore \angle OGF = 31^\circ \\[3ex] (ii) \\[3ex] \angle OGH = 90^\circ \\[3ex] ...radius\;OG\;\;\perp\;\;tangent\;SGH\;\;at\;\;point\;\;of\;\;contact\;G \\[3ex] \angle SGD + \angle DGO + \angle OGH = 180^\circ ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] 65 + \angle DGO + 90 = 180 \\[3ex] \angle DGO + 155 = 180 \\[3ex] \angle DGO = 180 - 155 \\[3ex] \angle DGO = 25^\circ \\[3ex] \underline{Cyclic\;\;Quadrilateral\;GDEF} \\[3ex] \angle G + \angle E = 180 \\[3ex] ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quad\;\;are\;\;supplementary \\[3ex] \angle G = \angle DGO + \angle OGF \\[3ex] \angle G = 25 + 31 \\[3ex] \angle G = 56^\circ \\[3ex] \angle E = \angle DEF \\[3ex] \implies \\[3ex] 56 + \angle E = 180 \\[3ex] \angle E = 180 - 56 \\[3ex] \angle E = 124 \\[3ex] \therefore \angle DEF = 124^\circ $

$ (i) \\[3ex] \angle OAE = 90^\circ ... radius\;OA \perp tangent\;AE\:\:at\:\:point\:\:of\:\:contact\;A \\[3ex] (ii) \\[3ex] \angle OBE = 90^\circ ... radius\;OB \perp tangent\:BE\:\:at\:\:point\:\:of\:\:contact\;B \\[3ex] \underline{Quadrilateral\:\:AOBE} \\[3ex] \angle A + \angle O + \angle B + \angle E = 360 ...sum\:\:of\:\:interior\:\:\angle s\:\:of\:\:a\:\:Quadrilateral \\[3ex] 90 + \angle O + 90 + 48 = 360 \\[3ex] 228 + \angle O = 360 \\[3ex] \angle O = 360 - 228 \\[3ex] \angle O = 132 \\[3ex] \angle O = \angle AOB = 132^\circ \\[3ex] (iii) \\[3ex] Obtuse\;\;\angle AOB = 2 * \angle ACB ...\angle \:\:at\:\:center\:\:is\:\:twice\:\:\angle\:\:at\:\:circumference \\[3ex] 132 = 2 * \angle ACB \\[3ex] 2 * \angle ACB = 132 \\[3ex] \angle ACB = \dfrac{132}{2} \\[5ex] \angle ACB = 66^\circ \\[3ex] (iv) \\[3ex] Reflex\;\;\angle AOB + Obtuse\;\;\angle AOB = 360 ...\angle s\:\:at\:\:a\:\:point \\[3ex] Reflex\;\;\angle AOB + 132 = 360 \\[3ex] Reflex\;\;\angle AOB = 360 - 132 \\[3ex] Reflex\;\;\angle AOB = 228 \\[3ex] Reflex\;\;\angle AOB = 2 * \angle ADB ...\angle \:\:at\:\:center\:\:is\:\:twice\:\:\angle\:\:at\:\:circumference \\[3ex] 228 = 2 * \angle ADB \\[3ex] 2 * \angle ADB = 228 \\[3ex] \angle ADB = \dfrac{228}{2} \\[5ex] \angle ADB = 114^\circ \\[3ex] OR \\[3ex] \underline{Cyclic\;\;Quadrilateral\:\:ACBD} \\[3ex] \angle C + \angle D = 180 ...sum\:\:of\:\:interior\:\:opposite\:\:\angle s\:\:of\:\:a\:\:Cyclic\:\:Quadrilateral \\[3ex] 66 + \angle D = 180 \\[3ex] \angle D = 180 - 66 \\[3ex] \angle D = 114 \\[3ex] \angle D = \angle ADB = 114^\circ $

$ \angle DBC = 58^\circ ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] \angle DCB = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle\;\;is\;\;a\;\;right\;\;angle \\[3ex] \underline{\triangle DCB} \\[3ex] \angle CDB + \angle DCB + \angle DBC = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle DCB \\[3ex] \angle CDB + 90 + 58 = 180 \\[3ex] \angle CDB + 148 = 180 \\[3ex] \angle CDB = 180 - 148 \\[3ex] \angle CDB = 32^\circ $

We can review the properties of angles between parallel lines to determine that SP is parallel to RQ
Of those properties, the easiest one to check for is the Alternate Interior Angles property...alternate interior angles between parallel lines are equal
Let us demonstrate it with a diagram

So, we need to show that $\angle RQS = \angle PSQ$ ...indicated by the angles marked in red
Once, we can show that $\angle RQS = \angle PSQ$, then we have shown one of the properties of angles between parallel lines and hence show that the two lines: SP and RQ are parallel

$ \angle PSQ = 41^\circ ...\angle \;\;between\;\;tangent\;TPV\;\;and\;\;chord\;PS = \angle\;\;in\;\;the\;\;alternate\;\;segment \\[3ex] \underline{\triangle PSQ} \\[3ex] \angle SPQ + \angle PSQ + \angle PQS = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle PSQ \\[3ex] \angle SPQ + 41 + 57 = 180 \\[3ex] \angle SPQ + 98 = 180 \\[3ex] \angle SPQ = 180 - 98 \\[3ex] \angle SPQ = 82^\circ \\[3ex] \underline{Cyclic\;\;Quadrilateral\;SRQP} \\[3ex] \angle P + \angle R = 180^\circ ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quad\;\;are\;\;supplementary \\[3ex] \angle P = \angle SPQ = 82^\circ \\[3ex] \implies \\[3ex] 82 + \angle R = 180 \\[3ex] \angle R = 180 - 82 \\[3ex] \angle R = 98^\circ \\[3ex] \underline{\triangle SRQ} \\[3ex] \angle RSQ = \angle RQS = p ...isosceles\;\;\triangle SRQ \\[3ex] \angle SRQ = \angle R = 98^\circ \\[3ex] \angle RSQ + \angle RQS + \angle SRQ = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle SRQ \\[3ex] p + p + 98 = 180 \\[3ex] 2p = 180 - 98 \\[3ex] 2p = 82 \\[3ex] p = 41 \\[3ex] \therefore \angle RQS = 41^\circ \\[3ex] Because\;\; \angle PSQ = \angle RQS = 41^\circ,\;\;SP\;\;||\;\;RQ $

$ \angle OPT = 90^\circ \\[3ex] ...radius\;OP\perp tangent\;PT\;\;at\;\;point\;\;of\;\;contact\;P \\[3ex] \underline{\triangle OPT} \\[3ex] \angle POT + \angle OPT + \angle OTP = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle OPT \\[3ex] \angle POT + 90 + 30 = 180 \\[3ex] \angle POT + 120 = 180 \\[3ex] \angle POT = 180 - 120 \\[3ex] \angle POT = 60^\circ \\[3ex] \underline{\triangle OPR} \\[3ex] \angle OPR = \angle ORP = k ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle OPR \\[3ex] \angle OPR + \angle ORP + \angle POR = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle OPR \\[3ex] \angle POR = \angle POT = 60^\circ \\[3ex] \implies \\[3ex] k + k + 60 = 180 \\[3ex] 2k = 180 - 60 \\[3ex] 2k = 120 \\[3ex] k = \dfrac{120}{2} \\[5ex] k = 60 \\[3ex] \therefore \angle OPR = \angle ORP = 60^\circ \\[3ex] \triangle OPR \;\;is\;\;an\;\;equilateral\;\;triangle \\[3ex] \angle QRP = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle\;\;is\;\;a\;\;right\;\;\angle \\[3ex] \underline{\triangle PQR} \\[3ex] \angle PQR + \angle QRP + \angle QPR = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle PQR \\[3ex] \angle QPR = \angle OPR = 60^\circ \\[3ex] \implies \\[3ex] x + 90 + 60 = 180 \\[3ex] x + 150 = 180 \\[3ex] x = 180 - 150 \\[3ex] x = 30^\circ $

$ (47.1) \\[3ex] \angle LNK = \hat{N_2} = y...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] (47.2) \\[3ex] (a) \\[3ex] \angle LKM = \hat{K_1} = 2y ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] \underline{\triangle LMK} \\[3ex] \angle KLM + \angle LKM + \angle LMK = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle LMK \\[3ex] 87 + 2y + y = 180 \\[3ex] 3y = 180 - 87 \\[3ex] 3y = 93 \\[3ex] y = \dfrac{93}{3} \\[5ex] y = 31^\circ \\[3ex] (b) \\[3ex] \underline{Cyclic\;\;Quadrilateral\;PMNL} \\[3ex] T\hat{P}L = \hat{K_1} = 2y ...exterior\;\;\angle\;\;of\;\;a\;\;cyclic\;\;quad = interior\;\;opposite\;\;\angle \\[3ex] T\hat{P}L = 2(31) \\[3ex] T\hat{P}L = 62^\circ $

$ (a) \\[3ex] r = 30^\circ ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] (b) \\[3ex] s + 68 = 180^\circ \\[3ex] ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quad\;\;are\;\;supplementary \\[3ex] s = 180 - 68 \\[3ex] s = 112^\circ \\[3ex] (c) \\[3ex] The\;\;third\;\;angle = 90^\circ... \angle\;\;in\;\;a\;\;semicircle \\[3ex] t + 63 + 90 = 180 ... sum\;\;of\;\;\angle s\;\;in\;\;\triangle \\[3ex] t + 153 = 180 \\[3ex] t = 180 - 153 \\[3ex] t = 27^\circ $

$ (a.) \\[3ex] 51 + w = 118 ...\angle \:\:between\:\:tangent:\:PQ\;\;and\:\:chord:\;BD = \angle\:\:in\:\:the\;\;alternate\:\:segment \\[3ex] w = 118 - 51 \\[3ex] w = 67^\circ \\[3ex] (b.) \\[3ex] For\;\;this\;\;question: \\[3ex] \underline{Given}: \\[3ex] E\hat{F}G = x + y ...\angle \;\;at\;\;circumference \\[3ex] Acute\;\; \angle EHG = \angle \;\;at\;\;centre \\[3ex] \underline{To\;\;Prove}: \\[3ex] \angle EHG = 2(x + y) \\[3ex] \underline{\triangle FEH} \\[3ex] E\hat{F}H = x...as\;\;shown\;\;in\;\;the\;\;diagram \\[3ex] F\hat{E}H = E\hat{F}H = x ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle FEH \\[3ex] F\hat{H}E + F\hat{E}H + H\hat{F}E = 180...sum\;\;of\;\;\angle s\;\;in\;\;\triangle FEH \\[3ex] F\hat{H}E + x + x = 180 \\[3ex] F\hat{H}E + 2x = 180 \\[3ex] F\hat{H}E = 180 - 2x ...eqn.(1) \\[3ex] \underline{\triangle FEG} \\[3ex] H\hat{F}G = y...as\;\;shown\;\;in\;\;the\;\;diagram \\[3ex] F\hat{G}H = H\hat{F}G = y ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle FEH \\[3ex] F\hat{H}G + F\hat{G}H + H\hat{F}G = 180^\circ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle FEH \\[3ex] F\hat{H}G + y + y = 180 \\[3ex] F\hat{H}G + 2y = 180 \\[3ex] F\hat{H}G = 180 - 2y ...eqn.(2) \\[3ex] Reflex\;\; \angle EHG = F\hat{H}E + F\hat{H}G \\[3ex] Reflex\;\; \angle EHG = (180 - 2x) + (180 - 2y) \\[3ex] Reflex\;\; \angle EHG = 180 - 2x + 180 - 2y \\[3ex] Reflex\;\; \angle EHG = 360 - 2x - 2y \\[3ex] Reflex\;\; \angle EHG + Acute\;\; \angle EHG = 360^\circ...sum\;\;of\;\;\angle s\;\;at\;\;a\;\;point \\[3ex] (360 - 2x - 2y) + Acute\;\; \angle EHG = 360 \\[3ex] Acute\;\;\angle EHG = 360 - (360 - 2x - 2y) \\[3ex] Acute\;\;\angle EHG = 360 - 360 + 2x + 2y \\[3ex] Acute\;\;\angle EHG = 2x + 2y \\[3ex] Acute\;\;\angle EHG = 2(x + y) \\[3ex] \implies \\[3ex] $ The angle at the centre is twice the angle at the circumference.

Construction: Draw the radii: from point P on the circumference to point O on the center; and from point Q on the circumference to point O on the center.

$ a) \\[3ex] \angle POQ = 2 * \angle PAQ ...\angle\;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] Also: \\[3ex] \angle POQ = 2 * \angle PBQ ...\angle\;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] \implies \\[3ex] 2 * \angle PAQ = 2 * \angle PBQ \\[3ex] Divide\;\;both\;\;sides\;\;by\;\;2 \\[3ex] \angle PAQ = \angle PBQ \\[5ex] b) \\[3ex] i) \\[3ex] p = 65^\circ ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] ii) \\[3ex] 80^\circ = \angle DAB ...exterior\;\;\angle\;\;of\;\;cyclic\;\;quadrilateral\;ABCD = interior\;\;opposite\;\;\angle \\[3ex] \angle DAB = 80^\circ \\[3ex] q + \angle DAB + 45 = 180^\circ ... sum\;\;of\;\;\angle s\;\;in\;\;\triangle DAB \\[3ex] q + 80 + 45 = 180 \\[3ex] q + 125 = 180 \\[3ex] q = 180 - 125 \\[3ex] q = 55^\circ $

Construction: Join the line from Point P to Point R

$ \angle PRS = 90^\circ...\angle \;\;in\;\;a\;\;semicircle \\[3ex] \underline{Cyclic\;\;Quadrilateral\;QPSR} \\[3ex] \angle Q + \angle S = 180^\circ ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quadrilateral\;\;are\;\;supplementary \\[3ex] \angle Q + 56 = 180 \\[3ex] \angle Q = 180 - 56 \\[3ex] \angle Q = 124^\circ \\[3ex] \underline{\triangle RQP} \\[3ex] \angle QRP = \angle QPR = k ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle RQP \\[3ex] \angle Q = \angle RQP = 124^\circ ...diagram \\[3ex ] \angle QRP + \angle QPR + \angle RQP = 180^\circ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle RQP \\[3ex] k + k + 124 = 180 \\[3ex] 2k = 180 - 124 \\[3ex] 2k = 56 \\[3ex] k = \dfrac{56}{2} \\[5ex] k = 28 \\[3ex] \therefore \angle QRP = \angle QPR = 28^\circ \\[3ex] \angle QRS = \angle QRP + \angle PRS ...diagram \\[3ex] \angle QRS = 28 + 90 \\[3ex] \angle QRS = 118^\circ $

$ \underline{\triangle PCR} \\[3ex] \angle RPC = \angle PRC = 25^\circ ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle PCR \\[3ex] \angle PCR + \angle RPC + \angle PRC = 180^\circ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle PCR \\[3ex] \angle PCR + 25 + 25 = 180 \\[3ex] \angle PCR + 50 = 180 \\[3ex] \angle PCR = 180 - 50 \\[3ex] \angle PCR = 130^\circ \\[3ex] \angle PCR = 2 * \angle PQR ...\angle \;\;at\;\;center = 2 * \angle \;\;at\;\;circumference \\[3ex] 130 = 2 * \angle PQR \\[3ex] 2 * \angle PQR = 130 \\[3ex] \angle PQR = \dfrac{130}{2} \\[5ex] \angle PQR = 65 \\[3ex] \angle PQR = a = 65^\circ $

$ \angle RMQ = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] \underline{\triangle RMQ} \\[3ex] \angle MRQ + \angle RMQ + \angle RQM = 180^\circ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle RMQ \\[3ex] \angle MRQ + 90 + 20 = 180 \\[3ex] \angle MRQ + 110 = 180 \\[3ex] \angle MRQ = 180 - 110 \\[3ex] \angle MRQ = 70^\circ \\[3ex] (ii) \\[3ex] \angle PMR = 20^\circ...\angle \;\;between\;\;tangent\;MP\;\;and\;\;chord\;MR = \angle\;\;in\;\;alternate\;\;segment \\[3ex] (iii) \\[3ex] \underline{\triangle PMN} \\[3ex] |MP| = |NP|...two\;\;tangents:\;\;MP\;\;and\;\;NP\;\;drawn\;\;from\;\;the\;\;same\;\;external\;\;point:\;P\;\;are\;\;equal\;\;in\;\;length \\[3ex] \implies \;\; \angle PMN = \angle PNM \\[3ex] \implies \;\; \triangle PMN \;\;is\;\;an\;\;isosceles\;\;\triangle \\[3ex] Let\;\; \angle PMN = \angle PNM = k \\[3ex] \angle PMN + \angle PNM + \angle MPN = 180^\circ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle PMN \\[3ex] k + k + 54 = 180 \\[3ex] 2k = 180 - 54 \\[3ex] 2k = 126 \\[3ex] k = \dfrac{126}{2} \\[5ex] k = 63 \\[3ex] \therefore \angle PMN = \angle PNM = 63^\circ $

The ACT is a timed test.
It is important to do this question within a minute. Remember what I wrote earlier
So, the faster approach for me is to use alphabets rather than angles
This is what I mean

$ Let: \\[3ex] \angle WCX = m \\[3ex] \angle XCY = n \\[3ex] \angle YCZ = p \\[3ex] Based\;\;on\;\;the\;\;question\;\;and\;\;modified\;\;diagram: \\[3ex] m + n = 100 ...eqn.(1) \\[3ex] n + p = 80...eqn.(2) \\[3ex] m + n + p = 150 ...eqn.(3) \\[3ex] Substitute\;\;eqn.(1)\;\;into\;\;eqn.(3) \implies \\[3ex] 100 + p = 150 \\[3ex] p = 150 - 100 \\[3ex] p = 50 ...eqn.(4) \\[3ex] Substitute\;\;eqn.(4)\;\;into\;\;eqn.(2) \implies \\[3ex] n + 50 = 80 \\[3ex] n = 80 - 50 \\[3ex] n = 30 \\[3ex] \therefore \angle XCY = 30^\circ $

$ (i.) \\[3ex] \angle QRS = 90^\circ...\angle s\;\;in\;\;a\;\;semicircle \\[3ex] \angle QST = \angle RQS + \angle QRS \\[3ex] ...exterior\;\;\angle\;\;of\;\;a\;\;\triangle = sum\;\;of\;\;the\;\;two\;\;interior\;\;opposite\;\;\angle s\\[3ex] \implies \\[3ex] (3x + 15) = x + 90 \\[3ex] 3x + 15 = x + 90 \\[3ex] 3x - x = 90 - 15 \\[3ex] 2x = 75 \\[3ex] x = \dfrac{75}{2} \\[5ex] x = 37.5^\circ \\[3ex] (ii.) \\[3ex] \angle RSQ + \angle QST = 180^\circ ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] \angle RSQ + (3x + 15) = 180 \\[3ex] \angle RSQ + [3(37.5) + 15] = 180 \\[3ex] \angle RSQ + (112.5 + 15) = 180 \\[3ex] \angle RSQ + 127.5 = 180 \\[3ex] \angle RSQ = 180 - 127.5 \\[3ex] \angle RSQ = 52.5^\circ $

O is the centre of the circle

$ \angle O = 2 * b ...\angle \;\;at\;\;centre = 2 * \angle \;\;at\;\;circumference \\[3ex] 72 = 2 * b \\[3ex] 2b = 72 \\[3ex] b = \dfrac{72}{2} \\[5ex] b = 36^\circ $

$ \angle S + \angle Q = 180^\circ ...opposite\;\;\angle s\;\;of\;\;Cyclic\;\;Quadrilateral\;PQRS\;\;are\;\;supplementary \\[3ex] \angle Q = \angle PQR = 117^\circ ...diagram \\[3ex] \angle S = \angle RSP ...diagram \\[3ex] \implies \\[3ex] \angle RSP + 117 = 180 \\[3ex] \angle RSP = 180 - 117 \\[3ex] \angle RSP = 63^\circ \\[3ex] \angle RPS = \angle RSP = 63^\circ ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle PRS \\[3ex] \angle PRS + \angle RPS + \angle RSP = 180^\circ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle PRS \\[3ex] \angle PRS + 63 + 63 = 180 \\[3ex] \angle PRS + 126 = 180 \\[3ex] \angle PRS = 180 - 126 \\[3ex] \angle PRS = 54^\circ \\[3ex] \angle PST = \angle PRS = 54^\circ ...\angle\;\;between\;\;tangent\;TS\;\;and\;\;chord\;SP = \angle\;\;in\;\;the\;\;alternate\;\;segment $

$ (i) \\[3ex] \angle ADC = 90^\circ ...\angle \;\;in\;\;a\;\;semicircle \\[3ex] (ii) \\[3ex] \angle AOD = 2 * \angle ACD ...\angle \;\;at\;\;centre = 2 * \angle \;\;at\;\;circumference \\[3ex] 72 = 2 * \angle ACD \\[3ex] 2 * \angle ACD = 72 \\[3ex] \angle ACD = \dfrac{72}{2} \\[5ex] \angle ACD = 36^\circ \\[3ex] OR \\[3ex] \underline{\triangle AOD} \\[3ex] \angle OAD = \angle ODA = p ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle \\[3ex] \angle OAD + \angle ODA + \angle AOD = 180^\circ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle AOD \\[3ex] p + p + 72 = 180 \\[3ex] 2p = 180 - 72 \\[3ex] 2p = 108 \\[3ex] p = \dfrac{108}{2} \\[5ex] p = 54 \\[3ex] \therefore \angle OAD = \angle ODA = 54^\circ \\[3ex] \underline{\triangle ACD} \\[3ex] \angle CAD = \angle OAD = 54^\circ \\[3ex] \angle CAD + \angle ACD + \angle ADC = 180^\circ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle ACD \\[3ex] 54 + \angle ACD + 90 = 180 \\[3ex] \angle ACD + 144 = 180 \\[3ex] \angle ACD = 180 - 144 \\[3ex] \angle ACD = 36^\circ \\[3ex] (iii) \\[3ex] \angle CAD = 54^\circ \\[3ex] (iv) \\[3ex] \angle OAE = 90^\circ \\[3ex] ... radius \;OA \; \perp \; tangent\;TAE\;\;at\;\;point\;\;of\;\;contact\;A \\[3ex] \underline{\triangle OEA} \\[3ex] \angle OEA + \angle AOE + \angle OAE = 180^\circ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle OEA \\[3ex] \angle OEA + 72 + 90 = 180 \\[3ex] \angle OEA + 162 = 180 \\[3ex] \angle OEA = 180 - 162 \\[3ex] \angle OEA = 18^\circ $

(57.1)
Angles subtended by a chord of a circle, in the same segment of the circle, are equal.

$ (57.2.1) \\[3ex] Reflex\;\;\angle O = 2 * \angle PNM ...\angle\;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] Reflex\;\;\angle O = 2 * 128 \\[3ex] Reflex\;\;\angle O = 256^\circ \\[3ex] Reflex\;\;\angle O + Obtuse\;\;\angle O = 360^\circ ...\angle s\;\;at\;\;a\;\;point \\[3ex] 256 + Obtuse\;\;\angle O = 360 \\[3ex] Obtuse\;\;\angle O = 360 - 256 \\[3ex] Obtuse\;\;\angle O = 104^\circ \\[3ex] \underline{\triangle POM} \\[3ex] \angle OPM = \angle OMP = x ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle POM \\[3ex] \angle OPM + \angle OMP + \angle POM = 180^\circ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle POM \\[3ex] \angle POM = \hat{O_2} = Obtuse\;\;\angle O = 104^\circ \\[3ex] \implies \\[3ex] x + x + 104 = 180 \\[3ex] 2x = 180 - 104 \\[3ex] 2x = 76 \\[3ex] x = \dfrac{76}{2} \\[5ex] x = 38 \\[3ex] \therefore \angle OPM = \angle OMP = 38^\circ \\[3ex] \angle KPM = 90^\circ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] \underline{\triangle KMP} \\[3ex] \angle PKM + \angle KPM + \angle KMP = 180^\circ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle KMP \\[3ex] \angle KMP = \hat{M_2} = \angle OMP = 38^\circ \\[3ex] \implies \\[3ex] \angle PKM + 90 + 38 = 180 \\[3ex] \angle PKM + 128 = 180 \\[3ex] \angle PKM = 180 - 128 \\[3ex] \angle PKM = 52 \\[3ex] \angle PKM = \hat{K_1} = 52^\circ \\[3ex] (57.2.2) \\[3ex] \hat{L_2} = \hat{M_2} = 38^\circ ...\angle s\;\;in\;\;the\;\;segment\;\;are\;\;equal \\[3ex] OR \\[3ex] \hat{L_1} = \hat{K_1} = 52^\circ ...\angle s\;\;in\;\;the\;\;segment\;\;are\;\;equal \\[3ex] \angle KLM = \hat{L} = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] \hat{L_1} + \hat{L_2} = \hat{L} ...diagram \\[3ex] 52 + \hat{L_2} = 90 \\[3ex] \hat{L_2} = 90 - 52 \\[3ex] \hat{L_2} = 38^\circ \\[3ex] (57.2.3) \\[3ex] \hat{P_2} + \hat{P_3} = \angle OPM ...diagram \\[3ex] \hat{P_2} + 29 = 38 \\[3ex] \hat{P_2} = 38 - 29 \\[3ex] \hat{P_2} = 9^\circ $

$ If\;\;|FSJ|\;\;\parallel\;\;|GCT|,\;\;then \\[3ex] \angle CSJ = 100^\circ ...alternate\;\;interior\;\;\angle s\;\;are\;\;equal \\[3ex] Let\;\;us\;\;calculate\;\;\angle CSJ \\[3ex] If\;\;\angle CSJ = 100^\circ,\;\;then\;\; |FSJ|\;\;\parallel\;\;|GCT| \\[3ex] If\;\;\angle CSJ \ne 100^\circ,\;\;then\;\; |FSJ|\;\;\nparallel\;\;|GCT| \\[3ex] \angle CSJ = 90^\circ \\[3ex] ...radius\;CS \;\;\perp\;\;tangent\;FSJ\;\;at\;\;point\;\;of\;\;contact\;S \\[3ex] \angle CSJ \ne 100^\circ \\[3ex] \therefore |FSJ|\;\;\nparallel\;\;|GCT| $

$ (i) \\[3ex] Obtuse\;\;S\hat{O}Q = 2 * \angle S\hat{P}Q ...\angle \;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] Obtuse\;\;S\hat{O}Q = 2 * 54 \\[3ex] Obtuse\;\;S\hat{O}Q = 108^\circ \\[3ex] (ii) \\[3ex] Obtuse\;\;S\hat{O}Q + Reflex\;\;S\hat{O}Q = 360^\circ ...\angle s\;\;at\;\;a\;\;point \\[3ex] 108 + Reflec\;\;S\hat{O}Q = 360 \\[3ex] Reflex\;\;S\hat{O}Q = 360 - 108 \\[3ex] Reflex\;\;S\hat{O}Q = 252^\circ \\[3ex] Reflex\;\;S\hat{O}Q = 2 * \angle S\hat{R}Q ...\angle \;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] 252 = 2 * S\hat{R}Q \\[3ex] 2 * S\hat{R}Q = 252 \\[3ex] S\hat{R}Q = \dfrac{252}{2} \\[5ex] S\hat{R}Q = 126^\circ \\[3ex] (iii) \\[3ex] O\hat{S}R : O\hat{Q}R = 2:1 \\[3ex] \dfrac{O\hat{S}R}{O\hat{Q}R} = \dfrac{2}{1} \\[5ex] 1 * O\hat{S}R = 2 * O\hat{Q}R \\[3ex] O\hat{S}R = 2 * O\hat{Q}R \\[3ex] Let\;\;O\hat{Q}{R} = m \\[3ex] \implies O\hat{S}{R} = 2m \\[3ex] \underline{Quadrilateral\;SOQR} \\[3ex] \angle S + \angle O + \angle Q + \angle R = 360^\circ ...sum\;\;of\;\;interior\;\;\angle s\;\;of\;\;a\;\;quadrilateral \\[3ex] \angle S = O\hat{S}{R} = 2m \\[3ex] \angle O = Obtuse\;\;S\hat{O}Q = 108^\circ \\[3ex] \angle Q = O\hat{Q}{R} = m \\[3ex] \angle R = S\hat{R}Q = 126^\circ \\[3ex] \implies \\[3ex] 2m + 108 + m + 126 = 360 \\[3ex] 3m + 234 = 360 \\[3ex] 3m = 360 - 234 \\[3ex] 3m = 126 \\[3ex] m = \dfrac{126}{3} \\[5ex] m = 42 \\[3ex] But: \\[3ex] \angle S = O\hat{S}{R} = 2m \\[3ex] \angle O\hat{S}R = 2(42) \\[3ex] \angle O\hat{S}R = 84^\circ $

Construction: Draw the isosceles triangles based on the question

$ \angle XWY = \angle XYW = \angle YZX = \angle YXZ = k \\[3ex] ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle s\;\;WXY\;\;and\;\;XYZ \\[3ex] \underline{\triangle WXY} \\[3ex] \angle XWY + \angle XYW + \angle WXY = 180^\circ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle WXY \\[3ex] k + k + 80 = 180 \\[3ex] 2k = 180 - 80 \\[3ex] 2k = 100 \\[3ex] k = \dfrac{100}{2} \\[5ex] k = 50 \\[3ex] \therefore \angle XWY = \angle XYW = \angle YZX = \angle YXZ = 50^\circ \\[3ex] \underline{\triangle XYZ} \\[3ex] \angle YZX + \angle YXZ + \angle XYZ = 180^\circ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle XYZ \\[3ex] 50 + 50 + \angle XYZ = 180 \\[3ex] 100 + \angle XYZ = 180 \\[3ex] \angle XYZ = 180 - 100 \\[3ex] \angle XYZ = 80^\circ \\[3ex] \underline{Cyclic\;\;Quadrilateral\;XWZY} \\[3ex] \angle W + \angle Y = 180^\circ \\[3ex] ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quad\;\;are\;\;supplementary \\[3ex] \angle Y = \angle XYZ = 80^\circ \\[3ex] \angle W = \angle XWZ \\[3ex] \implies \\[3ex] \angle XWZ + 80 = 180 \\[3ex] \angle XWZ = 180 - 80 \\[3ex] \angle XWZ = 100^\circ $

$ \underline{\triangle AFE} \\[3ex] \angle AEF = \angle EAF = x ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle AFE \\[3ex] \angle AEF + \angle EAF + \angle AFE = 180^\circ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle AFE \\[3ex] x + x + \angle AFE = 180 \\[3ex] 2x + \angle AFE = 180 \\[3ex] \angle AFE = 180 - 2x \\[3ex] \angle EBD = \angle AFE = 180 - 2x ...exterior\;\;\angle\;\;of\;\;cyclic\;\;quadrilateral\;AFEB = interior\;\;opposite\;\;\angle \\[3ex] \angle BED + \angle AEB + \angle AEF = 180 ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] \angle BED + w + x = 180 \\[3ex] \angle BED = 180 - w - x \\[3ex] \angle BDE = \angle ADF = y ...diagram \\[3ex] \underline{\triangle BED} \\[3ex] \angle EBD + \angle BED + \angle BDE = 180^\circ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle BED \\[3ex] (180 - 2x) + (180 - w - x) + y = 180 \\[3ex] 180 - 2x + 180 - w - x + y = 180 \\[3ex] 180 + 180 - 2x - x + y = 180 + w \\[3ex] 360 - 3x + y = 180 + w \\[3ex] 180 + w = 360 - 3x + y \\[3ex] w = 360 - 3x + y - 180 \\[3ex] w = 180 - 3x + y $

$ \underline{\triangle CAB} \\[3ex] \angle BAC = \angle ABC = p ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle CAB \\[3ex] \angle BAC + \angle ABC + \angle BCA = 180^\circ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle CAB \\[3ex] p + p + 110 = 180 \\[3ex] 2p + 110 = 180 \\[3ex] 2p = 180 - 110 \\[3ex] 2p = 70 \\[3ex] p = \dfrac{70}{2} \\[5ex] p = 35 \\[3ex] \therefore \angle ABC = 35^\circ $

$ \angle\;\;between\;\;tangent\;SAT\;\;and\;\;chord\;AG = p \\[3ex] \angle\;\;in\;\;the\;\;alternate\;\;segment\;GHBA = p \\[3ex] Because: \\[3ex] \angle\;\;between\;\;tangent\;SAT\;\;and\;\;chord\;AG = \angle\;\;in\;\;the\;\;alternate\;\;segment\;GHBA \\[3ex] p = p \\[3ex] Also: \\[3ex] \angle GHA = \angle GBA = p ...\angle s\;\;in\;\;the\;\;same\;\;major\;\;segment\;GHBA \\[3ex] \therefore p = p = p $

Let us analyze the options

$ If\;\;BE\;\;is\;\;a\;\;diameter: \\[3ex] then\;\; \angle BFE = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] But: \\[3ex] \angle BFE + \angle FBE + \angle BEF = 180^\circ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle BFE \\[3ex] \angle BFE + 5 + 95 = 180 \\[3ex] \angle BFE + 100 = 180 \\[3ex] \angle BFE = 180 - 100 \\[3ex] \angle BFE = 80^\circ \\[3ex] Because\;\;\angle BFE \ne 90^\circ;\;\;BE \;\;is\;\;NOT\;\;a\;\;diameter \\[3ex] If\;\;AD\;\;is\;\;a\;\;diameter: \\[3ex] then\;\; \angle ACD = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] But: \\[3ex] \angle ACD + \angle ADC + \angle CAD = 180^\circ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle BFE \\[3ex] \angle ACD + 80 + 10 = 180 \\[3ex] \angle ACD + 90 = 180 \\[3ex] \angle ACD = 180 - 90 \\[3ex] \angle ACD = 90^\circ \\[3ex] Because\;\;\angle ACD = 90^\circ;\;\;AD \;\;is\;\;a\;\;diameter \\[3ex] If\;\;AC\;\;is\;\;a\;\;diameter: \\[3ex] then\;\; \angle ADC = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] But\;\;\angle ADC = 80^\circ \\[3ex] Because\;\;\angle ADC \ne 90^\circ;\;\;AC \;\;is\;\;NOT\;\;a\;\;diameter \\[3ex] If\;\;BF\;\;is\;\;a\;\;diameter: \\[3ex] then\;\; \angle BEF = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] But\;\;\angle BEF = 95^\circ \\[3ex] Because\;\;\angle BEF \ne 90^\circ;\;\;BF \;\;is\;\;NOT\;\;a\;\;diameter $

(67.1) Complete the following theorem:
The angle between the tangent to a circle and the chord drawn from the point of contact is equal to the angle in the alternate segment.

$ (67.2.1) \\[3ex] First\;\;\angle = 42^\circ: \\[3ex] \angle QRV = 42^\circ ...\angle s\;\;in\;\;the\;\;segment\;\;are\;\;equal \\[3ex] Second\;\;\angle = 42^\circ: \\[3ex] \angle RVW = \hat{V_4} = 42^\circ ... alternate\;\;\angle s\;\;between\;\;parallel\;\;lines\;QR\;\;and\;\;TW\;\;are\;\;equal \\[3ex] Third\;\;\angle = 42^\circ: \\[3ex] \angle VQR = \hat{Q_1} = 42^\circ ... \angle \;\;between\;\;tangent\;TVW\;\;and\;\;chord\;VR = \angle \;\;in\;\;the\;\;alternate\;\;segment \\[3ex] Fourth\;\;\angle = 42^\circ: \\[3ex] \angle RQV = \hat{V_1} = 42^\circ ... alternate\;\;\angle s\;\;between\;\;parallel\;\;lines\;QR\;\;and\;\;TW\;\;are\;\;equal \\[3ex] (67.2.2) \\[3ex] $ If QR is a diameter of the circle, then $\angle QVR = 90^\circ...\angle \;\;in\;\;a\;\;semicircle\;\;is\;\;a\;\;right\;\;\angle$
If QR is not the diameter of the circle, then $\angle QVR \ne 90^\circ$
So, we need to find the value of $\angle QVR$

$ \underline{\triangle QVR} \\[3ex] \angle RQV + \angle QRV + \angle QVR = 180^\circ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle QVR \\[3ex] 42 + 42 + \angle QVR = 180 \\[3ex] 84 + \angle QVR = 180 \\[3ex] \angle QVR = 180 - 84 \\[3ex] \angle QVR = 96^\circ \\[3ex] Because\;\;\angle QVR \ne 90^\circ: \\[3ex] QR \;\;is\;\;not\;\;a\;\;diameter \\[3ex] (67.2.3) \\[3ex] \underline{\triangle VQP} \\[3ex] \angle QVP + \angle QPV + \angle VQP = 180^\circ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle VQP \\[3ex] \angle QVP = \hat{V_2} = 67^\circ \\[3ex] \angle QPV = \hat{P} = 42^\circ \\[3ex] \angle VQP = \hat{Q_1} + \hat{Q_2} \\[3ex] \hat{Q_1} = \angle RQV = 42^\circ \\[3ex] So,\;\; \angle VQP = 42 + \hat{Q_2} \\[3ex] \implies \\[3ex] 67 + 42 + (42 + \hat{Q_2}) = 180 \\[3ex] 109 + 42 + \hat{Q_2} = 180 \\[3ex] 151 + \hat{Q_2} = 180 \\[3ex] \hat{Q_2} = 180 - 151 \\[3ex] \hat{Q_2} = 29^\circ $

Let us analyze the options

$ \angle AOB = 60^\circ...implies\;\;an\;\;acute\;\;\triangle \\[3ex] $ An acute triangle is a triangle that has three angles less than $90^\circ$
Option A appears in the figure

$ \underline{\triangle AOB} \\[3ex] OA = OB ...same\;\;radii \\[3ex] \implies \triangle AOB\;\;is\;\;an\;\;isosceles\;\;\triangle \\[3ex] \angle OBA = \angle OAB ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle AOB \\[3ex] $ Option C appears in the figure

$ \angle OBA + \angle OAB + \angle AOB = 180^\circ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle ABC \\[3ex] Let\;\;\angle OBA = \angle OAB = p \\[3ex] \angle AOB = 60^\circ \\[3ex] \underline{\triangle AOB} \\[3ex] p + p + 60 = 180 \\[3ex] 2p = 180 - 60 \\[3ex] 2p = 120 \\[3ex] p = \dfrac{120}{2} \\[5ex] p = 60 \\[3ex] \therefore \angle OBA = \angle OAB = 60^\circ \\[3ex] $ Each angle in $\triangle AOB = 60^\circ$
This implies that $\triangle AOB$ is an equilateral triangle.
Option B appears in the figure

$ \underline{\triangle COD} \\[3ex] \angle COD = 90^\circ ...implies\;\;a\;\;right\;\;\triangle \\[3ex] $ A right triangle is a triangle that has one angle of $90^\circ$ and two angles less than $90^\circ$ Option D appears in the figure

The only remaining option is Option E

$ x = 65^\circ ...\angle\;\;between\;\;tangent\;CD\;\;and\;\;chord\;CA = \angle\;\;in\;\;the\;\;alternate\;\;segment $

We observe at least two circle theorems with this circle:
(1.) Angle in a semicircle: we will need to do some construction to join point B to point E before we can use this theorem
(2.) The figure inside the circle is a cyclic quadrilateral. Hence, we shall use at least one of the theorems dealing with cyclic quadrilaterals

Construction: Join point B to point E

$ \underline{\triangle DEB} \\[3ex] \angle DEB = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] Let\;\;\angle DBE = p \\[3ex] \angle BDE + \angle DEB + \angle DBE = 180^\circ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle DBE \\[3ex] \implies \\[3ex] x + 90 + p = 180 \\[3ex] p = 180 - x - 90 \\[3ex] p = 90 - x \\[3ex] \underline{Cyclic\;\;Quadrilateral\;DEAB} \\[3ex] \angle D + \angle A = 180^\circ \\[3ex] ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quad\;\;are\;\;supplementary \\[3ex] x + \angle A = 180 \\[3ex] \angle A = 180 - x \\[3ex] \underline{\triangle BEA} \\[3ex] \angle BEA = \angle EBA = k ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle BEA \\[3ex] \angle BAE = \angle A = 180 - x \\[3ex] \angle BEA + \angle EBA + \angle BEA = 180^\circ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle BEA \\[3ex] \implies \\[3ex] k + k + (180 - x) = 180 \\[3ex] 2k + 180 - x = 180 \\[3ex] 2k = 180 - 180 + x \\[3ex] 2k = x \\[3ex] k = \dfrac{x}{2} \\[5ex] \angle DBA = p + k ...diagram \\[3ex] \angle DBA = (90 - x) + \dfrac{x}{2} \\[5ex] \angle DBA = 90 - x + \dfrac{x}{2} \\[5ex] \angle DBA = \dfrac{90}{1} - \dfrac{x}{1} + \dfrac{x}{2} \\[5ex] \angle DBA = \dfrac{180 - 2x + x}{2} \\[5ex] \angle DBA = \dfrac{180 - x}{2} $

(a.)
(I.)
Based on the information, the diagram is drawn as shown:

$ (II.) \\[3ex] (i.) \\[3ex] \underline{\triangle PTS} \\[3ex] \angle PST = \angle RTS + 34 ...exterior\;\;\angle\;\;a\;\;\triangle = sum\;\;of\;\;two\;\;interior\;\;\angle s \\[3ex] 65 = \angle RTS + 34 \\[3ex] \angle RTS + 34 = 65 \\[3ex] \angle RTS = 65 - 34 \\[3ex] \angle RTS = 31^\circ \\[3ex] (ii.) \\[3ex] $ Construction: Join point S to point R

$ \angle TSQ = \angle SRT = 65^\circ ...\angle \;\;between\;\;tangent\;PSQ\;\;and\;\;chord\;ST = \angle\;\;in\;\;alternate\;\;segment \\[3ex] \angle SRP + \angle SRT = 180^\circ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] \angle SRP + 65 = 180 \\[3ex] \angle SRP = 180 - 65 \\[3ex] \angle SRP = 115^\circ \\[3ex] (b.) \\[3ex] $
$ \angle ZVY = \angle ZYV = 52^\circ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle ZVY \\[3ex] \underline{\triangle XYZ} \\[3ex] \angle YXZ + \angle ZYX + \angle XZY = 180^\circ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle XYZ \\[3ex] \angle ZYX = \angle ZYV = 52^\circ ...diagram \\[3ex] 20 + 52 + \angle XZY = 180 \\[3ex] 72 + \angle XZY = 180 \\[3ex] \angle XZY = 180 - 72 \\[3ex] \angle XZY = 108^\circ \\[3ex] \angle YWZ = \angle ZVY = 52^\circ ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] \underline{\triangle WYZ} \\[3ex] \angle WYZ + \angle YWZ + \angle WZY = 180^\circ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle WYZ \\[3ex] \angle WZY = \angle XZY = 108^\circ...diagram \\[3ex] \implies \\[3ex] \angle WYZ + 52 + 108 = 180 \\[3ex] \angle WYZ + 160 = 180 \\[3ex] \angle WYZ = 180 - 160 \\[3ex] \angle WYZ = 20^\circ $

(i)
BC is a chord
OA is a radius

$ (ii) \\[3ex] (a) \\[3ex] \angle BAO = \angle ABO = 37^\circ ...base\;\;angle s\;\;of\;\;isosceles\;\;\triangle AOB \\[3ex] \underline{\triangle AOB} \\[3ex] \angle AOB + \angle BAO + \angle ABO = 180^\circ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle AOB \\[3ex] \angle AOB = x...diagram \\[3ex] \implies \\[3ex] x + 37 + 37 = 180 \\[3ex] x + 74 = 180 \\[3ex] x = 180 - 74 \\[3ex] x = 106^\circ \\[3ex] (b) \\[3ex] \underline{\triangle BCD} \\[3ex] \angle BCD = 90^\circ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] \angle BDC + \angle BCD + \angle CBD = 180^\circ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle BCD \\[3ex] \angle BDC = y...diagram \\[3ex] \implies \\[3ex] y + 90 + 49 = 180 \\[3ex] y + 139 = 180 \\[3ex] y = 180 - 139 \\[3ex] y = 41^\circ \\[3ex] $ Common Mistake: $y = 37^\circ...\angle s\;\;in\;\;the\;\;segment$
No
$37^\circ$ and $y^\circ$ are NOT in the same segment because |OA| did not extend to point C. (There is no |AC|)

(a.)
The angle between a tangent and a chord through the point of contact is equal to the angle $\underline{subtended}$ by the chord in the alternate segment.


$ (b.) \\[3ex] (i.) \\[3ex] \angle \;\;between\;\;tangent\;BC\;\;and\;\;chord\;BP = \beta \\[3ex] \angle \;\;in\;\;the\;\;alternate\;\;segment = \angle PQB \\[3ex] \beta = \angle PQB... \angle \;\;between\;\;tangent\;BC\;\;and\;\;chord\;BP = \angle \;\;in\;\;the\;\;alternate\;\;segment \\[3ex] \angle PQB = \beta^\circ \\[3ex] (ii.) \\[3ex] \angle \;\;between\;\;tangent\;AC\;\;and\;\;chord\;AP = \alpha \\[3ex] \angle \;\;in\;\;the\;\;alternate\;\;segment = \angle PQA \\[3ex] \alpha = \angle PQA... \angle \;\;between\;\;tangent\;AC\;\;and\;\;chord\;AP = \angle \;\;in\;\;the\;\;alternate\;\;segment \\[3ex] \angle PQA = \alpha^\circ \\[3ex] (iii.) \\[3ex] \angle AQB = \angle PQA + \angle PQB \\[3ex] \angle AQB = \alpha + \beta $

$ l\;\;and\;\;m\;\;are\;\;not\;\;parallel \\[3ex] Eliminate\;\;Option\;I \\[3ex] \underline{Cyclic\;\;Quadrilateral\;ABCD} \\[3ex] \angle A + \angle C = 180^\circ \\[3ex] ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quad\;\;are\;\;supplementary \\[3ex] \angle A + 85 = 180 \\[3ex] \angle A = 180 - 85 \\[3ex] \angle A = 95^\circ \\[3ex] Also: \\[3ex] \angle B + \angle D = 180^\circ \\[3ex] ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quad\;\;are\;\;supplementary \\[3ex] 95 + \angle D = 180 \\[3ex] \angle D = 180 - 95 \\[3ex] \angle D = 85^\circ \\[3ex] Try\;\;Option\;II \\[3ex] $
$ If\;\;\overleftrightarrow{AB} \;||\; \overleftrightarrow{DC} \\[3ex] e = 85^\circ ...corresponding\;\;\angle s\;\;are\;\;congruent \\[3ex] But: \\[3ex] e + 95^\circ = 180^\circ ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] e = 180 - 95 \\[3ex] e = 85^\circ \\[3ex] 85^\circ = 85^\circ \\[3ex] \implies that\;\; \overleftrightarrow{AB} \;||\; \overleftrightarrow{DC} \\[3ex] Try\;\;Option\;III \\[3ex] $
$ If\;\;\overleftrightarrow{AD} \;||\; \overleftrightarrow{BC} \\[3ex] f = 95^\circ ...corresponding\;\;\angle s\;\;are\;\;congruent \\[3ex] But: \\[3ex] f + 95^\circ = 180^\circ ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] f = 180 - 95 \\[3ex] f = 85^\circ \\[3ex] 95^\circ \ne 85^\circ \\[3ex] \implies that\;\; \overleftrightarrow{AD} \;\not||\; \overleftrightarrow{BC} \\[3ex] Eliminate\;\;Option\;III \\[3ex] Correct\;\;Option:\;\;II\;\;only $

$ q = \angle VUR ...\angle \;\;between\;\;tangent\;SRT \;\;and\;\;chord\;VR = \angle\;\;in\;\;the\;\;second\;\;segment \\[3ex] \underline{\triangle UVR} \\[3ex] \angle VUR = \angle VRU = q...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle \\[3ex] \implies \\[3ex] \angle VUR + \angle VRU + \angle UVR = 180^\circ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle UVR \\[3ex] q + q + 38 = 180 \\[3ex] 2q + 38 = 180 \\[3ex] 2q = 180 - 38 \\[3ex] 2q = 142 \\[3ex] q = \dfrac{142}{2} \\[5ex] q = 71^\circ $

Construction: Join the radius from point O to point B

Intercepted arc = angle at center (central angle)

$ \angle AOB = \overset{\huge\frown}{AB} = 80^\circ \\[3ex] \underline{\triangle OAB} \\[3ex] \angle BAO = \angle ABO = p ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle OAB \\[3ex] \angle BAO + \angle ABO + \angle AOB = 180^\circ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle OAB \\[3ex] \implies \\[3ex] p + p + 80 = 180 \\[3ex] 2p = 180 - 80 \\[3ex] 2p = 100 \\[3ex] p = \dfrac{100}{2} \\[5ex] p = 50 \\[3ex] \therefore \angle BAO = \angle ABO = 50^\circ $

$ (a) \\[3ex] (i) \\[3ex] a = 90^\circ ... \angle\;\;in\;\;a\;\;semicircle \\[3ex] $
$ (ii) \\[3ex] b^\circ = c^\circ ...\angle\;\;between\;\;tangent\;EG\;\;and\;\;chord\;EF = \angle\;\;in\;\;alternate\;\;segment \\[3ex] Also: \\[3ex] \angle EFG = c^\circ ...\angle\;\;between\;\;tangent\;FG\;\;and\;\;chord\;EF = \angle\;\;in\;\;alternate\;\;segment \\[3ex] \underline{\triangle EFG} \\[3ex] \angle FEG + \angle EFG + \angle FGE = 180^\circ ... sum\;\;of\;\;\angle s\;\;in\;\;\triangle EFG \\[3ex] b + c + 52 = 180 \\[3ex] But\;\; b = c \\[3ex] c + c + 52 = 180 \\[3ex] 2c = 180 - 52 \\[3ex] 2c = 128 \\[3ex] c = \dfrac{128}{2} \\[5ex] c = 64^\circ \\[3ex] b = c = 64^\circ \\[5ex] (b) \\[3ex] $
$ (i) \\[3ex] If:\;\;\triangle ADE \sim \triangle CBE \\[3ex] Then:\;\; \dfrac{AE}{CE} = \dfrac{ED}{EB} ...Similarity\;\;Ratio \;(Scale\;\;Factor) \\[5ex] AE * EB = CE * ED ... Intersecting\;\;Chords\;\;Theorem \\[3ex] \implies \\[3ex] \dfrac{AE}{CE} = \dfrac{ED}{EB} \\[5ex] (ii) \\[3ex] AD : CB = 3 : 2 \\[3ex] AD : BC = 3 : 2 \\[3ex] \implies \\[3ex] \dfrac{AD}{BC} = \dfrac{3}{2} \\[5ex] BC = 4\;cm \\[3ex] AD = ? \\[3ex] \implies \\[3ex] \dfrac{AD}{BC} = \dfrac{3}{2} = \dfrac{AD}{4} \\[5ex] \dfrac{AD}{4} = \dfrac{3}{2} \\[5ex] AD = \dfrac{3 * 4}{2} \\[5ex] AD = 3(2) \\[3ex] AD = 6\;cm $

Construction: Join the chord from point B to point C

$ \angle BDC = \angle BAC = 24^\circ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] \angle AFE + \angle EFD + \angle DFC = 180^\circ ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] x + 2x + \angle DFC = 180 \\[3ex] 3x + \angle DFC = 180 \\[3ex] \angle DFC = 180 - 3x \\[3ex] \underline{\triangle DFC} \\[3ex] \angle DFC = \angle DCF = 180 - 3x ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle DFC \\[3ex] \angle DFC + \angle DCF + \angle FDC = 180^\circ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle DFC \\[3ex] \angle FDC = \angle BDC = 24^\circ ...diagram \\[3ex] \implies \\[3ex] (180 - 3x) + (180 - 3x) + 24 = 180 \\[3ex] 180 - 3x + 180 - 3x + 24 = 180 \\[3ex] 384 - 6x = 180 \\[3ex] 384 - 180 = 6x \\[3ex] 204 = 6x \\[3ex] 6x = 204 \\[3ex] x = \dfrac{204}{6} \\[5ex] x = 34^\circ $

$ (i) \\[3ex] \angle PTR = 75^\circ ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] (ii) \\[3ex] \underline{Cyclic\;\;Quadrilateral\;PTRQ} \\[3ex] $
$ \angle TPQ = 60^\circ ...exterior\;\;\angle\;\;of\;\;a\;\;cyclic\;\;quadrilateral\;PTRQ = interior\;\;opposite\;\;\angle \\[3ex] $ Construction:
(a.) Indicate the centre of the circle, O
(b.) Draw the line from point P to centre O
(c.) Draw the line from centre O to point R

$ Obtuse\;\;\angle POR = 2 * \angle PVR ...\angle \;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] \angle POR = 2 * 75 \\[3ex] \angle POR = 150^\circ $

$ \theta = 145^\circ...interior\;\;alternate\;\;\angle s\;\;are\;\;equal \\[3ex] \angle BEC + \theta = 180^\circ ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] \angle BEC + 145 = 180 \\[3ex] \angle BEC = 180 - 145 \\[3ex] \angle BEC = 35^\circ \\[3ex] \angle CBE = 90^\circ \\[3ex] ...radius\;CB\;\perp\;tangent\;qBE\;\;at\;\;point\;\;of\;\;contact\;B \\[3ex] \underline{\triangle CBE} \\[3ex] \angle BCE + \angle BEC + \angle CBE = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle CBE \\[3ex] \angle BCE + 35 + 90 = 180 \\[3ex] \angle BCE + 125 = 180 \\[3ex] \angle BCE = 180 - 125 \\[3ex] \angle BCE = 55^\circ $

$ (i) \\[3ex] \angle OCE = 90^\circ ...radius\;OC \; \perp\; tangent\;DCE \;\;at\;\;point\;\;of\;\;contact\;C \\[3ex] (ii) \\[3ex] \angle BCD = \angle BAC = 70^\circ ... \angle\;\;between\;\;chord\;BC\;\;and\;\;tangent\;DCE = \angle\;\;in\;\;the\;\;alternate\;\;segment \\[3ex] (iii) \\[3ex] \angle BOC = 2 * \angle BAC ...\angle \;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] \angle BOC = 2 * 70 \\[3ex] \angle BOC = 140^\circ \\[3ex] (iv) \\[3ex] |BD| = |DCE| ...two\;\;tangents:\;BD\;\;and\;\;DCE\;\;drawn\;\;from\;\;the\;\;same\;\;external\;\;point:\;A\;\;are\;\;equal\;\;in\;\;length \\[3ex] \implies \\[3ex] \underline{\triangle BDC} \\[3ex] \angle BCD = \angle CBD = 70^\circ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle BDC \\[3ex] \angle BCD + \angle CBD + \angle BDC = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle BDC \\[3ex] 70 + 70 + \angle BDC = 180 \\[3ex] 140 + \angle BDC = 180 \\[3ex] \angle BDC = 180 - 140 \\[3ex] \angle BDC = 40^\circ $

Construction: Join the radius from the centre, O to point B on the circumference


$ \angle ABO = \angle BAO = x ...base\;\;\angle s \;\;of\;\;isosceles\;\;\triangle AOB \\[3ex] \angle OBC = 90^\circ \\[3ex] ...radius\;OB\;\perp\;tangent\;BC\;\;at\;\;point\;\;of\;\;contact\;B \\[3ex] \angle BOC = x + x \\[3ex] ... exterior\;\;\angle\;\;of\;\;a\;\;\triangle = sum\;\;of\;\;two\;\;interior\;\;opposite\;\;\angle s \\[3ex] \angle BOC = 2x \\[3ex] \underline{\triangle OBC} \\[3ex] \angle BOC + \angle OBC + \angle OCB = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle OBC \\[3ex] 2x + 90 + \angle OCB = 180 \\[3ex] 2x + 90 + \angle OCB = 180 \\[3ex] \angle OCB = 180 - 2x - 90 \\[3ex] \angle OCB = 90 - 2x \\[3ex] \angle OCB = 2(45 - x) \\[3ex] \angle OCB = \angle ACB ...diagram \\[3ex] \therefore \angle ACB = 2(45 - x) $

Sam got the answer correctly.
However, the reasons are incorrect.
He is supposed to state his working as:

$ \angle ACB = 56^\circ...\angle \;\;between\;\;tangent\;PAQ\;\;and\;\;chord\;AB = \angle\;\;in\;\;the\;\;alternate\;\;segment \\[3ex] m = 180 - 56...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] m = 124^\circ $

Let us analyze the options

$ \angle TMU = 50^\circ ...vertical \angle s\;\;are\;\;equal \\[3ex] \underline{\triangle MTU} \\[3ex] \angle UTM = \angle TUM = p ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle MTU \\[3ex] \angle UTM + \angle TUM + \angle TMU = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle MTU \\[3ex] p + p + 50 = 180 \\[3ex] 2p + 50 = 180 \\[3ex] 2p = 180 - 50 \\[3ex] 2p = 130 \\[3ex] p = \dfrac{130}{2} \\[5ex] p = 65 \\[3ex] \therefore \angle UTM = \angle TUM = 65^\circ \\[3ex] Option\;\;F\;\;is\;\;true \\[3ex] \underline{\triangle MRS} \\[3ex] \angle SRM = \angle RSM = k ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle MRS \\[3ex] \angle SRM + \angle RSM + \angle RMS = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle MRS \\[3ex] k + k + 50 = 180 \\[3ex] 2k + 50 = 180 \\[3ex] 2k = 180 - 50 \\[3ex] 2k = 130 \\[3ex] k = \dfrac{130}{2} \\[5ex] k = 65 \\[3ex] \therefore \angle SRM = \angle RSM = 65^\circ \\[3ex] \angle SRM = \angle SRU = 65^\circ ...diagram \\[3ex] \angle TUM = \angle TUR = 65^\circ ...diagram \\[3ex] Because: \\[3ex] \angle SRU = \angle TUR = 65^\circ ...alternate \angle s\;\;are\;\;equal \\[3ex] \implies \;\;that\;\; \overline{RS} \;\;||\;\; \overline{TU} \\[3ex] Option\;\;G\;\;is\;\;true \\[3ex] \overset{\Huge\frown}{TXU} = 50^\circ ...vertical \angle s\;\;are\;\;equal \\[3ex] ...also\;\;\angle\;\;subtended\;\;by\;\;arc\;TXU \\[3ex] Option\;\;H\;\;is\;\;true \\[3ex] \overline{RM} = \overline{TM} ...same\;\;radii\;\;from\;\;center\;M \\[3ex] Option\;\;J\;\;is\;\;true \\[3ex] For\;\;the\;\;remaining\;\; Option\;\;K \\[3ex] \overline{RM} = \overline{SM} ...same\;\;radii\;\;from\;\;center\;M \\[3ex] \implies\;\; \triangle RMS\;\;is\;\;an\;\;isosceles\;\;\triangle \\[3ex] However: \\[3ex] \overline{RS} \ncong \overline{SM} \\[3ex] $ Because $\triangle RMS$ is not an equilateral triangle
An equilateral triangle must have each of the three angles as $60^\circ$ in order for all sides to be congruent

$ Option\;\;K\;\;is\;\;NOT\;\;true $

$ (i) \\[3ex] \angle LAD = \angle CAD = 33^\circ...diagram \\[3ex] \underline{\triangle ALD} \\[3ex] Because\;\;the\;\;climbing\;\;frame\;\;is\;\;symmetrical\;\;about\;\;FE: \\[3ex] \angle LDA = \angle LAD = 33^\circ \\[3ex] \angle LDA + \angle LAD + \angle ALD = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle ALD \\[3ex] 33 + 33 + x = 180 \\[3ex] 66 + x = 180 \\[3ex] x = 180 - 66 \\[3ex] x = 114^\circ \\[3ex] (ii) \\[3ex] \angle DBA = 90^\circ ...\angle \;\;in\;\;a\;\;semicircle \\[3ex] \angle ALD = \angle DBA + \angle BAC ... exterior\;\;\angle\;\;of\;\;a\;\;\triangle = sum\;\;of\;\;the\;\;two\;\;interior\;\;opposite\;\;\angle s \\[3ex] x = 90 + y \\[3ex] 90 + y = x \\[3ex] y = x - 90 \\[3ex] y = 114 - 90 \\[3ex] y = 24^\circ $

$ (i) \\[3ex] \angle SAW = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] \angle SAF + \angle FAW = \angle SAW ...diagram \\[3ex] 54 + \angle FAW = 90 \\[3ex] \angle FAW = 90 - 54 \\[3ex] \angle FAW = 36^\circ \\[3ex] (ii) \\[3ex] \underline{Cyclic\;\;Quadrilateral\;SKFA} \\[3ex] \angle A + \angle K = 180^\circ \\[3ex] ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quad\;\;are\;\;supplementary \\[3ex] \angle A = \angle SAF = 54^\circ \\[3ex] \angle K = \angle SKF \\[3ex] \implies \\[3ex] 54 + \angle SKF = 180 \\[3ex] \angle SKF = 180 - 54 \\[3ex] \angle SKF = 126^\circ \\[3ex] (iii) \\[3ex] $
$ \theta = 62^\circ ...alternate\;\;\angle s\;\;are\;\;equal \\[3ex] \angle ASB + 54 + \theta = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle SBA \\[3ex] \angle ASB + 54 + 62 = 180 \\[3ex] \angle ASB + 116 = 180 \\[3ex] \angle ASB = 180 - 116 \\[3ex] \angle ASB = 64^\circ \\[3ex] \angle ASW = \angle ASB ...diagram \\[3ex] \therefore \angle ASW = 64^\circ $

Construction: Join the chord from point N to point P

$ (i) \\[3ex] \angle QNP = 90^\circ...\angle \;\;in\;\;a\;\;semicircle \\[3ex] x = \angle QPN ...\angle\;\;between\;\;tangent\;MN\;\;and\;\;a\;\;chord\;NQ = \angle\;\;in\;\;alternate\;\;segment \\[3ex] \underline{\triangle NQP} \\[3ex] \angle QNP + \angle QPN + \angle NQP = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle NQP \\[3ex] 90 + x + 46 = 180 \\[3ex] x + 136 = 180 \\[3ex] x = 180 - 136 \\[3ex] x = 44^\circ \\[3ex] (ii) \\[3ex] \angle PQM = 90^\circ ...radius\;OQ\;\perp\;tangent\;MQ\;\;at\;\;point\;\;of\;\;contact\;Q \\[3ex] \angle NQM + \angle NQP = \angle PQM ...diagram \\[3ex] \angle NQM + 46 = 90 \\[3ex] \angle NQM = 90 - 46 \\[3ex] \angle NQM = 44^\circ \\[3ex] \underline{\triangle MNQ} \\[3ex] \angle MNQ + \angle NMQ + \angle NQM = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle MNQ \\[3ex] x + y + 44 = 180 \\[3ex] 44 + y + 44 = 180 \\[3ex] y + 88 = 180 \\[3ex] y = 180 - 88 \\[3ex] y = 92^\circ \\[3ex] OR \\[3ex] (i) \;\;and\;\; (ii) \\[3ex] \angle PQM = 90^\circ ...radius\;OQ\;\perp\;tangent\;MQ\;\;at\;\;point\;\;of\;\;contact\;Q \\[3ex] \angle NQM + \angle NQP = \angle PQM ...diagram \\[3ex] \angle NQM + 46 = 90 \\[3ex] \angle NQM = 90 - 46 \\[3ex] \angle NQM = 44^\circ \\[3ex] |MN| = |MQ|...two\;\;tangents\;\;MN\;\;and\;\;MQ\;\;drawn\;\;from\;\;the\;\;same\;\;external\;\;point\;M\;\;are\;\;equal\;\;in\;\;length \\[3ex] \implies \angle NQM = \angle MNQ = 44^\circ = x ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle MNQ \\[3ex] \underline{\triangle MNQ} \\[3ex] \angle MNQ + \angle NMQ + \angle NQM = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle MNQ \\[3ex] 44 + y + 44 = 180 \\[3ex] y + 88 = 180 \\[3ex] y = 180 - 88 \\[3ex] y = 92^\circ $

$ First\;\;Approach:\;\;Recommmended\;\;for\;\;ACT \\[3ex] \underline{\triangle PST} \\[3ex] \angle PST = \angle PTS = 30^\circ ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle PST \\[3ex] \overset{\huge\frown}{RS} = \angle PST + \angle PTS \\[3ex] ...exterior\;\;\angle\;\;a\;\;\triangle = sum\;\;of\;\;the\;\;two\;\;interior\;\;opposite\;\;\angle s\\[3ex] \overset{\huge\frown}{RS} = 30 + 30 \\[3ex] \overset{\huge\frown}{RS} = 60^\circ \\[3ex] OR \\[3ex] Second\;\;Approach: \\[3ex] \angle PST + \angle PTS + \angle TPS = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle PST \\[3ex] 30 + 30 + \angle TPS = 180 \\[3ex] 60 + \angle TPS = 180 \\[3ex] \angle TPS = 180 - 60 \\[3ex] \angle TPS = 120^\circ \\[3ex] \angle TPS + \overset{\huge\frown}{RS} = 180^\circ ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] 120 + \overset{\huge\frown}{RS} = 180 \\[3ex] \overset{\huge\frown}{RS} = 180 - 120 \\[3ex] \overset{\huge\frown}{RS} = 60^\circ $

$ \underline{\triangle AOD} \\[3ex] \angle OAD + \angle ODA + \angle AOD = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle AOD \\[3ex] \angle OAD = \angle OAB = 50^\circ \\[3ex] \angle ODA = \angle ODB = 22^\circ \\[3ex] \implies \\[3ex] 50 + 22 + \angle AOD = 180 \\[3ex] 72 + \angle AOD = 180 \\[3ex] \angle AOD = 180 - 72 \\[3ex] \angle AOD = 108^\circ \\[3ex] \angle AOD = \angle EOF = 108^\circ ...vertical\;\;\angle s\;\;are\;\;equal \\[3ex] \underline{\triangle EOF} \\[3ex] \angle OEF = \angle OFE = p ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle EOF \\[3ex] \angle OEF + \angle OFE + \angle EOF = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle EOF \\[3ex] p + p + 108 = 180 \\[3ex] 2p = 180 - 108 \\[3ex] 2p = 72 \\[3ex] p = \dfrac{72}{2} \\[5ex] p = 36 \\[3ex] \therefore \angle OEF = \angle OFE = 36^\circ \\[3ex] Also: \\[3ex] \angle OBD = 90^\circ ... radius\;OB \;\perp\;tangent\;ABD\;\;at\;\;a\;\;point\;\;of\;\;contact\;B \\[3ex] \underline{\triangle BOD} \\[3ex] \angle BOD + \angle OBD + \angle ODB = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle BOD \\[3ex] \angle BOD + 90 + 22 = 180 \\[3ex] \angle BOD + 112 = 180 \\[3ex] \angle BOD = 180 - 112 \\[3ex] \angle BOD = 68^\circ \\[3ex] \angle EOJ = \angle BOD = 68^\circ ...vertical\;\;\angle s\;\;are\;\;equal \\[3ex] \underline{\triangle EOJ} \\[3ex] \angle EOJ + \angle OEJ + \angle EJO = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle EOJ \\[3ex] \angle OEJ = \angle OEF = 36^\circ ...diagram \\[3ex] \implies \\[3ex] 68 + 36 + z = 180 \\[3ex] 104 + z = 180 \\[3ex] z = 180 - 104 \\[3ex] z = 76^\circ $

$ \underline{Cyclic\;\;Quadrilateral\;QRST} \\[3ex] \angle R + \angle T = 180^\circ \\[3ex] ... opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quad\;\;are\;\;supplementary \\[3ex] \angle R + 52 = 180 \\[3ex] \angle R = 180 - 52 \\[3ex] \angle R = 128^\circ \\[3ex] \underline{\triangle QRS} \\[3ex] \angle RQS = \angle RSQ = k ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle QRS \\[3ex] \angle RQS + \angle RSQ + \angle QRS = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle QRS \\[3ex] \angle QRS = \angle R = 128^\circ \\[3ex] k + k + 128 = 180 \\[3ex] 2k = 180 - 128 \\[3ex] 2k = 52 \\[3ex] k = \dfrac{52}{2} \\[5ex] k = 26 \\[3ex] \therefore \angle RQS = \angle RSQ = 26^\circ \\[3ex] (i) \\[3ex] \angle RSQ = \angle SQT = 26^\circ ...alternate \;\;\angle s\;\;are\;\;equal \\[3ex] (ii) \\[3ex] \angle PQS = 90^\circ ...\angle \;\;in\;\;a\;\;semicircle \\[3ex] \angle PQS = \angle PQT + \angle SQT ...diagram \\[3ex] 90 = \angle PQT + 26 \\[3ex] \angle PQT + 26 = 90 \\[3ex] \angle PQT = 90 - 26 \\[3ex] \angle PQT = 64^\circ $

Construction: Join the radius from the centre O to the point B on the circumference

$ (a) \\[3ex] \angle ODB = \angle OBD = x ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle BOD \\[3ex] \underline{\triangle BOD} \\[3ex] \angle ODB + Obtuse\;\angle OBD + \angle BOD = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle BOD \\[3ex] x + x + Obtuse\;\angle BOD = 180 \\[3ex] 2x + Obtuse\;\angle BOD = 180 \\[3ex] Obtuse\;\angle BOD = 180 - 2x \\[3ex] Obtuse\;\angle BOD + Reflex\;\angle BOD = 360^\circ ...\angle s\;\;at\;\;a\;\;point \\[3ex] (180 - 2x) + Reflex\;\angle BOD = 360 \\[3ex] Reflex\;\angle BOD = 360 - (180 - 2x) \\[3ex] Reflex\;\angle BOD = 360 - 180 + 2x \\[3ex] Reflex\;\angle BOD = 180 + 2x \\[3ex] Reflex\;\angle BOD = 2 * \angle BAD ...\angle \;\;at\;\;centre = 2 * \angle \;\;at\;\;circumference \\[3ex] 180 + 2x = 2 * y \\[3ex] 180 + 2x = 2y \\[3ex] 2y = 180 + 2x \\[3ex] 2y = 2(90 + x) \\[3ex] Divide\;\;both\;\;sides\;\;by\;\;2 \\[3ex] y = 90 + x \\[3ex] \therefore y - x = 90 \\[3ex] (b) \\[3ex] Based\;\;on\;\;Dylan's\;\;response: \\[3ex] If\;\;y = 200: \\[3ex] Then\;\; Reflex\;\;\angle BOD = 2y = 2(200) = 400 \\[3ex] 400\;\;is\;\;not\;\;a\;\;Reflex\;\;\angle \\[3ex] This\;\;is\;\;not\;\;possible \\[3ex] Dylan\;\;is\;\;not\;\;correct $

Construction: Join the chord from point A to point C


$ 190 = 2 * \angle BAC ...intercepted\;\;arc = 2 * inscribed\;\;\angle \\[3ex] 2 * \angle BAC = 190 \\[3ex] \angle BAC = \dfrac{190}{2} \\[5ex] \angle BAC = 95^\circ \\[3ex] 90 = 2 * \angle BCA ...intercepted\;\;arc = 2 * inscribed\;\;\angle \\[3ex] 2 * \angle BCA = 90 \\[3ex] \angle BCA = \dfrac{90}{2} \\[5ex] \angle BCA = 45^\circ \\[3ex] \underline{\triangle BAC} \\[3ex] \angle ABC + \angle BAC + \angle BCA = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle BAC \\[3ex] \angle ABC + 95 + 45 = 180 \\[3ex] \angle ABC + 140 = 180 \\[3ex] \angle ABC = 180 - 140 \\[3ex] \angle ABC = 40^\circ $

$ (i) \\[3ex] \angle OSP = 26^\circ ...alternate\;\;\angle s\;\;are\;\;equal \\[3ex] \angle OSP = \angle OPS = 26^\circ ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle POS \\[3ex] \angle OSP + \angle OPS + \angle POS = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle POS \\[3ex] 26 + 26 + \angle POS = 180 \\[3ex] 52 + \angle POS = 180 \\[3ex] \angle POS = 180 - 52 \\[3ex] \angle POS = 128^\circ \\[3ex] \angle POS = 2 * \angle PTS ...\angle \;\;at\;\;centre = 2 * \angle \;\;at\;\;circumference \\[3ex] 128 = 2 * \angle PTS \\[3ex] 2 * \angle PTS = 128 \\[3ex] \angle PTS = \dfrac{128}{2} \\[5ex] \angle PTS = 64^\circ \\[3ex] (ii) \\[3ex] \angle SPQ = \angle PTS = 64^\circ... \angle\;\;between\;\;tangent\;PQ\;\;and\;\;chord\;PS = \angle\;\;in\;\;the\;\;alternate\;\;segment \\[3ex] \angle SPQ = \angle SPR + \angle RPQ ...diagram \\[3ex] 64 = 26 + \angle RPQ \\[3ex] 26 + \angle RPQ = 64 \\[3ex] \angle RPQ = 64 - 26 \\[3ex] \angle RPQ = 38^\circ $

$ \angle UQR = \angle PSR = 68^\circ \\[3ex] ...exterior\;\;\angle\;\;of\;\;cyclic\;\;quad\;PQRS = interior\;\;opposite\;\;\angle \\[3ex] \angle QSR = \angle RPQ = 40^\circ ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] \angle TPS + \angle SPR + \angle RPQ = 180^\circ \\[3ex] ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] 74 + \angle SPR + 40 = 180 \\[3ex] 114 + \angle SPR = 180 \\[3ex] \angle SPR = 180 - 114 \\[3ex] \angle SPR = 66^\circ \\[3ex] \underline{\triangle PSR} \\[3ex] \angle PSR + \angle PRS + \angle SPR = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle PSR \\[3ex] 68 + \angle PRS + 66 = 180 \\[3ex] 134 + \angle PRS = 180 \\[3ex] \angle PRS = 180 - 134 \\[3ex] \angle PRS = 46^\circ \\[3ex] OR \\[3ex] \angle TPS = \angle QRS = 74^\circ \\[3ex] ...exterior\;\;\angle\;\;of\;\;cyclic\;\;quad\;PQRS = interior\;\;opposite\;\;\angle \\[3ex] Similarly: \\[3ex] \angle UQR = \angle PSR = 68^\circ \\[3ex] ...exterior\;\;\angle\;\;of\;\;cyclic\;\;quad\;PQRS = interior\;\;opposite\;\;\angle \\[3ex] But:\;\; \angle PSR = \angle PSQ + \angle QSR ...diagram \\[3ex] 68 = \angle PSQ + 40 \\[3ex] \angle PSQ + 40 = 68 \\[3ex] \angle PSQ = 68 - 40 \\[3ex] \angle PSQ = 28 \\[3ex] \angle PSQ = \angle PRQ = 28^\circ ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] But:\;\;\angle QRS = \angle PRQ + \angle PRS \\[3ex] 74 = 28 + \angle PRS \\[3ex] 28 + \angle PRS = 74 \\[3ex] \angle PRS = 74 - 28 \\[3ex] \angle PRS = 46^\circ $

$ (i) \\[3ex] \angle ACE = \angle ABC...\angle \;\;between\;\;tangent\;DCE\;\;and\;\;chord\;AC = \angle\;\;in\;\;the\;\;alternate\;\;segemnt \\[3ex] 48 = \angle ABC \\[3ex] \angle ABC = 48^\circ \\[3ex] (ii) \\[3ex] \angle AOC = 2 * \angle ABC ...\angle \;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] \angle AOC = 2 * 48 \\[3ex] \angle AOC = 96^\circ \\[3ex] (iii) \\[3ex] \angle OCD = 90^\circ ...radius\;OC\;\perp\;tangent\;DCE\;\;at\;\;point\;\;of\;\;contact\;C \\[3ex] But:\;\;\angle OCD = \angle OCB + \angle BCD ...diagram \\[3ex] 90 = 26 + \angle BCD \\[3ex] 26 + \angle BCD = 90 \\[3ex] \angle BCD = 90 - 26 \\[3ex] \angle BCD = 64^\circ \\[3ex] (iv) \\[3ex] \angle BCD = \angle BAC...\angle \;\;between\;\;tangent\;DCE\;\;and\;\;chord\;BC = \angle\;\;in\;\;the\;\;alternate\;\;segemnt \\[3ex] 64 = \angle BAC \\[3ex] \angle BAC = 64^\circ \\[3ex] (v) \\[3ex] \underline{\triangle AOC} \\[3ex] \angle OAC = \angle OCA = p ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle AOC \\[3ex] \angle OAC + \angle OCA + \angle AOC = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle AOC \\[3ex] p + p + 96 = 180 \\[3ex] 2p = 180 - 96 \\[3ex] 2p = 84 \\[3ex] p = \dfrac{84}{2} \\[5ex] p = 42 \\[3ex] \therefore \angle OAC = \angle OCA = 42^\circ \\[3ex] (vi) \\[3ex] \angle BAC = \angle OAB + \angle OAC ...diagram \\[3ex] 64 = \angle OAB + 42 \\[3ex] \angle OAB + 42 = 64 \\[3ex] \angle OAB = 64 - 42 \\[3ex] \angle OAB = 22^\circ $

$ \overline{AD}\;\;is\;\;a\;\;diameter \\[3ex] \angle ACD = 90^\circ...\angle \;\;in\;\;a\;\;semicircle \\[3ex] \underline{\triangle ACD} \\[3ex] \angle ADC + \angle ACD + \angle CAD = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle ACD \\[3ex] \angle ADC + 90 + 30 = 180 \\[3ex] \angle ADC + 120 = 180 \\[3ex] \angle ADC = 180 - 120 \\[3ex] \angle ADC = 60^\circ \\[3ex] \overset{\huge\frown}{AC} = 2 * \angle ADC ... intercepted\;\;arc = 2 * inscribed\;\;\angle \\[3ex] \overset{\huge\frown}{AC} = 2 * 60 \\[3ex] \overset{\huge\frown}{AC} = 120^\circ $

$ \angle SRQ = \angle SPQ = 50^\circ ...\angle \:\:between\:\:tangent\;RQ\:\:and\:\:chord\;QS = \angle\:\:in\;\;the\:\:alternate\:\:segment \\[3ex] \angle SPQ = \angle SQP = 50^\circ ...base\:\: \angle s\:\:of\:\:isosceles\:\: \triangle \\[3ex] \underline{\triangle SRQ} \\[3ex] \angle RSQ = \angle SPQ + \angle SQP ...exterior\:\: \angle \:\:of\:\:a\:\: \triangle = sum\;\;of\;\;the\;\;two\;\;interior\;\;opposite\;\;\angle s \\[3ex] \angle RSQ = 50 + 50 \\[3ex] \angle RSQ = 100^\circ \\[3ex] \angle SRQ + \angle RSQ + \angle SQR = 180^\circ ...sum\:\:of\:\:\angle s\:\:of\:\:\triangle SRQ \\[3ex] \angle SRQ + 100 + 50 = 180 \\[3ex] \angle SRQ + 150 = 180 \\[3ex] \angle SRQ = 180 - 150 \\[3ex] \angle SRQ = 30^\circ $

$ \angle ABC = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] \angle CAB + \angle ABC + \angle ACB = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle ACB \\[3ex] 25 + 90 + \angle ACB = 180 \\[3ex] 115 + \angle ACB = 180 \\[3ex] \angle ACB = 180 - 115 \\[3ex] \angle ACB = 65^\circ \\[3ex] \angle ADB = \angle ACB = 65^\circ ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] \angle AED + \angle DEC = 180^\circ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] \angle AED + 100 = 180 \\[3ex] \angle AED = 180 - 100 \\[3ex] \angle AED = 80^\circ \\[3ex] \underline{\triangle AED} \\[3ex] \angle DAE + \angle AED + \angle ADE = 180^\circ ...sum\:\:of\:\:\angle s\:\:of\:\:\triangle ADE \\[3ex] \angle ADE = \angle ADB = 65^\circ...diagram \\[3ex] \implies \\[3ex] \angle DAE + 80 + 65 = 180 \\[3ex] \angle DAE + 145 = 180 \\[3ex] \angle DAE = 180 - 145 \\[3ex] \angle DAE = 35^\circ \\[3ex] \angle DAC = \angle DAE = 35^\circ ...diagram $

$ \angle BCE = \angle BEC = x ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle BCE \\[3ex] \angle BEI = 90^\circ ...radius\;BE \perp tangent\;GEI\;\;at\;\;point\;\;of\;\;contact\;E \\[3ex] \angle CEG + \angle BEC + \angle BEI = 180^\circ ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] 38 + x + 90 = 180 \\[3ex] x + 128 = 180 \\[3ex] x = 180 - 128 \\[3ex] x = 52^\circ \\[3ex] $ Common mistake: $\angle CEG = \angle EBC = 38^\circ...\angle \;\;between\;\;tangent\;\;and\;\;chord = \angle\;\;in\;\;alternate\;\;segment$
This is not correct because $\angle EBC$ is not in the alternate segment.
The alternate segement theorem requires that the angle is formed by the intersection of two chords that touches the circumference of the circle (not the angle at the centre of the circle formed by the intersection of two radii)

$ (i) \\[3ex] \underline{\triangle WXY} \\[3ex] \angle XWY + \angle WXY + \angle XYW = 180^\circ ...sum\:\:of\:\:\angle s\:\:of\:\:\triangle WXY \\[3ex] \angle XWY + 50 + 90 = 180 \\[3ex] \angle XWY + 140 = 180 \\[3ex] \angle XWY = 180 - 140 \\[3ex] \angle XWY = 40^\circ \\[3ex] \angle XWY = \angle WYZ = 40^\circ ...alternate\;\;\angle s\;\;are\;\;equal \\[3ex] \angle WOZ = 2 * \angle WYZ...\angle \;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] \angle WOZ = 2 * 40 \\[3ex] \angle WOZ = 80^\circ \\[3ex] (ii) \\[3ex] \underline{\triangle OWE} \\[3ex] \angle WOE + \angle OWE + \angle OEW = 180^\circ ...sum\:\:of\:\:\angle s\:\:of\:\:\triangle OWE \\[3ex] \angle WOE = \angle WOZ = 80^\circ...diagram \\[3ex] \angle OWE = \angle XWY = 40^\circ \\[3ex] \implies \\[3ex] 80 + 40 + \angle OEW = 180 \\[3ex] 120 + \angle OEW = 180 \\[3ex] \angle OEW = 180 - 120 \\[3ex] \angle OEW = 60^\circ \\[3ex] \angle YEZ = \angle OEW = 60^\circ ...vertical\;\;\angle s\;\;are\;\;equal $

Construction: Draw the radius from the centre O to point P on the circumference

$ \angle OPS = \angle OSP = 48^\circ ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle POS \\[3ex] \angle OPS + \angle OSP + \angle POS = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle POS \\[3ex] 48 + 48 + \angle POS = 180 \\[3ex] 96 + \angle POS = 180 \\[3ex] \angle POS = 180 - 96 \\[3ex] \angle POS = 84^\circ \\[3ex] \angle POS = 2 * \angle PMS ...\angle\;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] 84 = 2 * \angle PMS \\[3ex] 2 * \angle PMS = 84 \\[3ex] \angle PMS = \dfrac{84}{2} \\[5ex] \angle PMS = 42^\circ \\[3ex] \angle MST = \angle PMS = 42^\circ ... \angle \;\;between\;\;tangent\;RST\;\;and\;\;chord\;SM = \angle \;\;in\;\;the\;\;alternate\;\;segment $

$ (i) \\[3ex] \angle OQM = \angle OMQ = 71^\circ ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle OQM \\[3ex] \angle MQN = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] \underline{\triangle MNQ} \\[3ex] \angle QMN + \angle MQN + \angle MNQ = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle MNQ \\[3ex] \angle QMN = \angle OMQ = 71^\circ ...diagram \\[3ex] \angle MNQ = p \\[3ex] \implies \\[3ex] 71 + 90 + p = 180 \\[3ex] 161 + p = 180 \\[3ex] p = 180 - 161 \\[3ex] p = 19^\circ \\[3ex] $ Construction: Join the radius from centre O to point S on the circumference


$ (ii) \\[3ex] \underline{Same\;\;Radius} \\[3ex] |OR| = |OM| = |OQ| = |ON| = |OS| \\[3ex] \implies \\[3ex] \underline{Base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle s} \\[3ex] \triangle ORM:\;\;\;\angle ORM = \angle OMR = e \\[3ex] \triangle OMQ:\;\;\;\angle OMQ = \angle OQM = 71^\circ \\[3ex] \triangle OQN:\;\;\;\angle OQN = \angle ONQ = p \\[3ex] \triangle ONS:\;\;\;\angle ONS = \angle OSN = 37^\circ \\[3ex] \triangle OSR:\;\;\;\angle OSR = \angle ORS = 75^\circ \\[3ex] \underline{Sum\;\;of\;\;\angle s\;\;in\;\;a\;\;\triangle = 180^\circ} \\[3ex] \triangle ORM:\;\;\;e + e + \angle ROM = 180 \\[3ex] 2e + \angle ROM = 180 \\[3ex] \angle ROM = 180 - 2e \\[3ex] \triangle OMQ:\;\;\; 71 + 71 + \angle MOQ = 180 \\[3ex] 142 + \angle MOQ = 180 \\[3ex] \angle MOQ = 180 - 142 \\[3ex] \angle MOQ = 38^\circ \\[3ex] \triangle OQN:\;\;\; p + p + \angle QON = 180 \\[3ex] 2p + \angle QON = 180 \\[3ex] 2(19) + \angle QON = 180 \\[3ex] 38 + \angle QON = 180 \\[3ex] \angle QON = 180 - 38 \\[3ex] \angle QON = 142^\circ \\[3ex] \triangle NOS:\;\;\; 37 + 37 + \angle NOS = 180 \\[3ex] 74 + \angle NOS = 180 \\[3ex] \angle NOS = 180 - 74 \\[3ex] \angle NOS = 106^\circ \\[3ex] \triangle SOR:\;\;\;75 + 75 + \angle SOR = 180 \\[3ex] 150 + \angle SOR = 180 \\[3ex] \angle SOR = 180 - 150 \\[3ex] \angle SOR = 30^\circ \\[3ex] \underline{\angle s\;\;at\;\;a\;\;point = 360^\circ} \\[3ex] \angle ROM + \angle MOQ + \angle QON + \angle NOS + \angle SOR = 360 \\[3ex] (180 - 2e) + 38 + 142 + 106 + 30 = 360 \\[3ex] 180 - 2e + 316 = 360 \\[3ex] 496 - 2e = 360 \\[3ex] 496 - 360 = 2e \\[3ex] 136 = 2e \\[3ex] 2e = 136 \\[3ex] e = \dfrac{136}{2} \\[5ex] e = 68^\circ $

$ \angle SRQ = 28^\circ ...\angle \;\;between\;\;tangent\;MRQ\;\;and\;\;chord\;RS = \angle \;\;in\;\;the\;\;alternate\;\;segment \\[3ex] \angle VRM + \angle VRS + \angle SRQ = 180^\circ ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] 46 + \angle VRS + 28 = 180 \\[3ex] 74 + \angle VRS = 180 \\[3ex] \angle VRS = 180 - 74 \\[3ex] \angle VRS = 106^\circ \\[3ex] \underline{Cyclic\;\;Quadrilateral\;VUSR} \\[3ex] \angle U + \angle R = 180^\circ ...opposite\;\;\angle s\;\;of\;\;acyclic\;\;quadrilateral\;\;are\;\;supplementary \\[3ex] \angle U = \angle VUS = x \\[3ex] \angle R = \angle VRS = 106^\circ \\[3ex] \implies \\[3ex] x + 106 = 180 \\[3ex] x = 180 - 106 \\[3ex] x = 74^\circ $

Construction: Join the radius from the centre O to point S


$ \underline{Same\;\;Radii} \\[3ex] |OR| = |OS| = |ON| \\[3ex] \implies \\[3ex] \underline{\triangle SOR} \\[3ex] \angle ORS = \angle OSR = 75^\circ ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle SOR \\[3ex] \angle ORS + \angle OSR + \angle SOR = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle SOR \\[3ex] 75 + 75 + \angle SOR = 180 \\[3ex] 150 + \angle SOR = 180 \\[3ex] \angle SOR = 180 - 150 \\[3ex] \angle SOR = 30^\circ \\[3ex] \angle RSN = \angle OSR + \angle OSN = z ...diagram \\[3ex] 75 + \angle OSN = z \\[3ex] \angle OSN = z - 75 \\[3ex] \underline{\triangle SON} \\[3ex] \angle OSN = \angle ONS = z - 75 ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle SON \\[3ex] \angle SOM = \angle OSN + \angle ONS \\[3ex] ...exterior\;\;\angle \;\;of\;\;a\;\;\triangle = sum\;\;of\;\;two\;\;interior\;\;opposite\;\;\angle s\\[3ex] \angle SOM = \angle SOR + \angle ROM \\[3ex] \angle ROM = y \\[3ex] \therefore \angle SOM = 30 + y \\[3ex] \implies \\[3ex] 30 + y = (z - 75) + (z - 75) \\[3ex] 30 + y = z - 75 + z - 75 \\[3ex] 30 + y = 2z - 150 \\[3ex] 2z - 150 = 30 + y \\[3ex] 2z = 30 + y + 150 \\[3ex] 2z = y + 180 \\[3ex] z = \dfrac{y + 180}{2} \\[5ex] OR \\[3ex] \angle OSN + \angle ONS + \angle SON = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle SON \\[3ex] (z - 75) + (z - 75) + \angle SON = 180 \\[3ex] z - 75 + z - 75 + \angle SON = 180 \\[3ex] 2z - 150 + \angle SON = 180 \\[3ex] \angle SON = 180 - 2z + 150 \\[3ex] \angle SON = 330 - 2z \\[3ex] \angle ROM + \angle SOR + \angle SON = 180^\circ ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] y + 30 + (330 - 2z) = 180 \\[3ex] y + 30 + 330 - 2z = 180 \\[3ex] y + 360 - 180 = 2z \\[3ex] y + 180 = 2z \\[3ex] 2z = y + 180 \\[3ex] z = \dfrac{y + 180}{2} $

$ (i) \\[3ex] \underline{\triangle BOE} \\[3ex] \angle BEO = \angle EBO = 20^\circ ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle BOE \\[3ex] \angle BEO + \angle EBO + \angle BOE = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle BOE \\[3ex] 20 + 20 + \angle BOE = 180 \\[3ex] 40 + \angle BOE = 180 \\[3ex] \angle BOE = 180 - 40 \\[3ex] \angle BOE = 140^\circ \\[3ex] (ii) \\[3ex] 42^\circ = \angle BED ...\angle\;\;between\;\;tangent\;BC\;\;and\;\;chord\;BD = \angle\;\;in\;\;the\;\;alternate\;\;segment \\[3ex] \angle BED = 42^\circ \\[3ex] Also: \\[3ex] \angle BED = \angle BEO + \angle OED ...diagram \\[3ex] 42 = 20 + \angle OED \\[3ex] 20 + \angle OED = 42 \\[3ex] \angle OED = 42 - 20 \\[3ex] \angle OED = 22^\circ \\[3ex] (iii) \\[3ex] Reflex\;\;\angle BOE + Obtuse\;\;\angle BOE = 360^\circ...\angle s\;\;at\;\;a\;\;point \\[3ex] Reflex\;\;\angle BOE + 140 = 360 \\[3ex] Reflex\;\;\angle BOE = 360 - 140 \\[3ex] Reflex\;\;\angle BOE = 220^\circ \\[3ex] Reflex\;\;\angle BOE = 2 * \angle BFE \\[3ex] 220 = 2 * \angle BFE \\[3ex] 2 * \angle BFE = 220 \\[3ex] \angle BFE = \dfrac{220}{2} \\[5ex] \angle BFE = 110^\circ $

$ Reflex\;\;\angle AOC + Obtuse\;\;\angle AOC = 360^\circ...\angle s\;\;at\;\;a\;\;point \\[3ex] Reflex\;\;\angle AOC + y = 360 \\[3ex] Reflex\;\;\angle AOC = 360 - y \\[3ex] Also: \\[3ex] Reflex\;\;\angle AOC = 2 * \angle ABC \\[3ex] ...\angle\;\;at\;\;center = 2 * \angle\;\;at\;\;circumference \\[3ex] \implies \\[3ex] 360 - y = 2 * \angle ABC \\[3ex] 2 * \angle ABC = 360 - y \\[3ex] \angle ABC = \dfrac{360 - y}{2} $

$ (i) \\[3ex] \angle OCP = 90^\circ... \\[3ex] radius\;OC \perp tangent\;PCQ \;\;at\;\;point\;\;of\;\;contact\;C \\[3ex] But: \\[3ex] \angle OCP = \angle ACO + \angle ACP ...diagram \\[3ex] 90 = \angle ACO + 40 \\[3ex] \angle ACO + 40 = 90 \\[3ex] \angle ACO = 90 - 40 \\[3ex] \angle ACO = 50^\circ \\[3ex] (ii) \\[3ex] \angle ACP = \angle ABC = 40^\circ ... \angle\;\;between\;\;tangent\;PCQ\;\;and\;\;chord\;AC = \angle\;\;in\;\;the\;\;alternate\;\;segment \\[3ex] \angle ABC = \angle OBA = 40^\circ ...diagram \\[3ex] Because\;\;|OA| = |OB|...same\;\;radius: \\[3ex] \angle OAB = \angle OBA = 40^\circ ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle AOB $

$ (i) \\[3ex] a) \\[3ex] \angle UOX = \angle OZU + \angle OUZ \\[3ex] ...exterior\;\;\angle\;\;a\;\;\triangle = sum\;\;of\;\;the\;\;two\;\;interior\;\;opposite\;\;\angle s \\[3ex] \angle OZU = \angle OUZ = p ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle ZOU \\[3ex] \implies \\[3ex] 70 = p + p \\[3ex] 70 = 2p \\[3ex] 2p = 70 \\[3ex] p = \dfrac{70}{2} \\[5ex] p = 35 \\[3ex] \therefore \angle OZU = \angle OUZ = 35^\circ \\[3ex] b) \\[3ex] \angle OZU = \angle OYX = 35^\circ ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] \angle OUW = 90^\circ \\[3ex] ...radius\;OU \perp tangent\;UVW\;\;at\;\;point\;\;of\;\;contact\;U \\[3ex] \underline{\triangle UYV} \\[3ex] \angle UYV = \angle OYX = 35^\circ ...diagram \\[3ex] \angle YUV = \angle OUW = 90^\circ...diagram \\[3ex] \angle UYV + \angle YUV + \angle UVY = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle UYV \\[3ex] \implies \\[3ex] 35 + 90 + \angle UVY = 180 \\[3ex] 125 + \angle UVY = 180 \\[3ex] \angle UVY = 180 - 125 \\[3ex] \angle UVY = 55^\circ \\[3ex] c) \\[3ex] \underline{\triangle UOW} \\[3ex] \angle UOW = \angle UOX = 70^\circ ...diagram \\[3ex] \angle UOW + \angle OUW + \angle UWO = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle UOW \\[3ex] \implies \\[3ex] 70 + 90 + \angle UWO = 180 \\[3ex] 160 + \angle UWO = 180 \\[3ex] \angle UWO = 180 - 160 \\[3ex] \angle UWO = 20^\circ \\[3ex] (ii) \\[3ex] a) \\[3ex] \underline{\triangle ZOU} \\[3ex] |OZ| = |OU|...same\;\;radius \\[3ex] \angle ZOU + \angle UOX = 180^\circ ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] \angle ZOU + 70 = 180 \\[3ex] \angle ZOU = 180 - 70 \\[3ex] \angle ZOU = 110^\circ \\[3ex] \underline{\triangle YOX} \\[3ex] |OY| = |OX|...same\;\;radius \\[3ex] \angle YOX + \angle UOX = 180^\circ ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] \angle YOX + 70 = 180 \\[3ex] \angle YOX = 180 - 70 \\[3ex] \angle YOX = 110^\circ \\[3ex] Because: \\[3ex] |ZO| = |YO| \;\;and \\[3ex] \angle ZOU = \angle YOX \;\;and \\[3ex] |OU| = |OX| \\[3ex] This\;\;implies\;\;that \\[3ex] \triangle YOX \cong \triangle ZOU ...Side-Angle-Side \\[3ex] \underline{\triangle YXU} \\[3ex] \angle UYX = \angle UYV = 35^\circ...diagram \\[3ex] \angle YXU = 90^\circ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] |YU| = diameter \\[3ex] \underline{\triangle ZUX} \\[3ex] \angle UZX = \angle UYX = 35^\circ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] \angle ZUX = 90^\circ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] |ZX| = diameter \\[3ex] Because: \\[3ex] \angle UYX = \angle UZX \;\;and \\[3ex] \angle ZUX = \angle YXU \;\;and \\[3ex] |YU| = |ZX| \\[3ex] This\;\;implies\;\;that \\[3ex] \triangle ZUX \cong \triangle YXU ...Angle-Angle-Side $

$ (i) \\[3ex] 46^\circ = \angle PTR ...\angle\;\;between\;\;tangent\;PQ\;\;and\;\;chord\;PR = \angle\;\;in\;\;the\;\;alternate\;\;segment \\[3ex] \angle PTR = 46^\circ \\[3ex] (ii) \\[3ex] \angle TRP = \angle RPQ + \angle RQP ...exterior\;\;\angle\;\;a\;\;\triangle = sum\;\;of\;\;the\;\;two\;\;interior\;\;opposite\;\;\angle s \\[3ex] \angle TRP = 46 + 32 \\[3ex] \angle TRP = 78^\circ \\[3ex] \underline{\triangle TPR} \\[3ex] \angle PTR + \angle TPR + \angle TRP = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle TPR \\[3ex] 46 + \angle TPR + 78 = 180 \\[3ex] 124 + \angle TPR = 180 \\[3ex] \angle TPR = 180 - 124 \\[3ex] \angle TPR = 56^\circ \\[3ex] (iii) \\[3ex] \underline{Cyclic\;\;Quadrilateral\;TPRS} \\[3ex] \angle P + \angle S = 180^\circ ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quadrilateral\;\;are\;\;supplementary \\[3ex] \angle P = \angle TPR = 56^\circ...diagram \\[3ex] \angle S = \angle TSR ...diagram \\[3ex] \implies \\[3ex] 56 + \angle TSR = 180 \\[3ex] \angle TSR = 180 - 56 \\[3ex] \angle TSR = 124^\circ $

$ Reflex\;\;\angle XOZ + Obtuse\;\;\angle O = 360^\circ ...\angle s\;\;at\;\;a\;\;point \\[3ex] Reflex\;\;\angle XOZ + 10m = 360 \\[3ex] Reflex\;\;\angle XOZ = 360 - 10m ...eqn.(1) \\[3ex] Also: \\[3ex] Reflex\;\;\angle XOZ = 2 * \angle XWZ \\[3ex] ...\angle\;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] Reflex\;\;\angle XOZ = 2 * m ...eqn.(2) \\[3ex] eqn.(1) = eqn.(2) \implies \\[3ex] 360 - 10m = 2 * m \\[3ex] 2 * m = 360 - 10m \\[3ex] 2m + 10m = 360 \\[3ex] 12m = 360 \\[3ex] m = \dfrac{360}{12} \\[5ex] m = 30^\circ $

$ (111.1) \\[3ex] \underline{Cyclic\;\;Quadrilateral\;PQRS} \\[3ex] \hat{Q} + \hat{S_2} = 180^\circ ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quadrilateral\;\;are\;\;supplementary \\[3ex] \hat{Q} + 108 = 180 \\[3ex] \hat{Q} = 180 - 108 \\[3ex] \hat{Q} = 72^\circ \\[3ex] (111.2) \\[3ex] \underline{\triangle PQR} \\[3ex] \hat{P_1} = \hat{R_2} = k ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle PQR \\[3ex] \hat{P_1} + \hat{Q} + \hat{R_2} = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle PQR \\[3ex] k + 72 + k = 180 \\[3ex] 2k + 72 = 180 \\[3ex] 2k = 180 - 72 \\[3ex] 2k = 108 \\[3ex] k = \dfrac{108}{2} \\[5ex] k = 54 \\[3ex] \therefore \hat{P_1} = \hat{R_2} = 54^\circ \\[3ex] (111.3) \\[3ex] 42^\circ = \hat{P_2} ...\angle\;\;between\;\;tangent\;ST\;\;and\;\;chord\;SR = \angle\;\;in\;\;the\;\;alternate\;\;segment \\[3ex] \hat{P_2} = 42^\circ \\[3ex] (111.4) \\[3ex] \hat{R_3} = \hat{P}...exterior\;\;\angle\;\;of\;\;a\;\;cyclic\;\;quadrilateral = interior\;\;opposite\;\;\angle \\[3ex] \hat{P} = \hat{P_1} + \hat{P_2} ...diagram \\[3ex] \implies \\[3ex] \hat{R_3} = \hat{P_1} + \hat{P_2} \\[3ex] \hat{R_3} = 54 + 42 \\[3ex] \hat{R_3} = 96^\circ $

Let us draw the diagram for this question

AB = AC ...the two tangents: AB and AC, drawn from the same external point:A are equal in length

$ \implies \triangle ABC\;\;is\;\;an\;\;isosceles\;\;triangle \\[3ex] \underline{\triangle ABC} \\[3ex] \angle ABC + \angle ACB + \angle BAC = 180^\circ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle ABC \\[3ex] Let\;\; \angle ABC = \angle ACB = p \\[3ex] \angle BAC = \angle A = 70^\circ \\[3ex] \implies \\[3ex] p + p + 70 = 180 \\[3ex] 2p = 180 - 70 \\[3ex] 2p = 110 \\[3ex] p = \dfrac{110}{2} \\[5ex] p = 55 \\[3ex] \therefore \angle ABC = 55^\circ $

$ (i) \\[3ex] \angle EBF + \angle OBF = 90^\circ ...radius\;OB \perp tangent\;EBC\;\;at\;\;point\;\;of\;\;contact\;B \\[3ex] \angle EBF + 35 = 90 \\[3ex] \angle EBF = 90 - 35 \\[3ex] \angle EBF = 55^\circ \\[3ex] (ii) \\[3ex] \underline{\triangle BOA} \\[3ex] \angle OBA = \angle OAB = 40^\circ ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle BOA \\[3ex] \angle OBA + \angle OAB + \angle BOA = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle BOA \\[3ex] 40 + 40 + \angle BOA = 180 \\[3ex] 80 + \angle BOA = 180 \\[3ex] \angle BOA = 180 - 80 \\[3ex] \angle BOA = 100^\circ \\[3ex] (iii) \\[3ex] \angle BOA = 2 * \angle AFB ...\angle\;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] 100 = 2 * \angle AFB \\[3ex] 2 * \angle AFB = 100 \\[3ex] \angle AFB = \dfrac{100}{2} \\[5ex] \angle AFB = 50^\circ \\[3ex] (iv) \\[3ex] \underline{\triangle BFA} \\[3ex] \angle ABF + \angle AFB + \angle BAF = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle BFA \\[3ex] \angle ABF = 35 + 40 ...diagram \\[3ex] \angle ABF = 75^\circ \\[3ex] \angle BAF = \angle OAB + \angle OAF ...diagram \\[3ex] \angle BAF = 40 + \angle OAF \\[3ex] \implies \\[3ex] 75 + 50 + (40 + \angle OAF) = 180 \\[3ex] 125 + 40 + \angle OAF = 180 \\[3ex] 165 + \angle OAF = 180 \\[3ex] \angle OAF = 180 - 165 \\[3ex] \angle OAF = 15^\circ $

$ \angle ZXW = \angle ZYW = 33^\circ ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] \angle WZX = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] \underline{\triangle WZX} \\[3ex] \angle ZWX + \angle WZX + \angle ZXW = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle WZX \\[3ex] \angle ZWX + 90 + 33 = 180 \\[3ex] \angle ZWX + 123 = 180 \\[3ex] \angle ZWX = 180 - 123 \\[3ex] \angle ZWX = 57^\circ $

$ (i) \\[3ex] 21^\circ = \angle SQP...\angle\;\;between\;\;tangent\;ST\;\;and\;\;chord\;SP = \angle\;\;in\;\;the\;\;alternate\;\;segment \\[3ex] \angle SQP = 21^\circ \\[3ex] \angle SRP = \angle SQP = 21^\circ ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] \angle S + \angle Q = 180^\circ...opposite\;\;\angle s\;\;of\;\;Cyclic\;\;Quad\;\;RQPS\;\;are\;\;supplementary \\[3ex] \angle S = \angle RSP ...diagram \\[3ex] \angle Q = \angle RQP = 100^\circ...diagram \\[3ex] \implies \\[3ex] \angle RSP + 100 = 180 \\[3ex] \angle RSP = 180 - 100 \\[3ex] \angle RSP = 80^\circ \\[3ex] \underline{\triangle SRP} \\[3ex] \angle RSP + \angle SRP + \angle SPR = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle SRP \\[3ex] 80 + 21 + \angle SPR = 180 \\[3ex] 101 + \angle SPR = 180 \\[3ex] \angle SPR = 180 - 101 \\[3ex] \angle SPR = 79^\circ \\[3ex] (ii) \\[3ex] \angle SQR = \angle SPR = 79^\circ ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] \angle SRQ = 90^\circ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] \underline{\triangle SRQ} \\[3ex] \angle QSR + \angle SQR + \angle SRQ = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle SRQ \\[3ex] \angle QSR + 79 + 90 = 180 \\[3ex] \angle QSR + 169 = 180 \\[3ex] \angle QSR = 180 - 169 \\[3ex] \angle QSR = 11^\circ $

$ (a) \\[3ex] 12(12 + x) = x(x + 4) ...Intersecting\;\;Secants\;\;Theorem \\[3ex] 144 + 12x = x^2 + 4x \\[3ex] x^2 + 4x = 144 + 12x \\[3ex] x^2 + 4x - 12x - 144 = 0 \\[3ex] x^2 - 8x - 144 = 0 \\[3ex] (b) \\[3ex] x^2 - 8x - 144 = 0 \\[3ex] Compare\;\;to\;\;general\;\;form:\;\;ax^2 + bx + c = 0 \\[3ex] a = 1 \\[3ex] b = -8 \\[3ex] c = -144 \\[3ex] x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} ...Quadratic\;\;Formula \\[5ex] x = \dfrac{-(-8) \pm \sqrt{(-8)^2 - 4(1)(-144)}}{2(1)} \\[5ex] x = \dfrac{8 \pm \sqrt{64 + 576}}{2} \\[5ex] x = \dfrac{8 \pm \sqrt{640}}{2} \\[5ex] x = \dfrac{8 \pm \sqrt{64} * \sqrt{10}}{2} \\[5ex] x = \dfrac{8 \pm 8 * \sqrt{10}}{2} \\[5ex] x = \dfrac{8(1 \pm \sqrt{10}}{2} \\[5ex] x = 4(1 \pm \sqrt{10}) \\[3ex] x = 4(1 \pm 3.16227766) \\[3ex] x = 4(1 + 3.16227766) \;\;\;OR\;\;\; x = 4(1 - 3.16227766) \\[3ex] x = 4(4.16227766) \;\;\;OR\;\;\; x = 4(-2.16227766) \\[3ex] $ Because the length cannot be negative, discard the negative value

$ x = 16.64911064 \\[3ex] |AC| = |AB| + |BC| ... diagram \\[3ex] = x + 4 \\[3ex] = 16.64911064 + 4 \\[3ex] = 20.64911064 \\[3ex] \approx 20.6\;cm ...to\;\;3\;\;significant\;\;figures $

$ \angle RSP = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] \angle PST = 24^\circ ...\angle\;\;between\;\;tangent\;ST\;\;and\;\;chord\;SP = \angle\;\;in\;\;the\;\;alternate\;\;segment \\[3ex] \angle RST = \angle RSP + \angle PST ...diagram \\[3ex] \angle RST = 90 + 24 \\[3ex] \angle RST = 114^\circ \\[3ex] \underline{\triangle RST} \\[3ex] \angle TRS + \angle RST + \angle STR = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle RST \\[3ex] \angle TRS = \angle PRS = 24^\circ ...diagram \\[3ex] \implies \\[3ex] 24 + 114 + \angle STR = 180 \\[3ex] 138 + \angle STR = 180 \\[3ex] \angle STR = 180 - 138 \\[3ex] \angle STR = 42^\circ $

$ (i) \\[3ex] \underline{Cyclic\;\;Quadrilateral\;SRQP} \\[3ex] \angle R + \angle P = 180^\circ \\[3ex] ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quad\;\;are\;\;supplementary \\[3ex] \angle R = \angle QRS = 58^\circ \\[3ex] \angle P = \angle SPQ ...diagram \\[3ex] \implies \\[3ex] 58 + \angle SPQ = 180 \\[3ex] \angle SPQ = 180 - 58 \\[3ex] \angle SPQ = 122^\circ \\[3ex] OR \\[3ex] Obtuse\;\;\angle SOQ = 2 * \angle SRQ \\[3ex] ...\angle\;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] Obtuse\;\;\angle SOQ = 2 * 58 \\[3ex] Obtuse\;\;\angle SOQ = 116^\circ \\[3ex] Reflex\;\;\angle SOQ + Obtuse\;\;\angle SOQ = 360^\circ ...\angle s\;\;at\;\;a\;\;point \\[3ex] Reflex\;\;\angle SOQ + 116 = 360 \\[3ex] Reflex\;\;\angle SOQ = 360 - 116 \\[3ex] Reflex\;\;\angle SOQ = 244^\circ \\[3ex] Also: \\[3ex] Reflex\;\;\angle SOQ = 2 * \angle SPQ \\[3ex] ...\angle\;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] 244 = 2 * \angle SPQ \\[3ex] 2 * \angle SPQ = 244 \\[3ex] \angle SPQ = \dfrac{244}{2} \\[5ex] \angle SPQ = 122^\circ \\[3ex] (ii) \\[3ex] \underline{\triangle SOQ} \\[3ex] \angle OSQ = \angle OQS = k ... base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle SOQ \\[3ex] \angle OSQ + \angle OQS + Obtuse\;\;\angle SOQ = 180^\circ \\[3ex] ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle SOQ \\[3ex] k + k + 116 = 180 \\[3ex] 2k = 180 - 116 \\[3ex] 2k = 64 \\[3ex] k = \dfrac{64}{2} \\[5ex] k = 32 \\[3ex] \therefore \angle OSQ = \angle OQS = 32^\circ $

$ (i) \\[3ex] Line\;\;OB\;\;is\;\;the\;\;radius\;\;of\;\;the\;\;circle \\[3ex] (ii) \\[3ex] \angle ABO = 90^\circ ...radius\;OB \perp tangent\;ABC\;\;at\;\;point\;\;of\;\;contact\;B \\[3ex] (iii) \\[3ex] \angle AOB + \angle OAB + \angle ABO = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle AOB \\[3ex] \angle AOB + 49 + 90 = 180 \\[3ex] \angle AOB + 139 = 180 \\[3ex] \angle AOB = 180 - 139 \\[3ex] \angle AOB = 41^\circ \\[3ex] (iv) \\[3ex] \angle ADC = \angle AOB = 41^\circ ... corresponding\;\;\angle s\;\;between\;\;parallel\;\;lines\;OB\;\;and\;\;DC,\;\;are\;\;equal \\[3ex] (v) \\[3ex] $ Triangle AOB is similar to triangle ADC

$ (vi) \\[3ex] (a) \\[3ex] SOHCAHTOA \\[3ex] opp = |OB| \\[3ex] adj = |AB| \\[3ex] hyp = |OA| \\[3ex] \tan 49 = \dfrac{|OB|}{5.4} \\[5ex] 5.4 * \tan 49 = |OB| \\[3ex] |OB| = 5.4 * \tan 49 \\[3ex] |OB| = 5.4(1.150368407) \\[3ex] |OB| = 6.211989399\;cm \\[3ex] (b) \\[3ex] \cos 49 = \dfrac{5.4}{|OA|} \\[5ex] |OA| * \cos 49 = 5.4 \\[3ex] |OA| = \dfrac{5.4}{\cos 49} \\[5ex] |OA| = \dfrac{5.4}{0.656059029} \\[5ex] |OA| = 8.230966668\;cm \\[3ex] OR \\[3ex] hyp^2 = opp^2 + adj^2 ...Pythagorean\;\;Theorem \\[3ex] |OA|^2 = |OB|^2 + |AB|^2 \\[3ex] |OA|^2 = 6.211989399^2 + 5.4^2 \\[3ex] |OA|^2 = 38.58881229 + 29.16 \\[3ex] |OA|^2 = 67.74881229 \\[3ex] |OA| = \sqrt{67.74881229} \\[3ex] |OA| = 8.230966668\;cm \\[3ex] (c) \\[3ex] Area\;\;of\;\;\triangle AOB \\[3ex] = \dfrac{1}{2} * |OA| * |AB| * \sin \angle OAB \\[5ex] = \dfrac{1}{2} * 8.230966668 * 5.4 * \sin 49 \\[5ex] = 0.5 * 8.230966668 * 5.4 * 0.7547095802 \\[3ex] = 16.77237138\;cm^2 \\[3ex] OR \\[3ex] Area\;\;of\;\;\triangle AOB \\[3ex] = \dfrac{1}{2} * |AB| * |OB| \\[5ex] = 0.5 * 5.4 * 6.211989399 \\[3ex] = 16.77237138\;cm^2 $

$ (i) \\[3ex] Yes,\;\;she\;\;is\;\;correct \\[3ex] \angle PAT = 90^\circ...\angle\;\;in\;\;a\;\;semicircle\;\;is\;\;a\;\;right\;\;\angle \\[3ex] $ Construction: Draw the radii: from point X on the centre to point E on the circumference; and from point X on the centre to point N on the circumference of the circle

$ (ii) \\[3ex] \angle XEP = \angle XPE = 52^\circ ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle PXE \\[3ex] \angle XEP + \angle XPE + \angle PXE = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle PXE \\[3ex] 52 + 52 + \angle PXE = 180 \\[3ex] 104 + \angle PXE = 180 \\[3ex] \angle PXE = 180 - 104 \\[3ex] \angle PXE = 76^\circ \\[3ex] \angle XEN = \angle PXE = 76^\circ ...alternate\;\;\angle s\;\;are\;\;equal \\[3ex] \angle XNE = \angle XEN = 76^\circ ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle XEN \\[3ex] \angle XNE + \angle XEN + \angle NXE = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle XEN \\[3ex] 76 + 76 + \angle NXE = 180 \\[3ex] 152 + \angle NXE = 180 \\[3ex] \angle NXE = 180 - 152 \\[3ex] \angle NXE = 28^\circ \\[3ex] (iii) \\[3ex] \angle XPE = w \\[3ex] \angle NXE = 4w - 180 \\[3ex] \angle NXE \gt 0 \\[3ex] \implies \\[3ex] 4w - 180 \gt 0 \\[3ex] 4w \gt 0 + 180 \\[3ex] 4w \gt 180 \\[3ex] w \gt \dfrac{180}{4} \\[5ex] w \gt 45^\circ $

$ \angle OTN = \angle ONT = k...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle TON \\[3ex] \angle OTN + \angle ONT + \angle TON = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle TON \\[3ex] k + k + 108 = 180 \\[3ex] 2k = 180 - 108 \\[3ex] 2k = 72 \\[3ex] k = \dfrac{72}{2} \\[5ex] k = 36 \\[3ex] \therefore \angle OTN = \angle ONT = 36^\circ \\[3ex] \angle OTP = 90^\circ...radius\;OT \perp tangent\;PTR\;\;at\;\;point\;\;of\;\;contact\;T \\[3ex] \angle PTN = \angle OTP + \angle OTN ...diagram \\[3ex] \angle PTN = 90 + 36 \\[3ex] \angle PTN = 126^\circ $

$ (a) \\[3ex] u = 35^\circ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] \angle DBC = \angle ADB = u \\[3ex] ...alternate\;\;\angle s\;\;between\;\;parallel\;\;lines\;AD\;\;and\;\;BC\;\;are\;\;equal \\[3ex] \angle DAX + \angle AXD + \angle ADX = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle AXD \\[3ex] \angle ADX = \angle ADB = u = 35^\circ \\[3ex] \implies \\[3ex] 35 + v + 35 = 180 \\[3ex] v + 70 = 180 \\[3ex] v = 180 - 70 \\[3ex] v = 110^\circ \\[3ex] (b) \\[3ex] Reflex\;\;\angle FOG + Obtuse\;\;\angle FOG = 360^\circ ...\angle s\;\;at\;\;a\;\;point \\[3ex] 210 + Obtuse\;\;\angle FOG = 360 \\[3ex] Obtuse\;\;\angle FOG = 360 - 210 \\[3ex] Obtuse\;\;\angle FOG = 150^\circ \\[3ex] Also: \\[3ex] Obtuse\;\;\angle FOG = 2 * \angle FHG \\[3ex] ...\angle\;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] \angle FHP = p ...diagram \\[3ex] 150 = 2 * p \\[3ex] \implies \\[3ex] 2p = 150 \\[3ex] p = \dfrac{150}{2} \\[5ex] p = 75^\circ $

$ a) \\[3ex] \angle TPQ = 90\circ ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] b) \\[3ex] \angle MTQ = 90^\circ ...radius\;OT \perp tangent\;LTM\;\;at\;\;point\;\;of\;\;contact\;T \\[3ex] $
$ c) \\[3ex] 23^\circ = \angle TQS ...\angle\;\;between\;\;tangent\;LTM\;\;and\;\;chord\;TS = \angle\;\;in\;\;the\;\;alternate\;\;segment \\[3ex] \angle TQS = 23^\circ \\[3ex] d) \\[3ex] \angle MTS + \angle STQ = \angle MTQ ...diagram \\[3ex] 23 + \angle STQ = 90 \\[3ex] \angle STQ = 90 - 23 \\[3ex] \angle STQ = 67^\circ \\[3ex] \underline{Cyclic\;\;Quadrilateral\;TQRS} \\[3ex] \angle T + \angle R = 180^\circ ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quadrilateral\;\;are\;\;supplementary \\[3ex] \angle T = \angle STQ = 67^\circ ...diagram \\[3ex] \angle R = \angle SRQ ...diagram \\[3ex] \implies \\[3ex] 67 + \angle SRQ = 180 \\[3ex] \angle SRQ = 180 - 67 \\[3ex] \angle SRQ = 113^\circ $

$ (i) \\[3ex] \angle OQR = \angle ORQ = 32^\circ ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle QOR \\[3ex] \angle OQR + \angle ORQ + \angle QOR = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle QOR \\[3ex] 32 + 32 + \angle QOR = 180 \\[3ex] 64 + \angle QOR = 180 \\[3ex] \angle QOR = 180 - 64 \\[3ex] \angle QOR = 116^\circ \\[3ex] \angle QOR = 2 * \angle QPR ...\angle\;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] 116 = 2 * \angle QPR \\[3ex] 2 * \angle QPR = 116 \\[3ex] \angle QPR = \dfrac{116}{2} \\[5ex] \angle QPR = 58^\circ \\[3ex] (ii) \\[3ex] $
$ \underline{Cyclic\;\;Quadrilateral\;TPRQ} \\[3ex] \angle Q + \angle P = 180^\circ \\[3ex] ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quad\;\;are\;\;supplementary \\[3ex] \angle Q = \angle TQO + \angle OQR ...diagram \\[3ex] \angle Q = \angle TQO + 32 \\[3ex] \angle P = \angle TPQ + \angle QPR ...diagram \\[3ex] \angle P = 15 + 58 \\[3ex] \angle P = 73^\circ \\[3ex] \implies \\[3ex] (\angle TOQ + 32) + 73 = 180 \\[3ex] \angle TOQ + 32 + 73 = 180 \\[3ex] \angle TOQ + 105 = 180 \\[3ex] \angle TOQ = 180 - 105 \\[3ex] \angle TOQ = 75^\circ $

(1.) Angle RCT = 47° because: two tangents: STR and BCR drawn from the same external point: R are equal in length.
Because |TR| = |CR|,
∠RCT = ∠RTC = 47°...base angles of isosceles triangle TRC

(2.) Angle RTC = Angle TAC = 47° because the angle between tangent STR and chord TC = angle in the alternate segment TAC

(3.) Angle ATC = 86° because the interior opposite angles of the cyclic quadrilateral ATCD are supplementary

$ \angle T + \angle D = 180^\circ ...opposite\;\;\angle s\;\;of\;\;cyclic\;\;quadrilateral\;ATCD\;\;are\;\;supplementary \\[3ex] \angle T = \angle ATC ...diagram \\[3ex] \angle D = \angle ADC = 94^\circ ... diagram \\[3ex] \implies \\[3ex] \angle ATC + 94 = 180 \\[3ex] \angle ATC = 180 - 94 \\[3ex] \angle ATC = 86^\circ \\[3ex] $ (4.) Angle STA = 47° because

$ \angle STA + \angle ATC + \angle RTC = 180^\circ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] \angle STA + 86 + 47 = 180 \\[3ex] \angle STA + 133 = 180 \\[3ex] \angle STA = 180 - 133 \\[3ex] \angle STA = 47^\circ \\[3ex] $ (5.) So the lines AC and SR are parallel because angle STA = angle TAC = 47°...alternate angles are equal
Alternate angles between the parallel lines: AC and SR, are equal.

$ \underline{\triangle PQS} \\[3ex] \angle PSQ + \angle QPS + \angle SQP = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle PQS \\[3ex] 60 + 100 + \angle SQP = 180 \\[3ex] 160 + \angle SQP = 180 \\[3ex] \angle SQP = 180 - 160 \\[3ex] \angle SQP = 20^\circ \\[3ex] \underline{Cyclic\;\;Quadrilateral\;PQRS} \\[3ex] \angle Q + \angle S = 180^\circ \\[3ex] ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quad\;\;are\;\;supplementary \\[3ex] \angle Q = \angle SQP + \angle SQR ...diagram \\[3ex] \angle Q = 20 + \angle SQR \\[3ex] \angle S = 60 + 80 \\[3ex] \angle S = 140^\circ \\[3ex] \implies \\[3ex] 20 + \angle SQR + 140 = 180 \\[3ex] 160 + \angle SQR = 180 \\[3ex] \angle SQR = 180 - 160 \\[3ex] \angle SQR = 20^\circ $

$ (i) \\[3ex] Obtuse\;\;\angle XOZ = 2 * \angle XWZ ...\angle\;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] Obtuse\;\;\angle XOZ = 2 * 64 \\[3ex] Obtuse\;\;\angle XOZ = 128^\circ \\[3ex] Reflex\;\;\angle XOZ + Obtuse\;\;\angle XOZ = 360^\circ...\angle s\;\;at\;\;a\;\;point \\[3ex] Reflex\;\;\angle XOZ + 128 = 360 \\[3ex] Reflex\;\;\angle XOZ = 360 - 128 \\[3ex] Reflex\;\;\angle XOZ = 232^\circ \\[3ex] Also: \\[3ex] Reflex\;\;\angle XOZ = 2 * \angle XYZ ...\angle\;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] 232 = 2 * \angle XYZ \\[3ex] 2 * \angle XYZ = 232 \\[3ex] \angle XYZ = \dfrac{232}{2} \\[5ex] \angle XYZ = 116^\circ \\[3ex] (ii) \\[3ex] 23^\circ = \angle YXZ ...\angle\;\;between\;\;tangent\;TYV\;\;and\;\;chord\;YZ = \angle\;\;in\;\;the\;\;alternate\;\;segment \\[3ex] \angle YXZ = 23^\circ \\[3ex] (iii) \\[3ex] \underline{\triangle XOZ} \\[3ex] \angle OXZ = \angle OZX = p ... base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle XOZ \\[3ex] \angle OXZ + \angle OZX + Obtuse\;\;\angle XOZ = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle XOZ \\[3ex] p + p + 128 = 180 \\[3ex] 2p = 180 - 128 \\[3ex] 2p = 52 \\[3ex] p = \dfrac{52}{2} \\[5ex] p = 26 \\[3ex] \therefore \angle OXZ = \angle OZX = 26^\circ $

$ \angle XZY = \angle XWY = 48^\circ ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] \angle ZXY = \angle XZY = 48^\circ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle XZY \\[3ex] \angle ZXY + \angle XZY + \angle XYZ = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle XZY \\[3ex] 48 + 48 + \angle XYZ = 180 \\[3ex] 96 + \angle XYZ = 180 \\[3ex] \angle XYZ = 180 - 96 \\[3ex] \angle XYZ = 84^\circ \\[3ex] But: \\[3ex] \angle XYZ = \angle WYX + \angle WYZ ...diagram \\[3ex] 84 = \angle WYX + 65 \\[3ex] \angle WYX + 65 = 84 \\[3ex] \angle WYX = 84 - 65 \\[3ex] \angle WYX = 19^\circ $

$ \angle VUX = 50^\circ ...\angle\;\;between\;\;tangent\;YXW\;\;chord\;XV = \angle\;\;in\;\;the\;\;alternate\;\;segment \\[3ex] \angle VXU = \angle VUX = 50^\circ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle XUV \\[3ex] \angle UXY + \angle VXU + \angle XWY = 180^\circ ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] \angle UXY + 50 + 50 = 180 \\[3ex] \angle UXY + 100 = 180 \\[3ex] \angle UXY = 180 - 100 \\[3ex] \angle UXY = 80^\circ $

$ \angle OST = 90^\circ \\[3ex] ...radius\;OS \perp tangent\;ST\;\;at\;\;point\;\;of\;\;contact\;S \\[3ex] \angle OQT = 90^\circ \\[3ex] ...radius\;OQ \perp tangent\;QT\;\;at\;\;point\;\;of\;\;contact\;Q \\[3ex] Obtuse\;\;\angle O + \angle OST + \angle T + \angle OQT = 360^\circ \\[3ex] ...sum\;\;of\;\;the\;\;interior\;\;\angle s\;\;of\;\;quadrilateral\;OSTQ \\[3ex] \implies \\[3ex] Obtuse\;\;\angle O + 90 + x + 90 = 360 \\[3ex] Obtuse\;\;\angle O = 360 - 90 - x - 90 \\[3ex] Obtuse\;\;\angle O = 180 - x \\[3ex] Reflex\;\;\angle O + Obtuse\;\;\angle O = 360^\circ ...\angle s\:\:at\:\:a\:\:point \\[3ex] Reflex\;\;\angle O + (180 - x) = 360 \\[3ex] Reflex\;\;\angle O + 180 - x = 360 \\[3ex] Reflex\;\;\angle O = 360 - 180 + x \\[3ex] Reflex\;\;\angle O = 180 + x \\[3ex] Also: \\[3ex] Reflex\;\;\angle O = 2 * \angle SPQ \\[3ex] ...\angle\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] 180 + x = 2 * \angle SPQ \\[3ex] 2 * \angle SPQ = 180 + x \\[3ex] \angle SPQ = \dfrac{180 + x}{2} $

Construction: Join the chord from point B to point C
$ (a) \\[3ex] \angle BOC = 2 * \angle CAB ...\angle\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] \angle BOC = 2 * x \\[3ex] \angle BOC = 2x \\[3ex] (b) \\[3ex] \angle BCD = \angle CAB = x ...\angle\;\;between\;\;tangent\;CD\;\;and\;\;chord\;CB = \angle\;\;in\;\;the\;\;alternate\;\;segment \\[3ex] |BD| = |CD|...two\;\;tangents\;\;drawn\;\;from\;\;the\;\;same\;\;external\;\;point\;D\;\;are\;\;equal\;\;in\;\;length \\[3ex] \angle CBD = \angle BCD = x \;\;base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle BDC \\[3ex] \angle CBD + \angle BCD + \angle BDC = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle BDC \\[3ex] x + x + \angle BDC = 180 \\[3ex] 2x + \angle BDC = 180 \\[3ex] \angle BDC = 180 - 2x \\[3ex] $
$ (c) \\[3ex] \angle OCB = \angle OBC \;\;base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle COB \\[3ex] \angle OCB + \angle OBC + \angle BOC = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle COB \\[3ex] \angle OBC + \angle OBC + 2x = 180 \\[3ex] 2\angle OBC = 180 - 2x \\[3ex] \angle OBC = \dfrac{180 - 2x}{2} \\[5ex] \angle OBC = \dfrac{2(90 - x)}{2} \\[5ex] \angle OBC = 90 - x \\[3ex] \angle ABC = \angle BCD = x ...alternate\;\;\angle s\;\;are\;\;equal \\[3ex] Also: \\[3ex] \angle ABC = \angle ABO + \angle OBC ...diagram \\[3ex] \implies \\[3ex] x = \angle ABO + (90 - x) \\[3ex] x = \angle ABO + 90 - x \\[3ex] \angle ABO + 90 - x = x \\[3ex] \angle ABO = x - 90 + x \\[3ex] \angle ABO = 2x - 90 \\[3ex] \angle ABO = 2(x - 45) \\[3ex] (d) \\[3ex] \angle OCE = 90^\circ ...radius\;OC \perp tangent\;EH\;\;at\;\;point\;\;of\;\;contact\;C \\[3ex] \angle ACE = \angle CAB = x ...alternate\;\;\angle s\;\;are\;\;equal \\[3ex] \angle OCE = \angle ACE + \angle ACO ...diagram \\[3ex] 90 = \angle = x + \angle ACO \\[3ex] x + \angle ACO = 90 \\[3ex] \angle ACO = 90 - x $

$ \angle ROT = 2 * \angle RST...\angle\;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] \angle ROT = 2 * 46 \\[3ex] \angle ROT = 92^\circ \\[3ex] \angle ORT = \angle OTR = p ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle ROT \\[3ex] \angle ORT + \angle OTR + \angle ROT = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle ROT \\[3ex] p + p + 92 = 180 \\[3ex] 2p = 180 - 92 \\[3ex] 2p = 88 \\[3ex] p = \dfrac{88}{2} \\[5ex] p = 44 \\[3ex] \therefore \angle ORT = \angle OTR = 44^\circ \\[3ex] \angle OTR = \angle STO + \angle STR ...diagram \\[3ex] 44 = \angle STO + 29 \\[3ex] \angle STO + 29 = 44 \\[3ex] \angle STO = 44 - 29 \\[3ex] \angle STO = 15^\circ $

$ (i) \\[3ex] |EH| = |EF| ...two\;\;tangents\;\;drawn\;\;from\;\;the\;\;same\;\;external\;\;point\;E\;\;are\;\;equal\;\;in\;\;length \\[3ex] \angle EHF = \angle EFH = p \;\;base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle FEH \\[3ex] \angle EHF + \angle EFH + \angle FEH = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle FEH \\[3ex] p + p + 44 = 180 \\[3ex] 2p + 44 = 180 \\[3ex] 2p = 180 - 44 \\[3ex] 2p = 136 \\[3ex] p = \dfrac{136}{2} \\[5ex] p = 68 \\[3ex] \therefore \angle EHF = \angle EFH = 68^\circ \\[3ex] (ii) \\[3ex] \angle EHF = \angle FGH = 68^\circ ...\angle\;\;between\;\;tangent\;EH\;\;and\;\;chord\;HF = \angle\;\;in\;\;the\;\;alternate\;\;segment \\[3ex] (iii) \\[3ex] \angle JHE = 90^\circ ...radius\;OH \perp tangent\;EH\;\;at\;\;point\;\;of\;\;contact\;H \\[3ex] (iv) \\[3ex] \angle JGH = 90^\circ...\angle\;\;in\;\;a\;\;semicircle $

$ (a) \\[3ex] O\hat{A}D = 90^\circ...radius\;OA \perp tangent\;AD\;\;at\;\;point\;\;of\;\;contact\;A \\[3ex] \underline{Quadrilateral\;DCOA} \\[3ex] \hat{D} + \hat{C} + \hat{O} + \hat{A} = 360^\circ \\[3ex] ...sum\;\;of\;\;the\;\;interior\;\;\angle s\;\;of\;\;a\;\;quadrilateral \\[3ex] \hat{D} = x \\[3ex] \hat{C} = O\hat{C}D \\[3ex] \hat{O} = 2x \\[3ex] \hat{A} = O\hat{A}D = 90^\circ \\[3ex] \implies \\[3ex] x + O\hat{C}D + 2x + 90 = 360 \\[3ex] O\hat{C}D = 360 - 90 - 2x - x \\[3ex] O\hat{C}D = 270 - 3x \\[3ex] (b) \\[3ex] C\hat{O}A = 2 * C\hat{B}A ...\angle\;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] 2x = 2 * C\hat{B}A \\[3ex] 2 * C\hat{B}A = 2x \\[3ex] C\hat{B}A = \dfrac{2x}{2} \\[5ex] C\hat{B}A = x \\[3ex] \underline{\triangle DBA} \\[3ex] B\hat{D}A + D\hat{B}A + D\hat{A}B = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle DBA \\[3ex] B\hat{D}A = x \\[3ex] D\hat{B}A = C\hat{B}A = x \\[3ex] D\hat{A}B = O\hat{A}D + O\hat{A}B ... diagram \\[3ex] D\hat{A}B = 90 + O\hat{A}B \\[3ex] \implies \\[3ex] x + x + (90 + O\hat{A}B) = 180 \\[3ex] 2x + 90 + O\hat{A}B = 180 \\[3ex] O\hat{A}B = 180 - 90 - 2x \\[3ex] O\hat{A}B = 90 - 2x $

Let us solve this question in at least two ways
First Method:
Construction: Draw the chord from point C to point A

$ \angle TAE = \angle ECA = x^\circ ...\angle\;\;between\;\;tangent\;SAT\;\;and\;\;chord\;AE = \angle\;\;in\;\;the\;\;alternate\;\;segment \\[3ex] \angle ECA = \angle CAB = x^\circ ...alternate\;\;\angle s\;\;between\;\;parallel\;\;lines\;\;CE\;\;and\;\;AB\;\;are\;\;equal \\[3ex] \angle BAS = \angle ADB = 60^\circ ... \angle\;\;in\;\;the\;\;alternate\;\;segment = \angle\;\;between\;\;tangent\;SAT\;\;and\;\;chord\;AB \\[3ex] \angle BAS + \angle CAB + \angle CAE + \angle TAE = 180^\circ ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] 60 + x + \angle CAE + x = 180 \\[3ex] \angle CAE + 60 + 2x = 180 \\[3ex] \angle CAE = 180 - 60 - 2x \\[3ex] \angle CAE = 120 - 2x \\[3ex] \underline{\triangle CAE} \\[3ex] \angle CAE + \angle AEC + \angle ECA = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle CAE \\[3ex] (120 - 2x) + 5x + x = 180 \\[3ex] 120 - 2x + 6x = 180 \\[3ex] 4x = 180 - 120 \\[3ex] 4x = 60 \\[3ex] x = \dfrac{60}{4} \\[5ex] x = 15^\circ \\[3ex] $ OR
Second Method:
Construction: Draw the straight line EA and extend to K

$ \angle KAB = 5x^\circ ... exterior\;\;\angle\;\;of\;\;cyclic\;\;quadrilateral\;AECB = interior\;\;opposite\;\;\angle \\[3ex] \underline{Straight\;\;Line\;\;KAE} \\[3ex] \angle KAB + \angle BAE = 180^\circ ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] 5x + \angle BAE = 180 \\[3ex] \angle BAE = 180 - 5x \\[3ex] \angle BAS = \angle ADB = 60^\circ ... \angle\;\;in\;\;the\;\;alternate\;\;segment = \angle\;\;between\;\;tangent\;SAT\;\;and\;\;chord\;AB \\[3ex] \underline{Straight\;\;Line\;\;SAT} \\[3ex] \angle BAS + \angle BAE + \angle TAE = 180^\circ ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] 60 + (180 - 5x) + x = 180 \\[3ex] 60 + 180 - 5x + x = 180 \\[3ex] 240 - 4x = 180 \\[3ex] 240 - 180 = 4x \\[3ex] 60 = 4x \\[3ex] 4x = 60 \\[3ex] x = \dfrac{60}{4} \\[5ex] x = 15^\circ $

$ \angle ABC = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] \angle BCD = \angle BAC + \angle ABC \\[3ex] ...exterior\;\;\angle\;\;of\;\;a\;\;\triangle = sum\;\;of\;\;two\;\;interior\;\;opposite\;\;\angle s \\[3ex] \angle BCD = 63 + 90 \\[3ex] \angle = 153^\circ $

$ (i) \\[3ex] A\hat{C}B = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] C\hat{A}B = 58^\circ ...diagram \\[3ex] C\hat{A}B + A\hat{B}C + A\hat{C}B = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle ABC \\[3ex] 58 + A\hat{B}C + 90 = 180 \\[3ex] 148 + A\hat{B}C = 180 \\[3ex] A\hat{B}C = 180 - 148 \\[3ex] A\hat{B}C = 32^\circ \\[3ex] (ii) \\[3ex] \underline{Cyclic\;\;Quadrilateral\;ABMC} \\[3ex] \angle A + \angle M = 180^\circ ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quadrilateral\;\;are\;\;supplementary \\[3ex] \angle A = C\hat{A}B = 58^\circ ...diagram \\[3ex] \angle M = C\hat{M}B ...diagram \\[3ex] \implies \\[3ex] 58 + C\hat{M}B = 180 \\[3ex] C\hat{M}B = 180 - 58 \\[3ex] C\hat{M}B = 122^\circ \\[3ex] (iii) \\[3ex] C\hat{M}N = C\hat{A}B = 58^\circ ...exterior\;\;\angle\;\;of\;\;cyclic\;\;quadrilateral\;ABMC = interior\;\;opposite\;\;\angle \\[3ex] OR \\[3ex] C\hat{M}N + C\hat{M}B = 180^\circ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] C\hat{M}N + 122 = 180 \\[3ex] C\hat{M}N = 180 - 122 \\[3ex] C\hat{M}N = 58^\circ \\[3ex] \underline{\triangle CMN} \\[3ex] N\hat{C}M + C\hat{M}N + C\hat{N}M = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle CMN \\[3ex] N\hat{C}M + 58 + 90 = 180 \\[3ex] N\hat{C}M + 148 = 180 \\[3ex] N\hat{C}M = 180 - 148 \\[3ex] N\hat{C}M = 32^\circ $

$ \underline{Cyclic\;\;Quadrilateral\;DCBE} \\[3ex] (a) \\[3ex] A\hat{B}E = C\hat{D}E = 110^\circ \\[3ex] ...exterior\;\;\angle\;\;of\;\;a\;\;cyclic\;\;quad = interior\;\;opposite\;\;\angle \\[3ex] (b) \\[3ex] B\hat{E}D + D\hat{C}B = 180^\circ \\[3ex] ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quad\;\;are\;\;supplementary \\[3ex] B\hat{E}D + 100 = 180 \\[3ex] B\hat{E}D = 180 - 100 \\[3ex] B\hat{E}D = 80^\circ $

$ (139.1) \\[3ex] (a) \\[3ex] \hat{N_2} = x...alternate\;\;\angle s\;\;are\;\;equal \\[3ex] (b) \\[3ex] x = M_2...\angle\;\;between\;\;tangent\;PR\;\;and\;\;chord\;PN = \angle\;\;in\;\;the\;\;alternate\;\;segment \\[3ex] M_2 = x \\[3ex] Q_2 = M_2 = x ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] (139.2) \\[3ex] $

$ Because: QN || PR,\;\; SN\;\;is\;\;also\;\;|| PR...diagram \\[3ex] \dfrac{MN}{NR} = \dfrac{MS}{SP} ...Side\;\;Splitter\;\;theorem \\[5ex] However: \\[3ex] \dfrac{MN}{NR} = \dfrac{MS}{SP} ...Given \\[5ex] So,\;\;let\;\;us\;\;just\;\;show\;\;that\;\;SP = SQ \\[3ex] \underline{\triangle QSP} \\[3ex] \hat{Q_2} = x ...already\;\;shown\;\;in\;\;(b) \\[3ex] \hat{N_1} = \hat{N_2} = x ...chord\;QN\;\;bisects\;\;M\hat{N}P \\[3ex] \hat{P_3} = \hat{N_1} = x ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] Because\;\;\hat{Q_2} = \hat{P_3} = x;\;\;\triangle QSR\;\;is\;\;isosceles \\[3ex] \implies \\[3ex] SQ = SP \\[3ex] \therefore \dfrac{MN}{NR} = \dfrac{MS}{SP} $

$ (i) \\[3ex] AC = diameter \\[3ex] \dfrac{AC}{2} = radius \\[5ex] \angle OAS = 90^\circ \\[3ex] ...radius \perp tangent\;SAT\;\;at\;\;point\;\;of\;\;contact\;A \\[3ex] \angle OAS = \angle SAB + \angle BAO ... diagram \\[3ex] 90 = 48 + \angle BAO \\[3ex] \angle BAO + 48 = 90 \\[3ex] \angle BAO = 90 - 48 \\[3ex] \angle BAO = 42^\circ \\[3ex] x = \angle BAO = 42^\circ ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] (ii) \\[3ex] \angle ACD = \angle ABD = 27^\circ ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] \underline{\triangle COD} \\[3ex] \angle OCD + \angle ODC + \angle COD = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle COD \\[3ex] \angle OCD = \angle ACD = 27^\circ ... diagram \\[3ex] \angle ODC = x^\circ \\[3ex] \angle COD = y^\circ \\[3ex] \implies \\[3ex] 27 + x + y = 180 \\[3ex] 27 + 42 + y = 180 \\[3ex] 69 + y = 180 \\[3ex] y = 180 - 69 \\[3ex] y = 111^\circ $

$ (i) \\[3ex] \angle ECD = 68^\circ ...alternate\;\;\angle s\;\;are\;\;equal \\[3ex] (ii) \\[3ex] \angle OEA = 90^\circ ...radius\;OE \perp tangent\;AEB\;\;at\;\;point\;\;of\;\;contact\;E \\[3ex] \angle OEA = \angle AEC + \angle CEG \\[3ex] 90 = 68 + \angle CEG \\[3ex] 68 + \angle CEG = 90 \\[3ex] \angle CEG = 90 - 68 \\[3ex] \angle CEG = 22^\circ \\[3ex] (iii) \\[3ex] \angle EFD = \angle ECD + \angle CEF ...exterior\;\;\angle\;\;of\;\;\triangle CEF = sum\;\;of\;\;the\;\;two\;\;interior\;\;\angle s \\[3ex] 106 = 68 + \angle CEF \\[3ex] 68 + \angle CEF = 106 \\[3ex] \angle CEF = 106 - 68 \\[3ex] \angle CEF = 38^\circ \\[3ex] \underline{Cyclic\;\;Quadrilateral\;ECGF} \\[3ex] \angle G + \angle E = 180^\circ ...interior\:\:opposite\:\:\angle s\:\:of\:\:a\:\:Cyclic\:\:Quadrilateral\;\;are\;\;supplementary \\[3ex] \angle G = \angle CGF ...diagram \\[3ex] \angle E = \angle CEF ... diagram \\[3ex] \implies \\[3ex] \angle CGF + \angle CEF = 180 \\[3ex] \angle CGF + 38 = 180 \\[3ex] \angle CGF = 180 - 38 \\[3ex] \angle CGF = 142^\circ $

$ (a) \\[3ex] Obtuse\;\;\angle COA = 2 * \angle CBA \\[3ex] ...\angle\;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] Obtuse\;\;\angle COA = 2 * 58 \\[3ex] Obtuse\;\;\angle COA = 116^\circ \\[3ex] \angle OCA = \angle OAC = p ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle COA \\[3ex] \angle OCA + \angle OAC + \angle COA = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle COA \\[3ex] p + p + 116 = 180 \\[3ex] 2p = 180 - 116 \\[3ex] 2p = 64 \\[3ex] p = \dfrac{64}{2} \\[5ex] p = 32 \\[3ex] \therefore \angle OCA = \angle OAC = 32^\circ \\[3ex] (b) \\[3ex] Reflex\;\;\angle COA + Obtuse\;\;\angle COA = 360^\circ ...\angle s\:\:at\:\:a\:\:point \\[3ex] Reflex\;\;\angle COA + 116 = 360 \\[3ex] Reflex\;\;\angle COA = 360 - 116 \\[3ex] Reflex\;\;\angle COA = 244^\circ \\[3ex] Also: \\[3ex] Reflex\;\;\angle COA = 2 * \angle CDA \\[3ex] ...\angle\;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] 244 = 2 * \angle CDA \\[3ex] 2 * \angle CDA = 244 \\[3ex] \angle CDA = \dfrac{244}{2} \\[5ex] \angle CDA = 122^\circ \\[3ex] \angle DCA + \angle CAD + \angle CDA = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle CDA \\[3ex] \angle DCA + 23 + 122 = 180 \\[3ex] \angle DCA + 145 = 180 \\[3ex] \angle DCA = 180 - 145 \\[3ex] \angle DCA = 35^\circ $

$ \angle PRT = 33^\circ...\angle s\;\;in\;\;same\;\;segment\;\;are\;\;equal \\[3ex] \angle TPR = 90^\circ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] \angle PTR + \angle TPR + \angle PRT = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle TPR \\[3ex] \angle PTR + 90 + 33 = 180 \\[3ex] \angle PTR + 123 = 180 \\[3ex] \angle PTR = 180 - 123 \\[3ex] \angle PTR = 57^\circ $

$ \angle OTS = \angle OST = k ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle TOS \\[3ex] \angle OTS + \angle OST + \angle TOS = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle TOS \\[3ex] k + k + 76 = 180 \\[3ex] 2k = 180 - 76 \\[3ex] 2k = 104 \\[3ex] k = \dfrac{104}{2} \\[5ex] k = 52 \\[3ex] \therefore \angle OTS = \angle OST = 52^\circ \\[3ex] \underline{Cyclic\;\;Quadrilateral\;PRST} \\[3ex] \angle T + \angle R = 180^\circ \\[3ex] ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quad\;\;are\;\;supplementary \\[3ex] \angle T = \angle PTR + \angle OTS ...diagram \\[3ex] \angle T = 57 + 52 \\[3ex] \angle T = 109^\circ \\[3ex] \angle PRS = \angle R ...diagram \\[3ex] \implies \\[3ex] 109 + \angle PRS = 180 \\[3ex] \angle PRS = 180 - 109 \\[3ex] \angle PRS = 71^\circ $

$ A\hat{O}C = 2 * A\hat{B}C ...\angle\;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] 6x = 2 * A\hat{B}C \\[3ex] 2 * A\hat{B}C = 6x \\[3ex] A\hat{B}C = \dfrac{6x}{2} \\[5ex] A\hat{B}C = 3x \\[3ex] Y\hat{A}C = A\hat{B}C = 3x ...\angle \:\:between\:\:tangent\;XAY\;\;and\:\:chord\;AC = \angle\:\:in\:\:the\;\;alternate\:\:segment $

$ (i) \\[3ex] Reflex\;\;\angle AOC = 2 * \angle ABC \\[3ex] ...\angle\;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] 190 = 2 * \angle ABC \\[3ex] 2 * \angle ABC = 190 \\[3ex] \angle ABC = \dfrac{190}{2} \\[5ex] \angle ABC = 95^\circ \\[3ex] (ii) \\[3ex] Reflex\;\;\angle AOC + Obtuse\;\;\angle AOC = 360^\circ ...\angle s\:\:at\:\:a\:\:point \\[3ex] 190 + Obtuse\;\;\angle AOC = 360 \\[3ex] Obtuse\;\;\angle AOC = 360 - 190 \\[3ex] Obtuse\;\;\angle AOC = 170^\circ \\[3ex] Also: \\[3ex] Obtuse\;\;\angle AOC = 2 * \angle ADC \\[3ex] ...\angle\;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] 170 = 2 * \angle ADC \\[3ex] 2 * \angle ADC = 170 \\[3ex] \angle ADC = \dfrac{170}{2} \\[5ex] \angle ADC = 85^\circ $

$ \angle SUT = \angle URS = 36^\circ ...\angle \:\:between\:\:tangent\;TU\;\;and\:\:chord\;US = \angle\:\:in\:\:the\;\;alternate\:\:segment \\[3ex] \angle UST = \angle V = 100^\circ ...exterior\;\;\angle\;\;of\;\;Cyclic\;\;Quadrilateral\;UVRS = interior\;\;opposite\;\;\angle \\[3ex] \underline{\triangle UST} \\[3ex] \angle STU + \angle UST + \angle SUT = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle TOS \\[3ex] \angle STU + 100 + 36 = 180 \\[3ex] \angle STU + 136 = 180 \\[3ex] \angle STU = 180 - 136 \\[3ex] \angle STU = 44^\circ $

$ (a) \\[3ex] x = 2 * 56 ...\angle\;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] x = 112^\circ \\[3ex] (b) \\[3ex] y = 56^\circ ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal $

$ a) \\[3ex] \angle CAO = 90^\circ ... radius\;OA \perp tangent\;AC \;\;at\;\;point\;\;of\;\;contact\; A \\[5ex] b) \\[3ex] At\;\;least\;\;three\;\;ways\;\;to\;\;show\;\;Congruency \\[3ex] Side:\;\; OA = OB ...radius \\[3ex] Angle:\;\; \angle OAC = \angle OBC = 90^\circ ... radius \perp tangent \;\;at\;\;point\;\;of\;\;contact \\[3ex] Side:\;\; AC = BC ... ...two\;\;tangents:\;\;AC\;\;and\;\;BC\;\;drawn\;\;from\;\;the\;\;same\;\;external\;\;point:\;C\;\;are\;\;equal\;\;in\;\;length \\[3ex] \implies \\[3ex] \triangle AOC \cong \triangle BOC...Side-Angle-Side \\[3ex] OR \\[3ex] Angle:\;\; \angle OAC = \angle OBC = 90^\circ ... radius \perp tangent \;\;at\;\;point\;\;of\;\;contact \\[3ex] Side:\;\; AC = BC ... ...two\;\;tangents:\;\;AC\;\;and\;\;BC\;\;drawn\;\;from\;\;the\;\;same\;\;external\;\;point:\;C\;\;are\;\;equal\;\;in\;\;length \\[3ex] Angle:\;\; \angle ACO = \angle BCO ... \\[3ex] $ For two tangents: AC and BC drawn from the same external point: C, the line joining the external point: C, and the centre of the circle: O; bisects the angle between the tangents.

$ \implies \\[3ex] \triangle AOC \cong \triangle BOC...Angle-Side-Angle \\[3ex] OR \\[3ex] Side:\;\; OA = OB ...radius \\[3ex] Side:\;\; AC = BC ...two\;\;tangents:\;\;AC\;\;and\;\;BC\;\;drawn\;\;from\;\;the\;\;same\;\;external\;\;point:\;C\;\;are\;\;equal\;\;in\;\;length \\[3ex] Side:\;\; OC = OC ... common\;\;side\;\;of\;\;the\;\;two\;\;\triangle s \\[3ex] \implies \\[3ex] \triangle AOC \cong \triangle BOC...Side-Side-Side \\[5ex] c) \\[3ex] AC = BC ... ...two\;\;tangents:\;\;AC\;\;and\;\;BC\;\;drawn\;\;from\;\;the\;\;same\;\;external\;\;point:\;C\;\;are\;\;equal\;\;in\;\;length $

$ (i) \\[3ex] Obtuse\;\;\angle COA = 2 * \angle CDA \\[3ex] ...\angle\;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] Obtuse\;\;\angle COA = 2 * 72 \\[3ex] Obtuse\;\;\angle COA = 144^\circ \\[3ex] Reflex\;\;\angle COA + Obtuse\;\;\angle COA = 360^\circ ...\angle s\:\:at\:\:a\:\:point \\[3ex] x + 144 = 360 \\[3ex] x = 360 - 144 \\[3ex] x = 216^\circ \\[3ex] $ Construction: Join the chord from point C to point A

$ (ii) \\[3ex] \angle OCA = \angle OAC = p \\[3ex] ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle BOC \\[3ex] \angle OCA + \angle OAC + Reflex\;\;\angle COA = 180^\circ \\[3ex] ...sum\:\:of\:\:\angle s\:\:of\:\;\triangle COA \\[3ex] \implies \\[3ex] p + p + 144 = 180 \\[3ex] 2p = 180 - 144 \\[3ex] 2p = 36 \\[3ex] p = \dfrac{36}{2} \\[5ex] p = 18^\circ \\[3ex] \therefore \angle OCA = \angle OAC = 18^\circ \\[3ex] \underline{\triangle DCA} \\[3ex] \angle DCA + \angle CAD + \angle ADC = 180^\circ \\[3ex] ...sum\:\:of\:\:\angle s\:\:of\:\;\triangle COA \\[3ex] \angle DCA = \angle DCO + \angle OCA ...diagram \\[3ex] \angle DCA = y + 18 \\[3ex] \angle CAD = \angle OAC + \angle OAA ...diagram \\[3ex] \angle CAD = 18 + 38 \\[3ex] \angle CAD = 56^\circ \\[3ex] \angle ADC = 72^\circ ...diagram \\[3ex] \implies \\[3ex] (y + 18) + 56 + 72 = 180 \\[3ex] y + 18 + 56 + 72 = 180 \\[3ex] y + 146 = 180 \\[3ex] y = 180 - 146 \\[3ex] y = 34^\circ $

$ (a) \\[3ex] \angle TBO = 90^\circ ... radius\;OB \perp tangent\;TBW\;\;at\;\;point\;\;of\;\;contact\;B \\[3ex] (b) \\[3ex] \underline{\triangle BOC} \\[3ex] \angle OBC = \angle OCB = p ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle BOC \\[3ex] \angle OBC + \angle OCB + \angle BOC = 180^\circ ...sum\:\:of\:\:\angle s\:\:of\:\;\triangle BOC \\[3ex] \implies \\[3ex] p + p + 72 = 180 \\[3ex] 2p = 180 - 72 \\[3ex] 2p = 108 \\[3ex] p = \dfrac{108}{2} \\[5ex] p = 54 \\[3ex] \therefore \angle OBC = \angle OCB = 54^\circ \\[3ex] \angle ABC = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] \underline{\triangle ABC} \\[3ex] \angle CAB + \angle ABC + \angle ACB = 180^\circ ...sum\:\:of\:\:\angle s\:\:of\:\;\triangle ABC \\[3ex] \angle ACB = \angle OCB = 54^\circ \\[3ex] \implies \\[3ex] \angle CAB + 90 + 54 = 180 \\[3ex] \angle CAB + 144 = 180 \\[3ex] \angle CAB = 180 - 144 \\[3ex] \angle CAB = 36^\circ \\[3ex] (c) \\[3ex] \angle CBW = \angle CAB = 36^\circ ...\angle \:\:between\:\:tangent\;TBW\;\;and\:\:chord\;BC = \angle\:\:in\:\:the\;\;alternate\:\:segment \\[3ex] (d) \\[3ex] \angle DBC = \angle OBC = 54^\circ ...diagram $

$ \angle RQK = 90^\circ \\[3ex] ... radius\;OQ \perp tangent\;PQK\;\;at\;\;point\;\;of\;\;contact\;Q \\[3ex] \angle SQK = \angle SQR + \angle RQK ...diagram \\[3ex] \angle SQK = 33 + 90 \\[3ex] \angle SQK = 123^\circ \\[3ex] \angle SQK = m + n \\[3ex] ...exterior\;\;\angle\;\;of\;\;a\;\;\triangle = sum\;\;of\;\;two\;\;interior\;\;opposite\;\;\angle s \\[3ex] m + n = \angle SQK \\[3ex] \therefore m + n = 123^\circ \\[3ex] $ OR

$ \angle RQP = 90^\circ \\[3ex] ... radius\;OQ \perp tangent\;PQ\;\;at\;\;point\;\;of\;\;contact\;Q \\[3ex] Also: \\[3ex] \angle RQP = \angle SQR + \angle SQP ...diagram \\[3ex] 90 = 33 + \angle SQP \\[3ex] 33 + \angle SQP = 90 \\[3ex] \angle SQP = 90 - 33 \\[3ex] \angle SQP = 57^\circ \\[3ex] \underline{\triangle PSQ} \\[3ex] \angle SPQ + \angle PSQ + \angle SQP = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle PSQ \\[3ex] m + n + 57 = 180 \\[3ex] m + n = 180 - 57 \\[3ex] m + n = 123^\circ $

$ a) \\[3ex] \angle DAB = 90^\circ ... \angle\;\;in\;\;a\;\;semicircle \\[3ex] a + \angle DAB + 65 = 180^\circ ... \angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] a + 90 + 65 = 180 \\[3ex] a + 155 = 180 \\[3ex] a = 180 - 155 \\[3ex] a = 25^\circ \\[3ex] 65^\circ = b ...\angle\;\;between\;\;tangent\;AP\;\;and\;\;chord\;AB = \angle\;\;in\;\;alternate\;\;segment \\[3ex] b = 65^\circ \\[3ex] c + (b + 72) = 180^\circ ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quadrilateral\;\;are\;\;supplementary \\[3ex] c + (65 + 72) = 180 \\[3ex] c + 137 = 180 \\[3ex] c = 180 - 137 \\[3ex] c = 43^\circ \\[3ex] OR \\[3ex] a = \angle DBA ...\angle\;\;between\;\;tangent\;AP\;\;and\;\;chord\;AD = \angle\;\;in\;\;alternate\;\;segment \\[3ex] \angle DBA = a = 25^\circ \\[3ex] \angle DCB = 90^\circ ... \angle\;\;in\;\;a\;\;semicircle \\[3ex] \angle CBD + \angle DCB + \angle CDB = 180^\circ... sum\;\;of\;\;\angle s\;\;in\;\;\triangle DCB \\[3ex] \angle CBD + 90 + 72 = 180 \\[3ex] \angle CBD + 162 = 180 \\[3ex] \angle CBD = 180 - 162 \\[3ex] \angle CBD = 18^\circ \\[3ex] c = \angle CBD + \angle DBA ... diagram \\[3ex] c = 18 + 25 \\[3ex] c = 43^\circ \\[5ex] b) \\[3ex] x:y = 1:2 \\[3ex] \implies \\[3ex] \dfrac{x}{y} = \dfrac{1}{2} \\[5ex] y(1) = x(2) \\[3ex] y = 2x ... eqn.(1) \\[3ex] \angle OAC = 90^\circ ... radius\;OA \perp tangent\;AC \;\;at\;\;point\;\;of\;\;contact\;A \\[3ex] \angle OAC + x + y = 180^\circ ... sum\;\;of\;\;\angle s\;\;in\;\;\triangle OAC \\[3ex] 90 + x + y = 180 \\[3ex] x + y = 180 - 90 \\[3ex] x + y = 90 \\[3ex] But:\;\; y = 2x...eqn.(1) \\[3ex] x + 2x = 90 \\[3ex] 3x = 90 \\[3ex] x = \dfrac{90}{3} \\[5ex] x = 30^\circ \\[3ex] From\;\;eqn.(1):\;\;y = 2x \\[3ex] y = 2(30) \\[3ex] y = 60^\circ \\[3ex] y = 2z ...\angle\;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] 2z = y \\[3ex] z = \dfrac{y}{2} \\[5ex] z = \dfrac{60}{2} \\[5ex] z = 30^\circ $

$ (i) \\[3ex] \angle DBC = \angle DAC = 27^\circ \\[3ex] ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] \angle ABC = \angle ABD + \angle DBC ...diagram \\[3ex] \angle ABC = 54 + 27 \\[3ex] \angle ABC = 81^\circ \\[3ex] \underline{\triangle ABC} \\[3ex] \angle CAB + \angle ABC + \angle ACB = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle ABC \\[3ex] \angle CAB + 81 + 63 = 180 \\[3ex] \angle CAB + 144 = 180 \\[3ex] \angle CAB = 180 - 144 \\[3ex] \angle CAB = 36^\circ \\[3ex] (ii) \\[3ex] \angle ADB = \angle ACB = 63^\circ ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] \angle AEB = \angle DAC + \angle ADB \\[3ex] ... exterior\;\;\angle\;\;of\;\;a\;\;\triangle = sum\;\;of\;\;two\;\;interior\;\;opposite\;\;\angle s \\[3ex] \angle AEB = 27 + 63 \\[3ex] \angle AEB = 90^\circ \\[3ex] OR \\[3ex] \underline{\triangle AEB} \\[3ex] \angle AEB + \angle EAB + \angle ABE = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle AEB \\[3ex] \angle EAB = \angle CAB = 36^\circ ...diagram \\[3ex] \angle ABE = \angle ABD = 54^\circ ...diagram \\[3ex] \implies \\[3ex] \angle AEB + 36 + 54 = 180 \\[3ex] \angle AEB + 90 = 180 \\[3ex] \angle AEB = 180 - 90 \\[3ex] \angle AEB = 90^\circ $

$ \angle PWR + y = 180^\circ ... \angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] \angle PWR = 180 - y \\[3ex] \underline{\triangle PRW} \\[3ex] \angle PRW = 90^\circ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] \angle RPW + \angle PRW + \angle PWR = 180^\circ ... sum\;\;of\;\;\angle s\;\;in\;\;\triangle PRW \\[3ex] \angle RPW + 90 + (180 - y) = 180 \\[3ex] \angle RPW + 90 + 180 - y = 180 \\[3ex] \angle RPW = 180 - 90 - 180 + y \\[3ex] \angle RPW = y - 90 \\[3ex] \angle RPS = \angle RPW = y - 90 ... diagram \\[3ex] \angle QPR = x ... diagram \\[3ex] \angle QPS = \angle QPR + \angle RPS...diagram \\[3ex] \angle QPS = x + (y - 90) \\[3ex] \angle QPS = x + y - 90 \\[3ex] \angle QSP = z ...diagram \\[3ex] y = \angle Q ...exterior\;\;\angle\;\;of\;\;Cyclic\;\;Quad\;\;PQRW = interior\;\;opposite\;\;\angle \\[3ex] \underline{\triangle PQS} \\[3ex] \angle PQS = \angle Q = y ...diagram \\[3ex] \angle QPS + \angle PQS + \angle QSP = 180^\circ ... sum\;\;of\;\;\angle s\;\;in\;\;\triangle PQS \\[3ex] (x + y - 90) + y + z = 180 \\[3ex] x + y - 90 + y + z = 180 \\[3ex] x + 2y - 90 + z = 180 \\[3ex] z = 180 - x - 2y + 90 \\[3ex] z = 270 - x - 2y $

$ \angle ABC = 90^\circ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] \angle BAC + \angle ABC + \angle ACB = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle ABC \\[3ex] 34 + 90 + \angle ACB = 180 \\[3ex] 124 + \angle ACB = 180 \\[3ex] \angle ACB = 180 - 124 \\[3ex] \angle ACB = 56^\circ $

$ \angle EOA + \angle AOD = 180^\circ ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] \angle EOA + 143 = 180 \\[3ex] \angle EOA = 180 - 143 \\[3ex] \angle EOA = 37^\circ \\[3ex] \angle CAO = \angle EOA = 37^\circ...alternate\;\;\angle s\;\;are\;\;equal \\[3ex] \angle OAB = 90^\circ...radius\;OA\;\;\perp\;\;tangent\;AB\;\;at\;\;point\;\;of\;\;contact\;A \\[3ex] \angle OAB = \angle CAO + \angle CAB ...diagram \\[3ex] 90 = 37 + \angle CAB \\[3ex] \angle CAB + 37 = 90 \\[3ex] \angle CAB = 90 - 37 \\[3ex] \angle CAB = 53^\circ \\[3ex] $
$ \angle OCA = \angle CAO = 37^\circ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle \\[3ex] \angle OCB = 90^\circ...radius\;OA\;\;\perp\;\;tangent\;AB\;\;at\;\;point\;\;of\;\;contact\;A \\[3ex] \angle OCB = \angle OCA + \angle ACB ...diagram \\[3ex] 90 = 37 + \angle ACB \\[3ex] \angle ACB + 37 = 90 \\[3ex] \angle ACB = 90 - 37 \\[3ex] \angle ACB = 53^\circ \\[3ex] \underline{\triangle BAC} \\[3ex] \angle ABC + \angle CAB + \angle ACB = 180^\circ...sum\;\;of\;\;\angle s\;\;in\;\;a\;\;\triangle \\[3ex] \angle ABC + 53 + 53 = 180 \\[3ex] \angle ABC + 106 = 180 \\[3ex] \angle ABC = 180 - 106 \\[3ex] \angle ABC = 74^\circ $

Construction: Draw two radii: one radius from point X to point O and the other radius from point Y to point O

$ \angle XOY = 2 * \angle XBY ...\angle\;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] \angle XOY = 2 * b \\[3ex] \angle XOY = 2b \\[3ex] \underline{Cyclic\;\;Quadrilateral\;AXOY} \\[3ex] \angle A + \angle O = 180^\circ \\[3ex] ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quad\;\;are\;\;supplementary \\[3ex] \angle A = a \\[3ex] \angle O = \angle XOY = 2b \\[3ex] \implies \\[3ex] a + 2b = 180 $

$ (a) \\[3ex] B\hat{A}C = 38^\circ ...\angle \:\:between\:\:tangent\;ECF\;\;and\:\:chord\;BC = \angle\:\:in\:\:the\;\;alternate\:\:segment \\[3ex] (b) \\[3ex] A\hat{B}C = A\hat{C}B = p ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle BAC \\[3ex] A\hat{B}C + A\hat{C}B + B\hat{A}C = 180^\circ ...sum\:\:of\:\:\angle s\:\:of\:\;\triangle BAC \\[3ex] \implies \\[3ex] p + p + 38 = 180 \\[3ex] 2p = 180 - 38 \\[3ex] 2p = 142 \\[3ex] p = \dfrac{142}{2} \\[5ex] p = 71 \\[3ex] \therefore A\hat{B}C = A\hat{C}B = 71^\circ \\[3ex] (c) \\[3ex] \underline{Cyclic\;\;Quadrilateral\:\:ABCD} \\[3ex] \hat{D} + \hat{B} = 180^\circ ...interior\:\:opposite\:\:\angle s\:\:of\:\:a\:\:Cyclic\:\:Quadrilateral\;\;are\;\;supplementary \\[3ex] \hat{D} = A\hat{D}C ...diagram \\[3ex] \hat{B} = A\hat{B}C = 71^\circ ...diagram \\[3ex] \implies \\[3ex] A\hat{D}C + 71 = 180 \\[3ex] A\hat{D}C = 180 - 71 \\[3ex] A\hat{D}C = 109^\circ \\[3ex] $
$ (d) \\[3ex] C\hat{O}B = 2 * B\hat{A}C ...\angle\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] C\hat{O}B = 2 * 38 \\[3ex] C\hat{O}B = 76^\circ $

$ \underline{Cyclic\;\;Quadrilateral\;\;DEFG} \\[3ex] \angle EDG + \angle EFG = 180^\circ \\[3ex] ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quad\;\;are\;\;supplementary \\[3ex] 82 + \angle EFG = 180 \\[3ex] \angle EFG = 180 - 82 \\[3ex] \angle EFG = 98^\circ \\[3ex] \angle HFJ = \angle EFG = 98^\circ ...vertical\;\;\angle s\;\;are\;\;equal $

$ a = 40^\circ ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[5ex] a + b = 90^\circ ... \angle\;\;in\;\;a\;\;semicircle \\[3ex] 40 + b = 90 \\[3ex] b = 90 - 40 \\[3ex] b = 50^\circ \\[5ex] (c + 40) + 115 = 180^\circ ...opposite\;\;\angle s\;\;of\;\;cyclic\;\;quadrilateral\;ABCD\;\;are\;\;supplementary \\[3ex] c + 40 + 115 = 180 \\[3ex] c + 155 = 180 \\[3ex] c = 180 - 155 \\[3ex] c = 25^\circ \\[5ex] d = 40^\circ ...\angle\;\;between\;\;tangent\;TD\;\;and\;\;chord\;CD = \angle\;\;in\;\;alternate\;\;segment \\[5ex] \angle ADC = 90^\circ ... \angle\;\;in\;\;a\;\;semicircle \\[3ex] \underline{\triangle ADT} \\[3ex] 40 + (\angle ADC + d) + e = 180^\circ ... sum\;\;of\;\;\angle s\;\;in\;\;\triangle ADT \\[3ex] 40 + (90 + 40) + e = 180 \\[3ex] 40 + 90 + 40 + e = 180 \\[3ex] 170 + e = 180 \\[3ex] e = 180 - 170 \\[3ex] e = 10^\circ $

$ (a) \\[3ex] \angle PQR = 90^\circ...\angle\;\;in\;\;\;a\;\;semicircle\;\;is\;\;a\;\;right\;\;\angle \\[3ex] (b) \\[3ex] |PR|^2 = |PQ|^2 + |QR|^2 ...Pythagorean\;\;theorem \\[3ex] |PR|^2 = 17^2 + 9^2 \\[3ex] |PR|^2 = 289 + 81 \\[3ex] |PR|^2 = 370 \\[3ex] |PR| = \sqrt{370} \\[3ex] |PR| = 19.23538406\;cm $

$ \angle\;\;in\;\;alternate\;\;segment = 32^\circ \\[3ex] \angle \;\;between\;\;chord\;YT\;\;and\;\;tangent\;TQ = \theta \\[3ex] \theta = 32^\circ ...\angle \;\;between\;\;chord\;YT\;\;and\;\;tangent\;TQ = \angle\;\;in\;\;alternate\;\;segment \\[3ex] \angle YTR = \phi \\[3ex] \theta + \phi + 40 = 180^\circ ...\angle s\;\;in\;\;a\;\;straight\;\;line \\[3ex] 32 + \phi + 40 = 180 \\[3ex] \phi = 180 - 72 \\[3ex] \phi = 108 \\[3ex] \therefore \angle YTR = 108^\circ \\[3ex] OR \\[3ex] \angle TXR = 40^\circ ...\angle \;\;between\;\;tangent\;TQ\;\;and\;\;chord\;TR = \angle\;\;in\;\;alternate\;\;segment \\[3ex] \angle YXR = 32 + 40 ...diagram \\[3ex] \angle YXR = 72^\circ \\[3ex] \underline{Cyclic\;\;Quadrilateral\;YXRT} \\[3ex] \angle YXR + \angle YTR = 180^\circ ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quad\;\;are\;\;supplementary \\[3ex] 72 + \angle YTR = 180 \\[3ex] \angle YTR = 180 - 72 \\[3ex] \angle YTR = 108^\circ $

$ (a) \\[3ex] A\hat{D}C = 90^\circ \\[3ex] ...diameter\;CD\;\;that\;\;bisects\;\;chord\;AB \;\;is\perp to\;\;the\;\;chord \\[3ex] (b) \\[3ex] \underline{\triangle ADC} \\[3ex] |CD| = diameter...Side \\[3ex] A\hat{D}C = 90^\circ...Angle \\[3ex] ...diameter\;CD\;\;that\;\;bisects\;\;chord\;AB \;\;is\perp to\;\;the\;\;chord \\[3ex] |AD| = |DB| ...D\;\;is\;\;the\;\;mid-point...Side \\[3ex] \underline{\triangle BDC} \\[3ex] |CD| = diameter...Side \\[3ex] B\hat{D}C = 90^\circ...Angle \\[3ex] ...diameter\;CD\;\;that\;\;bisects\;\;chord\;AB \;\;is\perp to\;\;the\;\;chord \\[3ex] |DB| = |AD| ...D\;\;is\;\;the\;\;mid-point...Side \\[3ex] \therefore \triangle ADC \cong \triangle BDC ...Side-Angle-Side \\[5ex] OR \\[3ex] \underline{\triangle ADC} \\[3ex] A\hat{C}D = B\hat{C}D ...Angle \\[3ex] ...diameter\;CD\;that\;\;bisects\;\;chord\;AB\;\;also\;\;bisects\;\;its\;\;arc \\[3ex] |CD| = diameter...Side \\[3ex] A\hat{D}C = 90^\circ...Angle \\[3ex] ...diameter\;CD\;\;that\;\;bisects\;\;chord\;AB \;\;is\perp to\;\;the\;\;chord \\[3ex] \underline{\triangle BDC} \\[3ex] B\hat{C}D = A\hat{C}D ...Angle \\[3ex] ...diameter\;CD\;that\;\;bisects\;\;chord\;AB\;\;also\;\;bisects\;\;its\;\;arc \\[3ex] |CD| = diameter...Side \\[3ex] B\hat{D}C = 90^\circ...Angle \\[3ex] ...diameter\;CD\;\;that\;\;bisects\;\;chord\;AB \;\;is\perp to\;\;the\;\;chord \\[3ex] \therefore \triangle ADC \cong \triangle BDC ...Angle-Side-Angle $

$ a = 60^\circ ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[5ex] a + b = 90^\circ ... radius\;OA \perp tangent\;PAQ \;\;at\;\;point\;\;of\;\;contact\;A \\[3ex] b = 90 - a \\[3ex] b = 90 - 60 \\[3ex] b = 30^\circ \\[5ex] b = c ...\angle\;\;between\;\;tangent\;PAQ\;\;and\;\;chord\;AB = \angle\;\;in\;\;alternate\;\;segment \\[3ex] c = b \\[3ex] c = 30^\circ \\[5ex] 110^\circ = d + 60^\circ ...exterior\;\;\angle\;\;of\;\;\triangle BCD = sum\;\;of\;\;two\;\;interior\;\;opposite\;\;\angle s \\[3ex] d + 60 = 110 \\[3ex] d = 110 - 60 \\[3ex] d = 50^\circ $

Construction: Extend the two parallel lines: AQ and NZ

$ \angle BAN = \angle ANZ = x^\circ ...alternate\;\;\angle s\;\;are\;\;equal \\[3ex] \angle NZQ = \angle BAN = x^\circ \\[3ex] ...exterior\:\: \angle \:\:of\:\:a\:\:cyclic\:\:quad = interior\:\:opposite\:\: \angle $

Construction: Join the line from the radius of the circle: O to the point: P.

$ (a) \\[3ex] \angle OAP = 90^\circ ... radius\;OA \perp tangent\;AC \;\;at\;\;point\;\;of\;\;contact\; A \\[5ex] (b) \\[3ex] At\;\;least\;\;three\;\;ways\;\;to\;\;prove\;\;Congruency \\[3ex] Side:\;\; OA = OB ...radius \\[3ex] Angle:\;\; \angle OBP = \angle OAP = 90^\circ ... radius \perp tangent \;\;at\;\;point\;\;of\;\;contact \\[3ex] Side:\;\; AP = BP ... ...two\;\;tangents:\;\;AP\;\;and\;\;BP\;\;drawn\;\;from\;\;the\;\;same\;\;external\;\;point:\;P\;\;are\;\;equal\;\;in\;\;length \\[3ex] \implies \\[3ex] \triangle AOP \cong \triangle BOP...Side-Angle-Side \\[3ex] OR \\[3ex] Angle:\;\; \angle OBP = \angle OAP = 90^\circ ... radius \perp tangent \;\;at\;\;point\;\;of\;\;contact \\[3ex] Side:\;\; AP = BP ... ...two\;\;tangents:\;\;AP\;\;and\;\;BP\;\;drawn\;\;from\;\;the\;\;same\;\;external\;\;point:\;P\;\;are\;\;equal\;\;in\;\;length \\[3ex] Angle:\;\; \angle APO = \angle BPO ... \\[3ex] $ For two tangents: AP and BP drawn from the same external point: P, the line joining the external point: P, and the centre of the circle: O; bisects the angle between the tangents.

$ \implies \\[3ex] \triangle AOP \cong \triangle BOP...Angle-Side-Angle \\[3ex] OR \\[3ex] Side:\;\; OA = OB ...radius \\[3ex] Side:\;\; AP = BP ...two\;\;tangents:\;\;AC\;\;and\;\;BC\;\;drawn\;\;from\;\;the\;\;same\;\;external\;\;point:\;P\;\;are\;\;equal\;\;in\;\;length \\[3ex] Side:\;\; OP = OP ... common\;\;side\;\;of\;\;the\;\;two\;\;\triangle s \\[3ex] \implies \\[3ex] \triangle AOP \cong \triangle BOP...Side-Side-Side $

$ \angle BDA = \angle DBA = x...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle DBA \\[3ex] \angle BDA + \angle DBA + \angle DAB = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle DBA \\[3ex] \implies \\[3ex] x + x + \angle DAB = 180 \\[3ex] 2x + \angle DAB = 180 \\[3ex] \angle DAB = 180 - 2x \\[3ex] \underline{Cyclic\;\;Quadrilateral\:\:CBAD} \\[3ex] \angle C + \angle A = 180^\circ \\[3ex] ...interior\:\:opposite\:\:\angle s\:\:of\:\:a\:\:cyclic\;\;quad\;\;are\;\;supplementary \\[3ex] \angle A = \angle DAB = 180 - 2x...diagram \\[3ex] \angle C = y^\circ...diagram \\[3ex] \implies \\[3ex] y + (180 - 2x) = 180 \\[3ex] y + 180 - 2x = 180 \\[3ex] y = 180 - 180 + 2x \\[3ex] y = 2x $

$ (2x + 18) + 84 = 180^\circ ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quad\;\;are\;\;supplementary \\[3ex] 2x + 18 + 84 = 180 \\[3ex] 2x + 102 = 180 \\[3ex] 2x = 180 - 102 \\[3ex] 2x = 78 \\[3ex] x = \dfrac{78}{2} \\[5ex] x = 39^\circ $

$ Obtuse\;\;A\hat{O}C = 2 * A\hat{B}C \\[3ex] ...\angle\;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] Obtuse\;\;A\hat{O}C = 2 * 2x \\[3ex] Obtuse\;\;A\hat{O}C = 4x \\[3ex] Reflex\;\;A\hat{O}C = 2 * A\hat{D}C \\[3ex] ...\angle\;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] Reflex\;\;A\hat{O}C = 2 * 3x \\[3ex] Reflex\;\;A\hat{O}C = 6x \\[3ex] Reflex\;\;A\hat{O}C + Obtuse\;\;A\hat{O}C = 360^\circ ...\angle s\:\:at\:\:a\:\:point \\[3ex] 6x + 4x = 130 \\[3ex] 10x = 130 \\[3ex] x = \dfrac{130}{10} \\[5ex] x = 13 \\[3ex] Obtuse\;\;A\hat{O}C \\[3ex] = 4x \\[3ex] = 4(13) \\[3ex] = 52^\circ $

$ (i) \\[3ex] x + 120 = 360^\circ ...\angle s\;\;at\;\;a\;\;point \\[3ex] x = 360 - 120 \\[3ex] x = 240^\circ \\[3ex] x = 2 * j ...\angle\;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] 2j = x \\[3ex] 2j = 240 \\[3ex] j = \dfrac{240}{2} \\[5ex] j = 120^\circ \\[3ex] (ii) \\[3ex] $ Construction: Extend the two parallel lines in order to use angle properties between parallel lines

$ h = \theta ...alternate\;\;\angle s\;\;are\;\;equal \\[3ex] \theta + 120 = 180^\circ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] \implies \\[3ex] h + 120 = 180 \\[3ex] h = 180 - 120 \\[3ex] h = 60^\circ \\[3ex] Also: \\[3ex] j = \phi ...alternate\;\;\angle s\;\;are\;\;equal \\[3ex] \phi + k = 180^\circ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] \implies \\[3ex] 120 + k = 180 \\[3ex] k = 180 - 120 \\[3ex] k = 60^\circ $

Based on Parts (i) and (ii) from Question (169.):

Because the opposite angles of the quadrilateral are equal, the quadrilateral is likely a rhombus.

$ \angle FGP = \angle FGE = 81°...diagram \\[3ex] \angle GFP + \angle FGP + \angle GPF = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle FGP \\[3ex] \angle GFP + 81 + 77 = 180 \\[3ex] \angle GFP + 158 = 180 \\[3ex] \angle GFP = 180 - 158 \\[3ex] \angle GFP = 22^\circ \\[3ex] (i) \\[3ex] x = \angle GFP = 22^\circ ... \angle s\;\;in\;\;same\;\;segment\;\;are\;\;equal \\[3ex] (ii) \\[3ex] Obtuse\;\;\angle FCE = 2 * \angle FGE ...\angle \;\;at\;\;center = 2 * \angle \;\;at\;\;circumference \\[3ex] Obtuse\;\;\angle FCE = 2 * 81 \\[3ex] Obtuse\;\;\angle FCE = 162^\circ \\[3ex] y + Obtuse\;\;\angle FCE = 360^\circ ... \angle s\;\;at\;\;a\;\;point \\[3ex] y + 162 = 360 \\[3ex] y = 360 - 162 \\[3ex] y = 198^\circ $

$ 1st: \\[3ex] a = 50^\circ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] 2nd: \\[3ex] O\hat{D}A = O\hat{A}D = 50^\circ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle AOD \\[3ex] O\hat{D}A + O\hat{A}D + A\hat{O}D = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle AOD \\[3ex] 50 + 50 + b = 180 \\[3ex] 100 + b = 180 \\[3ex] b = 180 - 100 \\[3ex] b = 80^\circ \\[3ex] 3rd: \\[3ex] b + c = 180^\circ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] 80 + c = 180 \\[3ex] c = 180 - 80 \\[3ex] c = 100^\circ \\[3ex] OR \\[3ex] c = O\hat{D}A + O\hat{A}D \\[3ex] ...exterior\;\;\angle\;\;of\;\;a\;\;\triangle = sum\;\;of\;\;two\;\;interior\;\;opposite\;\;\angle s \\[3ex] c = 50 + 50 \\[3ex] c = 100^\circ \\[3ex] 4th: \\[3ex] D\hat{O}C = A\hat{O}B ...vertical\;\;\angle s\;\;are\;\;equal \\[3ex] \implies \\[3ex] D\hat{O}C = c \\[3ex] D\hat{O}C = 100^\circ \\[3ex] OR \\[3ex] D\hat{O}C = O\hat{D}A + O\hat{A}D \\[3ex] ...exterior\;\;\angle\;\;of\;\;a\;\;\triangle = sum\;\;of\;\;two\;\;interior\;\;opposite\;\;\angle s \\[3ex] D\hat{O}C = 50 + 50 \\[3ex] D\hat{O}C = 100^\circ \\[3ex] O\hat{D}C = O\hat{C}D = p...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle AOD \\[3ex] O\hat{D}C + O\hat{C}D + D\hat{O}C = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle AOD \\[3ex] p + p + 100 = 180 \\[3ex] 2p = 180 - 100 \\[3ex] 2p = 80 \\[3ex] p = \dfrac{80}{2} \\[5ex] p = 40 \\[3ex] \therefore O\hat{D}C = O\hat{C}D = 40^\circ \\[3ex] \implies \\[3ex] O\hat{D}C = d = 40^\circ \\[3ex] 5th: \\[3ex] O\hat{C}D = e = 40^\circ $

$ \angle CQP = 90^\circ...radius\;CQ\;\;\perp\;\;tangent\;PQR\;\;at\;\;point\;\;of\;\;contact\;Q \\[3ex] \underline{\triangle PQW} \\[3ex] \angle QWP + \angle WQP + \angle WPQ = 180^\circ...sum\;\;of\;\;\angle s\;\;a\;\;\triangle \\[3ex] \angle WQP = 40^\circ ...diagram \\[3ex] \angle WQP = \angle CQP = 90^\circ...diagram \\[3ex] \implies \\[3ex] \angle QWP + 40 + 90 = 180 \\[3ex] \angle QWP + 130 = 180 \\[3ex] \angle QWP = 180 - 130 \\[3ex] \angle QWP = 50^\circ \\[3ex] \underline{\triangle CVW} \\[3ex] \angle CWV = \angle QWP = 50^\circ ... diagram \\[3ex] \angle CVW = \angle CWV = 50^\circ ...base\;\;\angle s\;\;of\;\;isosceles\;\triangle \\[3ex] e = \angle CVW + \angle CWV ...exterior\;\angle\;\;of\;\;a\;\;\triangle = sum\;\;of\;\;two\;\;interior\;\;opposite\;\;\angle s \\[3ex] e = 50 + 50 \\[3ex] e = 100^\circ $

$ \underline{Quadrilateral\:\:TUVW} \\[3ex] \angle U + \angle W = 180^\circ \\[3ex] ...opposite\:\:\angle s\:\:of\:\:a\:\:cyclic\:\:quad\;\;are\;\;supplementary \\[3ex] 3x + 20 + 88 = 180 \\[3ex] 3x + 108 = 180 \\[3ex] 3x = 180 - 108 \\[3ex] 3x = 72 \\[3ex] x = \dfrac{72}{3} \\[5ex] x = 24^\circ $

We can solve this question in at least two ways.
Use any approach you prefer.

$ \underline{First\;\;Method} \\[3ex] \angle ABD = \angle DEB = 77^\circ...\angle\;\;between\;\;tangent\;ABC\;\;and\;\;chord\;DB = \angle\;\;in\;\;alternate\;\;segment \\[3ex] \angle FEB = \angle DEF + \angle DEB ...diagram \\[3ex] \angle FEB = 64 + 77 \\[3ex] \angle FEB = 141^\circ \\[3ex] \angle BDF + \angle FEB = 180^\circ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quadrilateral\;\;are\;\;supplementary \\[3ex] \angle BDF + 141 = 180 \\[3ex] \angle BDF = 180 - 141 \\[3ex] \angle BDF = 39^\circ \\[5ex] \underline{Second\;\;Method} \\[3ex] \angle DBE = 90^\circ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] \angle ABD + \angle DBE + \angle EBC = 180^\circ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] 77 + 90 + \angle EBC = 180 \\[3ex] 167 + \angle EBC = 180 \\[3ex] \angle EBC = 180 - 167 \\[3ex] \angle EBC = 13^\circ \\[3ex] \angle EBC = \angle EDB = 13^\circ ... \angle\;\;between\;\;tangent\;ABC\;\;and\;\;chord\;BE = \angle\;\;in\;\;alternate\;\;segment \\[3ex] \angle DFE = 90^\circ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] \angle EDF + \angle DFE + \angle DEF = 180^\circ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle\;DFE \\[3ex] \angle EDF + 90 + 64 = 180 \\[3ex] \angle EDF + 154 = 180 \\[3ex] \angle EDF = 180 - 154 \\[3ex] \angle EDF = 26^\circ \\[3ex] \angle BDF = \angle EDB + \angle EDF ... diagram \\[3ex] \angle BDF = 13 + 26 \\[3ex] \angle BDF = 39^\circ $

$ 78 = 2 * y ...\angle\;\;at\;\;centre = 2 *\angle\;\;at\;\;circumference \\[3ex] 2y = 78 \\[3ex] y = \dfrac{78}{2} \\[5ex] y = 39^\circ $

We can solve this question in at least two ways.
Use any approach you prefer.
Construction is needed in either approach.

First Method
Construction: Join the radius from point T to center O
Extend tangent UT to J


$ \angle OTR = \angle ORT = 25^\circ ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle \\[3ex] \angle OTJ = 90^\circ...radius\;OT \;\;\perp\;\;tangent\;UTJ\;\;at\;\;point\;\;of\;\;contant\;T \\[3ex] \angle OTJ = \angle OTR + x ...diagram \\[3ex] 90 = 25 + x \\[3ex] 25 + x = 90 \\[3ex] x = 90 - 25 \\[3ex] x = 65^\circ \\[3ex] $ Second Method
Construction: Join the chord from point T to point S


$ \angle RTS = 90^\circ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] \angle RST = x^\circ ... \angle\;\;between\;\;tangent\;UT\;\;and\;\;chord\;RT = \angle\;\;in\;\;alternate\;\;segment \\[3ex] \angle RST + \angle RTS + \angle TRS = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle RTS \\[3ex] x + 90 + 25 = 180 \\[3ex] x + 115 = 180 \\[3ex] x = 180 - 115 \\[3ex] x = 65^\circ $

$ a) \\[3ex] p + 78 = 180^\circ \\[3ex] ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quad\;\;are\;\;supplementary \\[3ex] p = 180 - 78 \\[3ex] p = 102^\circ \\[3ex] q = p = 102^\circ \\[3ex] ... exterior\;\; \angle \:\:of\:\:a\:\:cyclic\:\:quadrilateral = interior\;\;opposite\;\;\angle \\[3ex] b) \\[3ex] i) \\[3ex] \angle BCD = 50^\circ ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] ii) \\[3ex] \angle ADC = 50^\circ ... alternate\;\;interior\;\;\angle s\;\;are\;\;equal \\[3ex] \angle PDC = \angle ADC = 50^\circ ... diagram \\[3ex] \angle PCD = \angle BCD = 50^\circ ... diagram \\[3ex] Because\;\; \angle PCD = 50^\circ \;\;and\;\; \angle PDC = 50^\circ \\[3ex] \triangle PCD \;\;is\;\;isosceles \\[3ex] Also: \\[3ex] \angle ABC = \angle ADC = 50^\circ ... alternate\;\;interior\;\;\angle s\;\;are\;\;equal \\[3ex] \angle ABP = \angle ABC = 50^\circ ... diagram \\[3ex] Because\;\; \angle ABP = 50^\circ \;\;and\;\; \angle BAP = 50^\circ \\[3ex] \triangle ABP \;\;is\;\;isosceles $

$ \angle MPQ = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] \angle QPN + \angle OMN = 180^\circ ... opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quad\;\;are\;\;supplementary \\[3ex] \angle QPN + 50 = 180 \\[3ex] \angle QPN = 180 - 50 \\[3ex] \angle QPN = 130^\circ \\[3ex] \angle MPN + \angle MPQ = \angle QPN ...diagram \\[3ex] \angle MPN + 90 = 130 \\[3ex] \angle MPN = 130 - 90 \\[3ex] \angle MPN = 40^\circ \\[3ex] \angle MPN = \angle NMP = 40^\circ ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle \\[3ex] (i) \\[3ex] \angle MNP + \angle MPN + \angle NMP = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle MNP \\[3ex] \angle MNP + 40 + 40 = 180 \\[3ex] \angle MNP + 80 = 180 \\[3ex] \angle MNP = 180 - 80 \\[3ex] \angle MNP = 100^\circ \\[5ex] \angle PQO + \angle MNP = 180^\circ ... opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quad\;\;are\;\;supplementary \\[3ex] \angle PQO + 100 = 180 \\[3ex] \angle PQO = 180 - 100 \\[3ex] \angle PQO = 80^\circ \\[3ex] \angle QPO = \angle PQO = 80^\circ ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle \\[3ex] (ii) \\[3ex] \angle POQ + \angle QPO + \angle PQO = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle POQ \\[3ex] \angle POQ + 80 + 80 = 180 \\[3ex] \angle POQ + 160 = 180 \\[3ex] \angle POQ = 180 - 160 \\[3ex] \angle POQ = 20^\circ $

$ (i) \\[3ex] \angle POQ = 2 * \angle PRQ...\angle\;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] 48 = 2 * \angle PRQ \\[3ex] 2 * \angle PRQ = 48 \\[3ex] \angle PRQ = \dfrac{48}{2} \\[5ex] \angle PRQ = 24^\circ \\[3ex] \angle OPR = \angle PRQ...alternate\;\;interior\;\;\angle s\;\;are\;\;equal \\[3ex] \therefore \angle OPR = 24^\circ \\[3ex] (ii) \\[3ex] radius = r = 5.4\;cm \\[3ex] Acute\;\angle POQ + Reflex\;\angle POQ = 360^\circ ...\angle s \;\;at\;\;a\;\;point \\[3ex] 48 + Reflex\;\angle POQ = 360 \\[3ex] Reflex\;\angle POQ = 360 - 48 \\[3ex] Reflex\;\angle POQ = 312^\circ \\[3ex] Length\;\;of\;\;major\;\;arc\;PQ = \dfrac{Reflex\;\angle POQ}{360} * 2\pi * r \\[5ex] = \dfrac{312}{360} * 2 * \dfrac{22}{7} * 5.4 \\[5ex] = \dfrac{74131.2}{2520} \\[5ex] = 29.41714286\;cm $

Construction: Join the chord from point S to point Q in order to use the isosceles triangle

$ \underline{\triangle POS} \\[3ex] \angle PSO = \angle SPO = 44^\circ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle POS \\[3ex] \underline{Cyclic\;\;Quadrilateral\;PQRS} \\[3ex] \angle P + \angle R = 180^\circ ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quadrilateral\;\;are\;\;supplementary \\[3ex] \angle P = \angle SPO = 44^\circ \\[3ex] \implies \\[3ex] 44 + \angle R = 180 \\[3ex] \angle R = 180 - 44 \\[3ex] \angle R = 136^\circ \\[3ex] \underline{\triangle SRQ} \\[3ex] \angle RSQ + \angle RQS + \angle SRQ = 180^\circ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle SRQ \\[3ex] \angle RSQ = \angle RQS = k ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle SRQ \\[3ex] \angle SRQ = \angle R = 136^\circ \\[3ex] \implies \\[3ex] k + k + 136 = 180 \\[3ex] 2k = 180 - 136 \\[3ex] 2k = 44 \\[3ex] k = \dfrac{44}{2} \\[5ex] k = 22 \\[3ex] \therefore \angle RSQ = \angle RQS = 22^\circ \\[3ex] \underline{\triangle SOQ} \\[3ex] \angle SOQ = \angle PSO + \angle SPO ...exterior\;\;\angle \;\;of\;\;a\;\;\triangle = sum\;\;of\;\;two\;\;interior\;\;opposite\;\;\angle s \\[3ex] \angle SOQ = 44 + 44 \\[3ex] \angle SOQ = 88^\circ \\[3ex] \angle OSQ + \angle OQS + \angle SOQ = 180^\circ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle SOQ \\[3ex] \angle OSQ = \angle OQS = m ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle SOQ \\[3ex] \implies \\[3ex] m + m + 88 = 180 \\[3ex] 2m = 180 - 88 \\[3ex] 2m = 92 \\[3ex] m = \dfrac{92}{2} \\[5ex] m = 46 \\[3ex] \therefore \angle OSQ = \angle OQS = 46^\circ \\[3ex] \angle OQR = \angle OQS + \angle RQS ...diagram \\[3ex] \angle OQR = 46 + 22 \\[3ex] \angle OQR = 68^\circ $

$ \underline{\triangle RPQ} \\[3ex] \angle RPQ = \angle RQP = k ... base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle RPQ \\[3ex] \angle RPQ + \angle RQP + \angle PRQ = 180^\circ ... sum\;\;of\;\;\angle s\;\;in\;\;\triangle RPQ \\[3ex] k + k + 80 = 180 \\[3ex] 2k = 180 - 80 \\[3ex] 2k = 100 \\[3ex] k = \dfrac{100}{2} \\[5ex] k = 50 \\[3ex] \therefore \angle RPQ = \angle RQP = 50^\circ \\[3ex] \angle SPQ = \angle RPQ = 50^\circ ... diagram \\[3ex] \angle STR = \angle SPQ = 50^\circ ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal $

$ \angle PSR + 117 = 180^\circ ... opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quad\;\;are\;\;supplementary \\[3ex] \angle PSR = 180 - 117 \\[3ex] \angle PSR = 63^\circ \\[3ex] \angle SPR = \angle PSR = 63^\circ ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle \\[3ex] \angle PRS + \angle SPR + \angle PSR = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle PRS \\[3ex] \angle PRS + 63 + 63 = 180 \\[3ex] \angle PRS + 126 = 180 \\[3ex] \angle PRS = 180 - 126 \\[3ex] \angle PRS = 54^\circ \\[3ex] \angle PST = \angle PRS = 54^\circ ... \angle\;\;between\;\;tangent\;TS\;\;and\;\;chord\;PS = \angle\;\;in\;\;alternate\;\;segment $

$ y = 2(130) ...\angle \:\:at\:\:center\:\:is\:\:twice\:\:\angle\:\:at\:\:circumference \\[3ex] y = 260^\circ \\[3ex] x + y = 360 ...\angle s \:\:at\:\:a\:\:point \\[3ex] x = 360 - y \\[3ex] x = 360 - 260 \\[3ex] x = 100^\circ $

$ \angle DCB = 58^\circ ...\angle\;\;between\;\;tangent\;ABE\;\;and\;\;chord\;BD = \angle\;\;in\;\;alternate\;\;segment \\[3ex] \angle OCB = \angle DCB = 58^\circ ...diagram \\[3ex] \angle OBC = \angle OCB = 58^\circ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle\;OCB \\[3ex] \angle OBA = 90^\circ ...radius\;OB\;\;\perp\;\;tangent\;ABE\;\;at\;\;point\;\;of\;\;contact\;B \\[3ex] But: \\[3ex] \angle OBA = \angle OBC + \angle CBA ...diagram \\[3ex] 90 = 58 + \angle CBA \\[3ex] \angle CBA + 58 = 90 \\[3ex] \angle CBA = 90 - 58 \\[3ex] \angle CBA = 32^\circ \\[3ex] \angle OCB = \angle CAB + \angle CBA ...exterior\;\;\angle\;\;of\;\;a\;\;\triangle = sum\;\;of\;\;two\;\;interior\;\;opposite\;\;\angle s \\[3ex] 58 = \angle CAB + 32 \\[3ex] \angle CAB + 32 = 58 \\[3ex] \angle CAB = 58 - 32 \\[3ex] \angle CAB = 26^\circ $

$ 2 * x = 7 * 5 ...Intersecting\;\;Chords\;\;Theorem \\[3ex] x = \dfrac{7 * 5}{2} \\[5ex] x = \dfrac{35}{2} \\[5ex] x = 17.5 $

$ \angle PRS = 20 + 35 ...exterior\;\;\angle\;\;of\;\;a\;\;\triangle = sum\;\;of\;\;two\;\;interior\;\;opposite\;\;\angle s \\[3ex] \angle PRS = 55^\circ \\[3ex] \angle POS = 2 * \angle PRS ...\angle\;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] \angle POS = 2 * 55 \\[3ex] \angle POS = 110^\circ \\[3ex] \angle PSO = \angle SPO = k ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle \\[3ex] \angle PSO + \angle SPO + \angle POS = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle POS \\[3ex] k + k + 110 = 180 \\[3ex] 2k = 180 - 110 \\[3ex] 2k = 70 \\[3ex] k = \dfrac{70}{2} \\[5ex] k = 35 \\[3ex] \therefore \angle PSO = 35^\circ $

$ 2m = n ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] m = \dfrac{n}{2} $

It is very necessary to draw the diagram

$ \angle KRQ = 31^\circ...\angle s\;\;in\;\;the\;\;same\;\;segment \\[3ex] \angle RKQ + 42 = 180^\circ ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] \angle RKQ = 180 - 42 \\[3ex] \angle RKQ = 138^\circ \\[3ex] \angle KQR + \angle RKQ + \angle KRQ = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle RKQ \\[3ex] \angle KQR + 31 + 138 = 180 \\[3ex] \angle KQR + 169 = 180 \\[3ex] \angle KQR = 180 - 169 \\[3ex] \angle KQR = 11^\circ $

$ \angle ECD = \angle ABE = 40^\circ ...alternate\;\;\angle s\;\;are\;\;equal \\[3ex] \angle OCD = \angle ECD = 40^\circ ...diagram \\[3ex] \angle ODC = \angle OCD = 40^\circ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle \\[3ex] \angle CDE = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] \angle CDE = \angle ODC + \angle ODE ...diagram \\[3ex] 90 = 40 + \angle ODE \\[3ex] \angle ODE + 40 = 90 \\[3ex] \angle ODE = 90 - 40 \\[3ex] \angle ODE = 50^\circ $

$ \angle PRS = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] a + \angle PRS + \angle PSR = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle PRS \\[3ex] a + 90 + 34 = 180 \\[3ex] a + 124 = 180 \\[3ex] a = 180 - 124 \\[3ex] a = 56^\circ \\[5ex] \angle QOP = \angle RSP = 34^\circ.. corresponding\;\;\angle s\;\;are\;\;equal \\[3ex] b + \angle QOP = 180^\circ ... \angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] b + 34 = 180 \\[3ex] b = 180 - 34 \\[3ex] b = 146^\circ \\[3ex] c + 34 = 180^\circ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quad\;\;are\;\;supplementary \\[3ex] c = 180 - 34 \\[3ex] c = 146^\circ \\[3ex] d = a = 56^\circ ...\angle\;\;between\;\;tangent\;RT\;\;and\;\;chord\;RS = \angle\;\;in\;\;alternate\;\;segment $

$ \angle OUT = \angle OTU = p ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle \\[3ex] \angle OUT + \angle OTU + \angle UOT = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;a\;\;\triangle \\[3ex] p + p + 70 = 180 \\[3ex] 2p = 180 - 70 \\[3ex] 2p = 110 \\[3ex] p = \dfrac{110}{2} \\[5ex] p = 55^\circ \\[3ex] \angle RUT + \angle RST = 180^\circ \\[3ex] ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quad\;\;are\;\;supplementary \\[3ex] \angle RUT + 100 = 180 \\[3ex] \angle RUT = 180 - 100 \\[3ex] \angle RUT = 80^\circ \\[3ex] \angle RUT = \angle RUO + \angle OUT ...diagram \\[3ex] 80 = \angle RUO + 55 \\[3ex] \angle RUO + 55 = 80 \\[3ex] \angle RUO = 80 - 55 \\[3ex] \angle RUO = 25^\circ $

$ \angle POQ = 2 * \angle PRQ ...\angle\;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] 62 = 2 * \angle PRQ \\[3ex] 2 * \angle PRQ = 62 \\[3ex] \angle PRQ = \dfrac{62}{2} \\[5ex] \angle PRQ = 31^\circ \\[3ex] \angle PQR + \angle RPQ + \angle PRQ = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle PRQ \\[3ex] \angle PQR + 73 + 31 = 180 \\[3ex] \angle PQR + 104 = 180 \\[3ex] \angle PQR = 180 - 104 \\[3ex] \angle PQR = 76^\circ \\[3ex] \angle PQO = \angle QOP = k ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle OPQ \\[3ex] \angle PQO + \angle QOP + \angle POQ = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle OPQ \\[3ex] k + k + 62 = 180 \\[3ex] 2k = 180 - 62 \\[3ex] 2k = 118 \\[3ex] k = \dfrac{118}{2} \\[5ex] k = 59 \\[3ex] \therefore \angle PQO = 59^\circ \\[3ex] \angle PQR = \angle PQO + \angle RQO ...diagram \\[3ex] 76 = 59 + \angle RQO \\[3ex] 59 + \angle RQO = 76 \\[3ex] \angle RQO = 76 - 59 \\[3ex] \angle RQO = 17^\circ $

$ \underline{First\;\;Approach} \\[3ex] \angle SRP = \angle SPR = 40^\circ ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle \\[3ex] \angle PSR + \angle SPR + \angle SRP = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle PSR \\[3ex] \angle PSR + 40 + 40 = 180 \\[3ex] \angle PSR + 80 = 180 \\[3ex] \angle PSR = 180 - 80 \\[3ex] \angle PSR = 100^\circ \\[3ex] \angle QSR = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] \angle PSR = \angle PSQ + \angle QSR ...diagram \\[3ex] 100 = \angle PSQ + 90 \\[3ex] \angle PSQ + 90 = 100 \\[3ex] \angle PSQ = 100 - 90 \\[3ex] \angle PSQ = 10^\circ \\[5ex] \underline{Second\;\;Approach} \\[3ex] \angle SPQ = \angle SPR = 40^\circ ... diagram \\[3ex] \angle SRP = \angle SPR = 40^\circ ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle \\[3ex] \angle SPQ = \angle SPR = 40^\circ \\[3ex] \angle QSR = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] \angle PQS = \angle QSR + \angle QRS \\[3ex] ...exterior\;\;\angle\;\;of\;\;a\;\;\triangle = sum\;\;of\;\;two\;\;interior\;\;opposite\;\;\angle s \\[3ex] \angle PQS = 90 + 40 \\[3ex] \angle PQS = 130^\circ \\[3ex] \angle PSQ + \angle PQS + \angle SPQ = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle PSR \\[3ex] \angle PSQ + 130 + 40 = 180 \\[3ex] \angle PSQ + 170 = 180 \\[3ex] \angle PSQ = 180 - 170 \\[3ex] \angle PSQ = 10^\circ $

We can solve this question using at least two approaches.
Choose any approach you prefer.

$ \underline{First\;\;Approach} \\[3ex] \angle ROS = 2 * \angle RPS ...\angle\;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] 110 = 2 * \angle RPS \\[3ex] 2 * \angle RPS = 110 \\[3ex] \angle RPS = \dfrac{110}{2} \\[5ex] \angle RPS = 55^\circ \\[3ex] \angle RPQ = \angle RPS = 55^\circ ...diagram \\[3ex] \angle PQR + \angle QRP + \angle RPQ = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle PQR \\[3ex] x + 90 + 55 = 180 \\[3ex] x + 145 = 180 \\[3ex] x = 180 - 145 \\[3ex] x = 35^\circ \\[5ex] \underline{Second\;\;Approach} \\[3ex] \angle ROS + \angle POS = 180^\circ ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] 110 + \angle POS = 180 \\[3ex] \angle POS = 180 - 110 \\[3ex] \angle POS = 70^\circ \\[3ex] \angle OPS = \angle OSP = k ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle OPS \\[3ex] \angle OPS + \angle OSP + \angle POS = 180^\circ ..sum\;\;of\;\;\angle s\;\;in\;\;\triangle OPS \\[3ex] k + k + 70 = 180 \\[3ex] 2k = 180 - 70 \\[3ex] 2k = 110 \\[3ex] k = \dfrac{110}{2} \\[5ex] k = 55 \\[3ex] \therefore \angle OPS = 55^\circ \\[3ex] \angle RPQ = \angle OPS = 55^\circ ...diagram \\[3ex] \angle PQR + \angle QRP + \angle RPQ = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle PQR \\[3ex] x + 90 + 55 = 180 \\[3ex] x + 145 = 180 \\[3ex] x = 180 - 145 \\[3ex] x = 35^\circ $

$ \angle PQR = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] \angle QRP + \angle QPR + \angle PQR = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;a\;\;\triangle \\[3ex] \angle QRP + x + 90 = 180 \\[3ex] \angle QRP = 180 - x - 90 \\[3ex] \angle QRP = 90 - x \\[3ex] \angle QRP = (90 - x)^\circ $

$ Sum\;\;of\;\;interior\;\;\angle s\;\;of\;\;regular\;\;Pentagon\;\;ABCDE \\[3ex] = 180(5 - 2) \\[3ex] = 180(3) \\[3ex] = 540^\circ \\[3ex] Each\;\;\angle \\[3ex] = \dfrac{540}{5} \\[5ex] = 108^\circ \\[3ex] \implies \\[3ex] \angle A = 108^\circ \\[3ex] \underline{Second\;\;Diagram} \\[3ex] Reflex\;\;\angle EOB = 2 * \angle A ... \angle\;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] Reflex\;\;\angle EOB = 2(108) = 216^\circ \\[3ex] Acute\;\;\angle EOB + Reflex\;\;\angle EOB = 360^\circ ...\angle s\;\;at\;\;a\;\;point \\[3ex] Acute\;\;\angle EOB + 216 = 360^\circ \\[3ex] Acute\;\;\angle EOB = 360 - 216 = 144 \\[3ex] \angle OFE = \angle OEF = \angle OFB = \angle OBF = p ...base\;\;\angle s\;\;of\;\;isosceles\;\; \triangle s\;\;OEF\;\;and\;\;OBF \\[3ex] \angle EFB = \angle OFE + \angle OFB ...diagram \\[3ex] Acute\;\;\angle EOB = 2 * \angle EFB ... \angle\;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] \implies \\[3ex] 144 = 2 * (p + p) \\[3ex] 144 = 2(2p) \\[3ex] 144 = 4p \\[3ex] 4p = 144 \\[3ex] p = \dfrac{144}{4} \\[5ex] p = 36 \\[3ex] \therefore \angle OFB = 36^\circ $

Construction: Draw the radius from center: M to circumference: C
Because MC and MA are radii (same length):
$\angle MCA = \angle MAC$...base angles of isosceles $\triangle MCA$

We can solve this question in at least two ways.
Use any approach you prefer

$ \underline{First\;\;Approach} \\[3ex] \angle TOR = 2 * \angle TQR ...\angle \;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] 82 = 2 * \angle TQR \\[3ex] 2 * \angle TQR = 82 \\[3ex] \angle TQR = \dfrac{82}{2} \\[5ex] \angle TQR = 41^\circ \\[3ex] \angle TQS = \angle TQR = 41^\circ ...diagram \\[3ex] \angle OTS = 90^\circ...radius\;OT \perp \;\;tangent\;PTS\;\;at\;\;point\;\;of\;\;contact\;T \\[3ex] \angle QTS = \angle OTS = 90^\circ ... diagram \\[3ex] \angle QST + \angle QTS + \angle TQS = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle QTS \\[3ex] \angle QST + 90 + 41 = 180 \\[3ex] \angle QST + 131 = 180 \\[3ex] \angle QST = 180 - 131 \\[3ex] \angle QST = 49^\circ \\[3ex] \angle RST = \angle QST = 49^\circ \\[5ex] \underline{Second\;\;Approach} \\[3ex] \angle OTS = 90^\circ...radius\;OT \perp \;\;tangent\;PTS\;\;at\;\;point\;\;of\;\;contact\;T \\[3ex] \angle OTR = \angle ORT = p ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle \\[3ex] \angle OTR + \angle ORT + \angle TOR = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle TOR \\[3ex] p + p + 82 = 180 \\[3ex] 2p = 180 - 82 \\[3ex] 2p = 98 \\[3ex] p = \dfrac{98}{2} \\[5ex] p = 49 \\[3ex] \therefore \angle OTR = 49^\circ \\[3ex] \angle OTS = \angle OTR + \angle RTS ...diagram \\[3ex] 90 = 49 + \angle RTS \\[3ex] 49 + \angle RTS = 90 \\[3ex] \angle RTS = 90 - 49 \\[3ex] \angle RTS = 41^\circ \\[3ex] \angle QRT = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] \angle QRT + \angle TRS = 180^\circ ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] 90 + \angle TRS = 180 \\[3ex] \angle TRS = 180 - 90 \\[3ex] \angle TRS = 90^\circ \\[3ex] \angle RTS + \angle TRS + \angle RST = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle TRS \\[3ex] 41 + 90 + \angle RST = 180 \\[3ex] 131 + \angle RST = 180 \\[3ex] \angle RST = 180 - 131 \\[3ex] \angle RST = 49^\circ $

Construction: Join the chord from point B to point D

$ \angle CBD = \angle CDB = k ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle \\[3ex] \angle CBD + \angle CDB + \angle BCD = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle BCD \\[3ex] k + k + 108 = 180 \\[3ex] 2k = 180 - 108 \\[3ex] 2k = 72 \\[3ex] k = \dfrac{72}{2} \\[5ex] k = 36 \\[3ex] \therefore \angle CDB = 36^\circ \\[3ex] \angle BDE = 90^\circ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] \angle CDE = \angle CDB + \angle BDE ...diagram \\[3ex] \angle CDE = 36 + 90 \\[3ex] \angle CDE = 126^\circ $

$ (i) \\[3ex] Obtuse\;\angle AEC = 2 * m = 2m...\angle\;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] Reflex\;\angle AEC + Obtuse\;\angle AEC = 360^\circ ...\angle s\;\;at\;\;a\;\;point \\[3ex] Reflex\;\angle AEC + 2m = 360 \\[3ex] Reflex\;\angle AEC = 360 - 2m \\[5ex] Reflex\;\angle AEC = 2 * \angle ADC ...\angle\;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] \implies \\[3ex] 360 - 2m = 2 * \angle ADC \\[3ex] 2 * \angle ADC = 360 - 2m \\[3ex] \angle ADC = \dfrac{360 - 2m}{2} \\[5ex] \angle ADC = \dfrac{2(180 - m)}{2} \\[5ex] \angle ADC = 180 - m \\[5ex] Obtuse\;\angle AEC = \angle ADC ... opposite\;\;\angle s\;\;of\;\;Rhombus\;EADC\;\;are\;\;equal \\[3ex] \implies \\[3ex] 2m = 180 - m \\[3ex] 2m + m = 180 \\[3ex] 3m = 180 \\[3ex] m = \dfrac{180}{3} \\[3ex] m = 60^\circ \\[3ex] (ii) \\[3ex] \angle EAC = \angle ECA = k ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle EAC \\[3ex] \angle EAC + \angle ECA + \angle AEC = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle EAC \\[3ex] k + k + 2m = 180 \\[3ex] 2k = 180 - 2m \\[3ex] k = \dfrac{180 - 2m}{2} \\[5ex] k = \dfrac{2(90 - m)}{2} \\[5ex] k = 90 - m \\[3ex] \therefore \angle EAC = \angle ECA = 90 - m \\[3ex] \underline{\triangle ABC} \\[3ex] \angle EAC + n + \angle ECA + n + m = 180^\circ ... sum\;\;of\;\;\angle s\;\;in\;\;a\;\;\triangle \\[3ex] (90 - m) + n + (90 - m) + n + m = 180 \\[3ex] 90 - m + 90 - m + m + 2n = 180 \\[3ex] 2n - m + 180 = 180 \\[3ex] 2n - m = 180 - 180 \\[3ex] 2n - m = 0 \\[3ex] 2n = m \\[3ex] n = \dfrac{m}{2} \\[5ex] n = \dfrac{60}{2} \\[5ex] n = 30^\circ $

It is necessary to draw the diagram

$ \angle QRS + \angle QPS = 180^\circ \\[3ex] ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quad\;\;are\;\;supplementary \\[3ex] \angle QRS + 75 = 180 \\[3ex] \angle QRS = 180 - 75 \\[3ex] \angle QRS = 105^\circ $

$ \angle XUV = 50^\circ... \angle\;\;between\;\;tangent\;YXW\;\;and\;\;chord\;XV = \angle\;\;in\;\;alternate\;\;segment \\[3ex] \angle UXV = \angle XUV = 50^\circ ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle \\[3ex] \angle UXY + \angle UXV + 50 = 180^\circ ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] \angle UXY + 50 + 50 = 180 \\[3ex] \angle UXY + 100 = 180 \\[3ex] \angle UXY = 180 - 100 \\[3ex] \angle UXY = 80^\circ $

$ Reflex\;\;\angle XOZ = 2 * m \\[3ex] ...\angle\;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] Reflex\;\;\angle XOZ = 2m \\[3ex] Reflex\;\;\angle XOZ + Obtuse\;\;\angle XOZ = 360^\circ...\angle s\;\;at\;\;a\;\;point \\[3ex] 2m + 10m = 360 \\[3ex] 12m = 360 \\[3ex] m = \dfrac{360}{12} \\[5ex] m = 30 $

$ (a.) \\[3ex] m + 120 = 180^\circ ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quad\;\;are\;\;supplementary \\[3ex] m = 180 - 120 \\[3ex] m = 60^\circ \\[5ex] (b.) \\[3ex] x = \angle YWZ ...\angle\;\;between\;\;tangent\;PQ\;\;and\;\;chord\;YZ = \angle\;\;in\;\;alternate\;\;segment \\[3ex] y = x ...alternate\;\;\angle s\;\;are\;\;equal \\[3ex] \angle YWZ + \angle WYZ + \angle WZY = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle WYZ \\[3ex] x + y + 120 = 180 \\[3ex] x + x = 180 - 120 \\[3ex] 2x = 60 \\[3ex] x = \dfrac{60}{2} \\[5ex] x = 30^\circ $

$ RT * TS = PT * TQ ...Intersecting\;\;Chords\;\;Theorem $

$ (i.) \\[3ex] \angle RSP = 90^\circ ...\angle \;\;in\;\;a\;\;semicircle \\[3ex] \angle RSP = \angle RSQ + \angle QSP ...diagram \\[3ex] \implies \\[3ex] 90 = \angle QSR + 54 \\[3ex] \angle QSR + 54 = 90 \\[3ex] \angle QSR = 90 - 54 \\[3ex] \angle QSR = 36^\circ \\[3ex] \angle QPR = \angle QSR = 36^\circ ... \angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] \implies \\[3ex] 2x = 36 \\[3ex] x = \dfrac{36}{2} \\[5ex] x = 18^\circ \\[3ex] (ii.) \\[3ex] \angle SRP + \angle RSP + \angle RPS = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle SRP \\[3ex] y + 90 + x = 180 \\[3ex] y + 90 + 18 = 180 \\[3ex] y + 108 = 180 \\[3ex] y = 180 - 108 \\[3ex] y = 72^\circ \\[3ex] (iii.) \\[3ex] \angle SQP = \angle SRP ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] \implies \\[3ex] z = y \\[3ex] z = 72^\circ $

$ \angle QRS + \angle QPS = 180^\circ \\[3ex] ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quad\;\;are\;\;supplementary \\[3ex] \angle QRS + 100 = 180 \\[3ex] \angle QRS = 180 - 100 \\[3ex] \angle QRS = 80^\circ \\[3ex] \angle SQR + \angle QSR + \angle QRS = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle QRS \\[3ex] \angle SQR + 80 + 80 = 180 \\[3ex] \angle SQR + 160 = 180 \\[3ex] \angle SQR = 180 - 160 \\[3ex] \angle SQR = 20^\circ $

Construction: Join radius $\overline{KZ}$ from center K to point Z


$ m\overset{\huge\frown}{LZ} = \angle PKZ = 56^\circ \\[3ex] \angle KZP = 90^\circ ...radius\;KZ \perp tangent\;PZ\;\;at\;\;point\;\;of\;\;contact\;Z \\[3ex] \underline{\triangle KZP} \\[3ex] \angle KPZ + \angle PKZ + \angle KZP = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;a\;\;\triangle \\[3ex] \angle KPZ + 56 + 90 = 180 \\[3ex] \angle KPZ + 146 = 180 \\[3ex] \angle KPZ = 180 - 146 \\[3ex] \angle KPZ = 34 \\[3ex] \angle P = \angle KPZ = 34^\circ ... diagram $

$ 4x + 2x = 180^\circ \\[3ex] ...interior\:\:opposite\;\;\angle s\:\:of\:\:a\:\:cyclic\:\:quad\;\;are\;\;supplementary \\[3ex] 6x = 180 \\[3ex] x = \dfrac{180}{6} \\[5ex] x = 30^\circ $

$ PA = PB ...two\;\;tangents:\;\;PA\;\;and\;\;PB\;\;drawn\;\;from\;\;the\;\;same\;\;external\;\;point:\;P\;\;are\;\;equal\;\;in\;\;length \\[3ex] \implies \\[3ex] \angle PBA = \angle PAB = 71^\circ ...base \angle s\;\;of\;\;isosceles\;\triangle PAB \\[3ex] \angle APB + \angle PBA + \angle PAB = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle PAB \\[3ex] \angle APB + 71 + 71 = 180 \\[3ex] \angle APB + 142 = 180 \\[3ex] \angle APB = 180 - 142 \\[3ex] \angle APB = 38^\circ $

$ \angle LMN = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] \angle LNM + \angle NLM + \angle LMN = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle LMN \\[3ex] (x + 15) + (x - 5) + 90 = 180 \\[3ex] x + 15 + x - 5 + 90 = 180 \\[3ex] 2x + 100 = 180 \\[3ex] 2x = 180 - 100 \\[3ex] 2x = 80 \\[3ex] x = \dfrac{80}{2} \\[5ex] x = 40^\circ $

Let us draw the circle
O is the center of the circle


$ \underline{Central\;\;\angle = Intercepted\;\; \overset{\huge\frown}{}} \\[3ex] \overset{\huge\frown}{AC} = \angle AOC = 72^\circ \\[3ex] \overset{\huge\frown}{BD} = \angle BOD = 34^\circ \\[3ex] \underline{\angle\;\;of\;\;Intersecting\;\;Secants\;(Outside\;\;the\;\;Circle)} \\[3ex] \angle P = \dfrac{72 - 34}{2} \\[5ex] \angle P = \dfrac{38}{2} \\[5ex] \angle P = 19^\circ $

$ 2y + 3x = 180^\circ \\[3ex] ...interior\:\:opposite\;\;\angle s\:\:of\:\:a\:\:cyclic\:\:quad\;\;are\;\;supplementary \\[3ex] 2y = 180 - 3x \\[3ex] 2y = 3(60 - x) \\[3ex] y = \dfrac{3(60 - x)}{2} $

$ \underline{1st\;\;Case} \\[3ex] \angle X + 204 = 360^\circ ...\angle s \;\;at\;\;a\;\;point \\[3ex] \angle X = 360 - 204 \\[3ex] \angle X = 156^\circ \\[3ex] \underline{2nd\;\;Case} \\[3ex] \angle X = 2(78^\circ) ... \angle \;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] \angle X = 156^\circ \\[3ex] Because\;\;\angle X\;\;in\;\;2nd\;\;Case = \angle X\;\;in\;\;1st\;\;Case: \\[3ex] \angle X \;\;is\;\;the\;\;centre\;\;of\;\;the\;\;circle $

$ \angle SQR = \angle QSR = 37^\circ ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle RSQ \\[3ex] \angle SRQ + \angle SQR + \angle QSR = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle RSQ \\[3ex] \angle SRQ + 37 + 37 = 180 \\[3ex] \angle SRQ + 74 = 180 \\[3ex] \angle SRQ = 180 - 74 \\[3ex] \angle SRQ = 106^\circ \\[3ex] x = 106^\circ \\[3ex] ... exterior\;\;\angle\;\;of\;\;a\;\;cyclic\;\;quad = interior\;\;opposite\;\;\angle $

Construction: Draw the chord from point S to point P

$ \angle RSP = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] \angle PST = 24^\circ ...\angle\;\;between\;\;tangent\;ST\;\;and\;\;chord\;SP = \angle\;\;in\;\;alternate\;\;segment \\[3ex] \angle RST = \angle RSP + \angle PST ... diagram \\[3ex] \angle RST = 90 + 24 \\[3ex] \angle RST = 114^\circ \\[3ex] \angle TRS = \angle PRS = 24^\circ ... diagram \\[3ex] \angle STR + \angle RST + \angle TRS = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle RST \\[3ex] \angle STR + 114 + 24 = 180 \\[3ex] \angle STR + 138 = 180 \\[3ex] \angle STR = 180 - 138 \\[3ex] \angle STR = 42^\circ $

$ (220.1.1) \\[3ex] \hat{R_1} + \hat{R_2} = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] 69 + \hat{R_2} = 90 \\[3ex] \hat{R_2} = 90 - 69 \\[3ex] \hat{R_2} = 21^\circ \\[3ex] (220.1.2) \\[3ex] \hat{O_1} = 2 * \hat{R_1} ... \angle \;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] \hat{O_1} = 2 * 69 \\[3ex] \hat{O_1} = 138^\circ \\[3ex] OR \\[3ex] \hat{M_1} = \hat{R_2} = 21^\circ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] \hat{N_1} = \hat{M_1} = 21^\circ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle OMN \\[3ex] \hat{O_1} + \hat{M_1} + \hat{N_1} = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle OMN \\[3ex] \hat{O_1} + 21 + 21 = 180 \\[3ex] \hat{O_1} + 42 = 180 \\[3ex] \hat{O_1} = 180 - 42 \\[3ex] \hat{O_1} = 138^\circ \\[3ex] (220.1.3) \\[3ex] \hat{M_1} = \hat{R_2} = 21^\circ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] (220.1.4) \\[3ex] \hat{O_2} + \hat{O_1} = 180^\circ ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] \hat{O_2} + 138 = 180 \\[3ex] \hat{O_2} = 180 - 138 \\[3ex] \hat{O_2} = 42^\circ \\[3ex] \angle MPR = \hat{O_2} = 42^\circ ...alternate\;\;\angle s\;\;are\;\;equal \\[3ex] \angle PMR + \angle MRP + \angle MPR = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle MRP \\[3ex] \hat{M_2} + (\hat{R_1} + \hat{R_2}) + 42 = 180 \\[3ex] \hat{M_2} + 90 + 42 = 180 \\[3ex] \hat{M_2} + 132 = 180 \\[3ex] \hat{M_2} = 180 - 132 \\[3ex] \hat{M_2} = 48^\circ $

$ Major\;\;arc = \overset{\huge\frown}{RS} = 121^\circ \\[3ex] Minor\;\;arc = \overset{\huge\frown}{WT} = ? \\[3ex] \underline{\angle\;\;of\;\;Intersecting\;\;Secants\;(Outside\;\;the\;\;Circle)} \\[3ex] \angle RPS = \dfrac{Major\;\;arc - Minor\;\;arc}{2} \\[5ex] 35 = \dfrac{121 - \overset{\huge\frown}{WT}}{2} \\[5ex] 35(2) = 121 - \overset{\huge\frown}{WT} \\[3ex] 70 = 121 - \overset{\huge\frown}{WT} \\[3ex] \overset{\huge\frown}{WT} = 121 - 70 \\[3ex] \overset{\huge\frown}{WT} = 51 \\[3ex] m\overset{\huge\frown}{WT} = 51^\circ $

(a) There are at least three ways to show congruency
Use any approach you prefer

$ \underline{First\;\;Approach:\;\;HL} \\[3ex] \angle ABC = \angle PQR = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] \therefore \triangle ABC \;\;and\;\;\triangle PQR \;\;are\;\;right\;\;\triangle s \\[3ex] \implies \\[3ex] |AC| = |PR| = diameter = hypotenuse \;(H) \\[3ex] |AB| = |PQ| ...diagram = leg\;(L) \\[3ex] \implies \\[3ex] HL\;\;Congruency \\[5ex] \underline{Second\;\;Approach:\;\;SSS} \\[3ex] |AC| = |PR| = diameter = hypotenuse = side \;(S) \\[3ex] |AB| = |PQ| ...diagram = side\;(S) \\[3ex] |BC|^2 = |AC|^2 - |AB|^2 ...Pythagorean\;\;Theorem \\[3ex] Similarly \\[3ex] |QR|^2 = |PR|^2 - |PQ|^2 ...Pythagorean\;\;Theorem \\[3ex] \implies \\[3ex] |BC|^2 = |QR|^2 \\[3ex] |BC| = |QR| ...side\;(S) \\[3ex] \implies \\[3ex] SSS\;\;Congruency \\[5ex] \underline{Third\;\;Approach:\;\;SAS} \\[3ex] |AB| = |PQ| ...diagram = side\;(S) \\[3ex] \angle ABC = \angle PQR = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] |BC|^2 = |AC|^2 - |AB|^2 ...Pythagorean\;\;Theorem \\[3ex] Similarly \\[3ex] |QR|^2 = |PR|^2 - |PQ|^2 ...Pythagorean\;\;Theorem \\[3ex] \implies \\[3ex] |BC|^2 = |QR|^2 \\[3ex] |BC| = |QR| ...side\;(S) \\[3ex] \implies \\[3ex] SAS\;\;Congruency \\[3ex] $ (b) William is correct.
A kite is a quadrilateral.
Any circle can be drawn around a quadrilateral (the four vertices of a quadrilateral).
That circle would be a cyclic quadrilateral where the opposite angles are right angles.
The sum of those opposite angles are supplementary...opposite angles of a cyclic quadrilateral are supplementary (sum of 180°)

We have a chord and a tangent, so we may need the Alternate Segment Theorem
Let us do some construction
Construction:
(1.) Draw a chord from Point B to Point D ...so we can use the Alternate Segment Theorem
(2.) Draw a chord from Point C to Point D ...we may need to use another Theorem


$ \angle BED = 73^\circ ...alternate\;\;\angle s\;\;are\;\;equal \\[3ex] \angle BDE = 73^\circ ...\angle\;\;between\;\;tangent\;BA\;\;and\;\;chord\;RL = \angle\;\;in\;\;alternate\;\;segment \\[3ex] \angle DBE + \angle BDE + \angle BED = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle BED \\[3ex] \angle DBE + 73 + 73 = 180 \\[3ex] \angle DBE + 146 = 180 \\[3ex] \angle DBE = 34^\circ \\[3ex] \angle DCE = \angle DBE = 34^\circ ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal $

$ \angle ROT = 2 * \angle RST \\[3ex] ...\angle\;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] \angle ROT = 2 * 46 \\[3ex] \angle ROT = 92^\circ \\[3ex] \angle ORT = \angle OTR = p ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle OTR \\[3ex] \angle ORT + \angle OTR + \angle ROT = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle ORT \\[3ex] p + p + 92 = 180 \\[3ex] 2p = 180 - 92 \\[3ex] 2p = 88 \\[3ex] p = \dfrac{88}{2} \\[5ex] p = 44^\circ \\[3ex] \therefore \angle OTR = 44^\circ \\[3ex] \angle OTR = \angle STR + \angle STO ...diagram \\[3ex] 44 = 29 + \angle STO \\[3ex] 29 + \angle STO = 44 \\[3ex] \angle STO = 44 - 29 \\[3ex] \angle STO = 15^\circ $

(225.1)
Two tangents drawn to a circle from the same point outside the circle are equal in length.

$ (225.2.1) \\[3ex] E\hat{M}O = 90^\circ ...radius\;OM \perp tangent\;EM\;\;at\;\;point\;\;of\;\;contact\;M \\[3ex] (225.2.2) \\[3ex] (a) \\[3ex] |EV| = |EM| ...two\;\;tangents:\;\;EV\;\;and\;\;EM\;\;drawn\;\;from\;\;the\;\;same\;\;external\;\;point:\;E\;\;are\;\;equal\;\;in\;\;length \\[3ex] \hat{V_1} = \hat{M_1} = p ...base\;\;\angle s\;\;of\;\;isosceles\;\; \triangle EMV \\[3ex] p + p + 52 = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle EMV \\[3ex] 2p = 180 - 52 \\[3ex] 2p = 128 \\[3ex] p = \dfrac{128}{2} \\[5ex] p = 64 \\[3ex] \therefore \hat{V_1} = \hat{M_1} = 64^\circ \\[3ex] (b) \\[3ex] \hat{M_1} + \hat{M_2} = E\hat{M}O = 90^\circ ... diagram \\[3ex] 64 + \hat{M_2} = 90 \\[3ex] \hat{M_2} = 90 - 64 \\[3ex] \hat{M_2} = 26^\circ \\[3ex] \angle SMV = \hat{M_2} = 26^\circ ... diagram \\[3ex] \angle MVS = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] \angle MSV + \angle SMV + \angle MVS = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle MSV \\[3ex] \angle MSV + 26 + 90 = 180 \\[3ex] \angle MSV + 116 = 180 \\[3ex] \angle MSV = 180 - 116 \\[3ex] \angle MSV = 64 \\[3ex] \hat{S_2} = \angle MSV = 64^\circ ... diagram \\[3ex] (c) \\[3ex] \hat{V_3} = \hat{S_2} = 64^\circ ...base\;\;\angle s\;\;of\;\;isosceles\;\; \triangle OVS \\[3ex] \hat{O_1} + \hat{V_3} + \hat{S_2} = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle OVS \\[3ex] \hat{O_1} + 64 + 64 = 180 \\[3ex] \hat{O_1} + 128 = 180 \\[3ex] \hat{O_1} = 180 - 128 \\[3ex] \hat{O_1} = 52^\circ \\[3ex] (d) \\[3ex] \angle MVT = \hat{E} + \hat{M_1} ...exterior\;\;\angle\;\;of\;\;a\;\;\triangle = sum\;\;of\;\;the\;\;two\;\;interior\;\;opposite\;\;\angle s \\[3ex] \angle MVT = 52 + 64 \\[3ex] \angle MVT = 116 \\[3ex] \angle MVT = \angle MVS + \hat{V_4} ... diagram \\[3ex] \implies \\[3ex] \angle MVS + \hat{V_4} = 116 \\[3ex] 90 + \hat{V_4} = 116 \\[3ex] \hat{V_4} = 116 - 90 \\[3ex] \hat{V_4} = 26^\circ \\[3ex] (e) \\[3ex] \hat{S_2} = \hat{V_4} + \hat{T} ...exterior\;\;\angle\;\;of\;\;a\;\;\triangle = sum\;\;of\;\;the\;\;two\;\;interior\;\;opposite\;\;\angle s \\[3ex] 64 = 26 + \hat{T} \\[3ex] \hat{T} + 26 = 64 \\[3ex] \hat{T} = 64 - 26 \\[3ex] \hat{T} = 38^\circ $

(a) Line AB is a tangent because it touches the circle (circumference of the circle) at only one point.

(b)

$|OP|$ = r = radius

$ (c) \\[3ex] x + 130 = 360^\circ ...\angle s\;\;at\;\;a\;\;point \\[3ex] x = 360 - 130 \\[3ex] x = 230^\circ $

$ \angle CBF = \angle BDF = 40^\circ ...\angle\;\;between\;\;tangent\;ABC\;\;and\;\;chord\;BF = \angle\;\;in\;\;alternate\;\;segment \\[3ex] \angle DBF + \angle DEF = 180^\circ ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quadrilateral\;\;are\;\;supplementary \\[3ex] \angle DBF + 100 = 180 \\[3ex] \angle DBF = 180 - 100 \\[3ex] \angle DBF = 80^\circ \\[3ex] \angle ABD + \angle DBF + \angle CBF = 180^\circ ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] \angle ABD + 80 + 40 = 180 \\[3ex] \angle ABD + 120 = 180 \\[3ex] \angle ABD = 180 - 120 \\[3ex] \angle ABD = 60^\circ $

$ \angle ZXW = \angle ZYW = 33^\circ \\[3ex] ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] \angle WZX = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] \angle ZWX + \angle WZX + \angle ZXW = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle WZX \\[3ex] \angle ZWX + 90 + 33 = 180 \\[3ex] \angle ZWX + 123 = 180 \\[3ex] \angle ZWX = 180 - 123 \\[3ex] \angle ZWX = 57^\circ $

$ (229.1) \\[3ex] ST = TR ...two\;\;tangents:\;\;ST\;\;and\;\;TR\;\;drawn\;\;from\;\;the\;\;same\;\;external\;\;point:\;T\;\;are\;\;equal\;\;in\;\;length \\[3ex] (229.2) \\[3ex] (229.2.1) \\[3ex] \hat{S_2} = \hat{R_2} = k ...base\;\;\angle s\;\;of\;\;isosceles\;\; \triangle TRS \\[3ex] \hat{S_2} + \hat{R_2} + \hat{T} = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle TRS \\[3ex] k + k + 78 = 180 \\[3ex] 2k = 180 - 78 \\[3ex] 2k = 102 \\[3ex] k = \dfrac{102}{2} \\[5ex] k = 51 \\[3ex] \therefore \hat{S_2} = \hat{R_2} = 51^\circ \\[3ex] (229.2.2) \\[3ex] \hat{S_2} + \hat{S_3} = \hat{Q} ... exterior\;\; \angle \:\:of\:\:a\:\:cyclic\:\:quadrilateral = interior\;\;opposite\;\;\angle \\[3ex] 51 + \hat{S_3} = 93 \\[3ex] \hat{S_3} = 93 - 51 \\[3ex] \hat{S_3} = 42^\circ $

$ (a)\;\;(i) \\[3ex] A\hat{B}C = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] (ii) \\[3ex] \underline{\triangle ABC} \\[3ex] SOHCAHTOA \\[3ex] \tan x = \dfrac{opp}{adj} \\[5ex] \tan x = \dfrac{7.5}{4.7} \\[5ex] \tan x = 1.595744681 \\[3ex] x = \tan^{-1} (1.595744681) \\[3ex] x = 57.92599912 \\[3ex] x \approx 57.9^\circ \\[3ex] (b) \\[3ex] y = x = 57.92599912 ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] y \approx 57.9^\circ $

Construction: Draw a chord from Point C to Point A


$ \angle BAC = x^\circ ...\angle\;\;between\;\;tangent\;PCQ\;\;and\;\;chord\;RL = \angle\;\;in\;\;alternate\;\;segment \\[3ex] \angle ACB = \angle BAC = x^\circ ...base\;\;\angle s\;\;of\;\;isosceles\;\; \triangle CBA \\[3ex] \angle CBA + \angle ACB + \angle BAC = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle CBA \\[3ex] \angle CBA + x + x = 180 \\[3ex] \angle CBA + 2x = 180 \\[3ex] \angle CBA = 180 - 2x \\[5ex] \angle CDA + \angle CBA = 180^\circ ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quadrilateral\;\;are\;\;supplementary \\[3ex] \angle CDA = 180 - \angle CBA \\[3ex] \angle CDA = 180 - (180 - 2x) \\[3ex] \angle CDA = 180 - 180 + 2x \\[3ex] \angle CDA = 2x^\circ $

We can solve this question using at least two approaches
Use whatever approach you prefer.

$ \underline{First\;\;Approach} \\[3ex] $ Join PQ to K

$ \angle OQK = 90^\circ \\[3ex] ...radius\;OQ \perp tangent\;PQ\;\;at\;\;point\;\;of\;\;contact\;Q \\[3ex] \angle SQK = 33 + 90 ...diagram \\[3ex] \angle SQK = 123^\circ \\[3ex] m + n = \angle SQK = 123^\circ \\[3ex] ...exterior\;\;\angle\;\;of\;\;a\;\;\triangle = sum\;\;of\;\;two\;\;interior\;\;opposite\;\;\angle s \\[5ex] \underline{Second\;\;Approach} \\[3ex] \angle OQP = 90^\circ \\[3ex] ...radius\;OQ \perp tangent\;PQ\;\;at\;\;point\;\;of\;\;contact\;Q \\[3ex] \angle SQP + \angle SQR = \angle OQP ...diagram \\[3ex] \angle SQP + 33 = 90 \\[3ex] \angle SQP = 90 - 33 \\[3ex] \angle SQP = 57^\circ \\[3ex] m + n + \angle SQP = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle PSQ \\[3ex] m + n + 57 = 180 \\[3ex] m + n = 180 - 57 \\[3ex] m + n = 123^\circ $

$ \angle MRL = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] x = \angle RML ...\angle\;\;between\;\;tangent\;PRQ\;\;and\;\;chord\;RL = \angle\;\;in\;\;alternate\;\;segment \\[3ex] \angle RML + \angle RLM + \angle MRL = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle MRL \\[3ex] x + 35 + 90 = 180 \\[3ex] x + 125 = 180 \\[3ex] x = 180 - 125 \\[3ex] x = 55^\circ $

$ \underline{Cyclic\;\;Quadrilateral\;BFDC} \\[3ex] x + \angle ADC = 180^\circ \\[3ex] ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quad\;\;are\;\;supplementary \\[3ex] \angle ADC = 180 - x \\[5ex] \underline{\triangle BEC} \\[3ex] \angle BCE + x + 32 = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle BEC \\[3ex] \angle BCE = 180 - x - 32 \\[3ex] \angle BCE = 148 - x \\[3ex] \angle ACD = \angle BCE = 148 - x ...diagram \\[5ex] \underline{\triangle ACD} \\[3ex] 54 + \angle ADC + \angle ACD = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle ACD \\[3ex] 54 + (180 - x) + (148 - x) = 180 \\[3ex] 54 + 180 - x + 148 - x = 180 \\[3ex] 382 - 2x = 180 \\[3ex] 382 - 180 = 2x \\[3ex] 202 = 2x \\[3ex] 2x = 202 \\[3ex] x = \dfrac{202}{2} \\[5ex] x = 101^\circ $

Construction: Join the radii: from centre O to the point D, and from centre O to point F
Radii: |OD| and |OF|
This forms a triangle.
Label the angles in the triangle.


$ \angle ODF = \theta \\[3ex] \angle OFD = \theta \\[3ex] \angle DOF = \psi \\[5ex] \angle ODF = \angle OFD = \theta ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle DOF \\[3ex] \theta + \theta + \psi = 180^\circ ...sum\;\;of\;\;\angle s\;\;of\;\;\triangle DOF \\[3ex] 2\theta + \psi = 180 \\[3ex] \psi = 180 - 2\theta ...eqn.(1) \\[5ex] \psi = 2 * \hat{E} ...\angle \;\;at\;\;centre = 2 * \angle \;\;at\;\;circumference \\[3ex] \psi = 2\hat{E} ...eqn.(2) \\[5ex] \psi = \psi \implies eqn.(2) = eqn.(1) \\[3ex] 2\hat{E} = 180 - 2\theta \\[3ex] 2\hat{E} = 2(90 - \theta) \\[3ex] \hat{E} = 90 - \theta \\[3ex] \theta = 90 - \hat{E} ... eqn.(3) \\[5ex] \theta + D\hat{F}K = 90^\circ ...radius\;OF \perp tangent\;KFC\;\;at\;\;point\;\;contact\;F \\[3ex] \theta + D\hat{F}K = 90 \\[3ex] \theta = 90 - D\hat{F}K ...eqn.(4) \\[5ex] \theta = \theta \implies eqn.(3) = eqn.(4) \\[3ex] 90 - \hat{E} = 90 - D\hat{F}K \\[3ex] D\hat{F}K = 90 - 90 + \hat{E} \\[3ex] D\hat{F}K = \hat{E} \\[3ex] $ Therefore, the angle between the tangent KFC and the chord DF (angle DFK) is equal to the angle in the alternate segment (angle E)

$ \angle MTS = 40^\circ ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] \angle PTR + \angle MTS = 180^\circ ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] \angle PTR + 40 = 180 \\[3ex] \angle PTR = 180 - 40 \\[3ex] \angle PTR = 140^\circ \\[3ex] \angle TPR = \angle TRP ...base\;\; \angle s\;\;of\;\;isosceles\;\;\triangle TPR \\[3ex] y = \angle TRP + \angle TRP \\[3ex] ...exterior\;\;\angle\;\;of\;\;a\;\;\triangle = sum\;\;of\;\;two\;\;interior\;\;opposite\;\;\angle s \\[3ex] \angle TPR + \angle TRP + \angle PTR = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle TPR \\[3ex] \implies \\[3ex] y + 140 = 180 \\[3ex] y = 180 - 140 \\[3ex] y = 40^\circ $

(i) The diagram for the information is:

$ (ii) \\[3ex] \angle MLN = 65^\circ ...\angle\;\;between\;\;tangent\;XMY\;\;and\;\;chord\;MN = \angle\;\;in\;\;alternate\;\;segment \\[3ex] (\alpha.) \\[3ex] \angle MLX + \angle MLN = 180^\circ \\[3ex] \angle MLX + 65 = 180 \\[3ex] \angle MLX = 180 - 65 \\[3ex] \angle MLX = 115^\circ \\[3ex] (\beta.) \\[3ex] \angle NMX + \angle NMY = 180^\circ ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] \angle NMX + 65 = 180 \\[3ex] \angle NMX = 180 - 65 \\[3ex] \angle NMX = 115^\circ \\[3ex] \angle XNM + \angle NXM + \angle NMX = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle XNM \\[3ex] \angle XNM + 34 + 115 = 180 \\[3ex] \angle XNM + 149 = 180 \\[3ex] \angle XNM = 180 - 149 \\[3ex] \angle XNM = 31^\circ \\[3ex] \angle LNM = \angle XNM = 31^\circ...diagram $

$ |MN|^2 = |MP|^2 + |PN|^2 ...Pythagoras\;\;Theorem \\[3ex] |MN|^2 = 3.5^2 + 9.7^2 \\[3ex] |MN|^2 = 12.25 + 94.09 \\[3ex] |MN|^2 = 106.34 \\[3ex] |MN| = \sqrt{106.34} \\[3ex] |MN| = 10.31212878 \\[3ex] diameter = d = |MN| = 10.31212878\;cm \\[3ex] circumference = C \\[3ex] C = \pi d \\[3ex] C = \dfrac{22}{7} * 10.31212878 \\[5ex] C = \dfrac{226.8668332}{7} \\[5ex] C = 32.4095476 \\[3ex] C \approx 32.4\;cm $

We can solve this question using at least two approaches
Use any approach you prefer.

First Approach

$ \angle OAC = 90^\circ ...radius\;OA \perp tangent\;AC\;\;at\;\;point\;\;of\;\;contact\;A \\[3ex] \angle OAC = \angle OAB + \angle BAC ...diagram \\[3ex] 90 = \angle OAB + 2x \\[3ex] \angle OAB + 2x = 90 \\[3ex] \angle OAB = 90 - 2x \\[3ex] \angle OBA = \angle OAB = 90 - 2x ...base\;\;\angle s\;\;of\;\;isosceles\;\; \triangle \\[3ex] \angle AOB + \angle OAB + \angle OBA = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle AOB \\[3ex] \angle AOB + (90 - 2x) + (90 - 2x) = 180 \\[3ex] \angle AOB + 90 - 2x + 90 - 2x = 180 \\[3ex] \angle AOB + 180 - 4x = 180 \\[3ex] \angle AOB = 180 - 180 + 4x \\[3ex] \angle AOB = 4x^\circ \\[3ex] $ Second Approach
Let us do a little work on the diagram so we can solve the question.
Construction: Draw a chord from point A and another chord from point B, to meet at a point say E on the circumference


$ \angle AEB = \angle BAC = 2x ...\angle\;\;between\;\;tangent\;AC\;\;and\;\;chord\;AB = \angle\;\;in\;\;alternate\;\;segment \\[3ex] \angle AOB = 2 * \angle AEB ... \angle \;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] \angle AOB = 2 * 2x \\[3ex] \angle AOB = 4x^\circ $

$ (i) \\[3ex] \angle OAC = \angle OCA = p ...base\;\;\angle s\;\;of\;\;isosceles\;\; \triangle AOC \\[3ex] \angle OAC + \angle OCA + \angle AOC = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle AOC \\[3ex] p + p + 72 = 180 \\[3ex] 2p = 180 - 72 \\[3ex] 2p = 108 \\[3ex] p = \dfrac{108}{2} \\[5ex] p = 54 \\[3ex] \therefore \angle OAC = \angle OCA = 54^\circ \\[3ex] \angle BAC = \angle OAC = 54^\circ ... diagram \\[3ex] \angle BCA = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] \angle ABC + \angle BAC + \angle BCA = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle ABC \\[3ex] \angle ABC + 54 + 90 = 180 \\[3ex] \angle ABC + 144 = 180 \\[3ex] \angle ABC = 180 - 144 \\[3ex] \angle ABC = 36^\circ \\[3ex] (ii) \\[3ex] \angle ACB = \angle BDA = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] \implies \\[3ex] \triangle ACB \;\;and\;\; \triangle BDA \;\;are\;\;right\;\;\triangle s \\[3ex] |AC| = |BD| ...given...leg\;(L) \\[3ex] |AB| = |AB| ...diameter = hypotenuse\;(H) \\[3ex] \therefore \triangle ACB \cong \triangle BDA ... HL\;\;Congruency \\[3ex] \implies \\[3ex] \angle BAD = \angle ABC = 36^\circ ... \triangle ACB \cong \triangle BDA \\[3ex] (iii) \\[3ex] \angle ABD + \angle BAD + \angle BDA = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle ABD \\[3ex] \angle ABD + 36 + 90 = 180 \\[3ex] \angle ABD + 126 = 180 \\[3ex] \angle ABD = 180 - 126 \\[3ex] \angle ABD = 54^\circ $

Construction: Draw a chord from point A to point C


$ (i) \\[3ex] \angle OAT = 90^\circ ...radius\;OA \perp tangent\;PAT\;\;at\;\;point\;\;of\;\;contact\;A \\[3ex] \angle BAP + \angle OAB + \angle OAT = 180^\circ ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] 60 + \angle OAB + 90 = 180 \\[3ex] \angle OAB + 150 = 180 \\[3ex] \angle OAB = 180 - 150 \\[3ex] \angle OAB = 30^\circ \\[3ex] \angle OBA = \angle OAB = 30^\circ ...base\;\;\angle s\;\;of\;\;isosceles\;\; \triangle OBA \\[5ex] Also: \\[3ex] \angle OCT = 90^\circ ...radius\;OC \perp tangent\;QCT\;\;at\;\;point\;\;of\;\;contact\;C \\[3ex] \angle OCT + \angle OCB + \angle BCQ = 180^\circ ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] 90 + \angle OCB + 55 = 180 \\[3ex] \angle OCB + 145 = 180 \\[3ex] \angle OCB = 180 - 145 \\[3ex] \angle OCB = 35^\circ \\[3ex] \angle OBC = \angle OCB = 35^\circ ...base\;\;\angle s\;\;of\;\;isosceles\;\; \triangle OBC \\[3ex] (ii) \\[3ex] \angle ABC = \angle OBA + \angle OBC ... diagram \\[3ex] \angle ABC = 30 + 35 \\[3ex] \angle ABC = 65^\circ \\[3ex] \angle AOC = 2 * \angle ABC ... \angle \;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] \angle AOC = 2 * 65 \\[3ex] \angle AOC = 130^\circ \\[3ex] (iii) \\[3ex] We\;\;can\;\;solve\;\;this\;\;question\;\;using\;\;at\;\;least\;\;two\;\;approaches \\[3ex] Use\;\;any\;\;approach\;\;you\;\;prefer \\[3ex] \underline{First\;\;Approach} \\[3ex] \angle OAC = \angle OCA = m ...base\;\;\angle s\;\;of\;\;isosceles\;\; \triangle AOC \\[3ex] \angle OAC + \angle OCA + \angle AOC = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle AOC \\[3ex] m + m + 130 = 180 \\[3ex] 2m = 180 - 130 \\[3ex] 2m = 50 \\[3ex] m = \dfrac{50}{2} \\[5ex] m = 25 \\[3ex] \therefore \angle OAC = \angle OCA = 25^\circ \\[3ex] \angle OAT = \angle OAC + \angle CAT ... diagram \\[3ex] 90 = 25 + \angle CAT \\[3ex] 25 + \angle CAT = 90 \\[3ex] \angle CAT = 90 - 25 \\[3ex] \angle CAT = 65^\circ \\[3ex] |AT| = |CT| ...two\;\;tangents:\;\;AT\;\;and\;\;CT\;\;drawn\;\;from\;\;the\;\;same\;\;external\;\;point:\;T\;\;are\;\;equal\;\;in\;\;length \\[3ex] \implies \\[3ex] \angle ACT = \angle CAT = 65^\circ ...base\;\;\angle s\;\;of\;\;isosceles\;\; \triangle ATC \\[3ex] \angle ATC + \angle ACT + \angle CAT = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle ATC \\[3ex] \angle ATC + 65 + 65 = 180 \\[3ex] \angle ATC + 130 = 180 \\[3ex] \angle ATC = 180 - 130 \\[3ex] \angle ATC = 50^\circ \\[3ex] \underline{Second\;\;Approach} \\[3ex] \angle ATC + \angle BAT + \angle ABC + \angle BCT = 360^\circ ...sum\;\;of\;\;interior\;\;\angle s\;\;of\;\;Quadrilateral\;BCTA \\[3ex] \angle BAT = \angle OAB + \angle OAT ... diagram \\[3ex] \angle BAT = 30 + 90 \\[3ex] \angle BAT = 120^\circ \\[3ex] \angle BCT = \angle OCB + \angle OCQ ... diagram \\[3ex] \angle BAT = 35 + 90 \\[3ex] \angle BAT = 125^\circ \\[3ex] \implies \\[3ex] \angle ATC + 120 + 65 + 125 = 360 \\[3ex] \angle ATC + 310 = 360 \\[3ex] \angle ATC = 360 - 310 \\[3ex] \angle ATC = 50^\circ $

$ (a) \\[3ex] \underline{Intersecting\;\;Secant-Tangent\;\;Theorem} \\[3ex] \overline{TD} * 8 = 10 * 10 \\[3ex] \overline{TD} = \dfrac{10 * 10}{8} \\[5ex] \overline{TD} = 12.5 \\[3ex] But: \\[3ex] \overline{TD} = \overline{CO} + \overline{OD} + \overline{TC}...diagram \\[3ex] \overline{CO} = \overline{OD} = radius = r ...diagram \\[3ex] \implies \\[3ex] \overline{DT} = r + r + 8 \\[3ex] 12.5 = 2r + 8 \\[3ex] 2r + 8 = 12.5 \\[3ex] 2r = 12.5 - 8 \\[3ex] 2r = 4.5 \\[3ex] r = \dfrac{4.5}{2} \\[5ex] r = 2.25cm \\[3ex] $ (b)
Let us visualize what we need to find
First, we have to find the angle, &theta
Then, we use the formula for the length of an arc to find it


$ \angle OAT = 90^\circ \\[3ex] ...radius\;OA \perp tangent\;AT\;\;at\;\;point\;\;of\;\;contact\;A \\[3ex] SOHCAHTOA \\[3ex] \sin \theta = \dfrac{opp}{hyp} = \dfrac{|AT|}{|TO|} \\[5ex] |TO| = |TC| + |CO| \\[3ex] |TO| = 8 + 2.25 = 10.25cm \\[3ex] \implies \\[3ex] \sin \theta = \dfrac{10}{10.25} \\[5ex] \sin \theta = 0.9756097561 \\[3ex] \theta = \sin^{-1}(0.9756097561) \\[3ex] \theta = 77.31961651^\circ \\[3ex] length\;\;of\;\;arc = L \\[3ex] L = \dfrac{\theta}{360} * 2\pi r \\[5ex] L = \dfrac{77.31961651}{360} * 2 * \dfrac{22}{7} * 2.25 \\[5ex] L = \dfrac{7654.642034}{2520} \\[5ex] L = 3.037556363 \\[3ex] L \approx 3.04cm...to\;\;3\;\;significant\;\;figures $

$ (a) \\[3ex] \angle ACB = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] (b) \\[3ex] \angle BAC + 32 = 90^\circ ...radius\;OA \perp tangent\;DAE\;\;at\;\;point\;\;of\;\;contact\;A \\[3ex] \angle BAC = 90 - 32 \\[3ex] \angle BAC = 58^\circ \\[3ex] radius = |OA| = |OB| = 8\;cm \\[3ex] diameter = |AB| = 2(8) = 16\;cm \\[3ex] \dfrac{|BC|}{\sin \angle BAC} = \dfrac{|AB|}{\sin \angle ACB} ... Sine\;\;Rule \\[5ex] \dfrac{|BC|}{\sin 58} = \dfrac{16}{\sin 90} \\[5ex] |BC| = \dfrac{16 \sin 58}{\sin 90} \\[5ex] |BC| = \dfrac{16(0.8480480962)}{1} \\[5ex] |BC| = 13.56876954 \\[3ex] |BC| \approx 13.6\;cm $

$ \angle PRS = 37^\circ ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] \angle RSP = 90^\circ...\angle\;in\;\;a\;\;semicircle \\[3ex] \angle SPR + \angle PRS + \angle RSP = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle PRS \\[3ex] \angle SPR + 37 + 90 = 180 \\[3ex] \angle SPR + 127 = 180 \\[3ex] \angle SPR = 180 - 127 \\[3ex] \angle SPR = 53^\circ $

$ \angle ORQ = 90^\circ ...radius\;OR \perp tangent\;RQ\;\;at\;\;point\;\;of\;\;contact\;R \\[3ex] \angle OPQ = 90^\circ ...radius\;OP \perp tangent\;PQ\;\;at\;\;point\;\;of\;\;contact\;P \\[3ex] \angle OQR = \dfrac{80}{2} = 40^\circ ... two\;\;tangents\;\;drawn\;\;from\;\;same\;\;external\;\;point\;\;bisects\;\;\angle\;\;formed\;\;by\;\;the\;\;tangents \\[5ex] \angle OQP = \dfrac{80}{2} = 40^\circ ... two\;\;tangents\;\;drawn\;\;from\;\;same\;\;external\;\;point\;\;bisects\;\;\angle\;\;formed\;\;by\;\;the\;\;tangents \\[5ex] \angle ROQ + \angle ORQ + \angle OQR = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle ROQ \\[3ex] \angle ROQ + 90 + 40 = 180 \\[3ex] \angle ROQ + 130 = 180 \\[3ex] \angle ROQ = 180 - 130 \\[3ex] \angle ROQ = 50^\circ \\[3ex] \angle POQ = \angle ROQ = 50^\circ ... two\;\;tangents\;\;drawn\;\;from\;\;same\;\;external\;\;point\;\;bisects\;\;\angle\;\;formed\;\;by\;\;the\;\;radii \\[5ex] Obtuse\;\angle POR = \angle POQ + \angle ROQ ... diagram \\[3ex] = 50 + 50 \\[3ex] = 100^\circ \\[3ex] Reflex\;\angle POR + Obtuse\;\angle POR = 360^\circ...\angle s\;\;at\;\;a\;\;point \\[3ex] Reflex\;\angle POR + 100 = 360 \\[3ex] = 360 - 100 \\[3ex] = 260^\circ $

$ \underline{\triangle BAF} \\[3ex] \hat{B_2} + 3x + 2x = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;a\;\;\triangle \\[3ex] \hat{B_2} + 5x = 180 \\[3ex] \hat{B_2} = 180 - 5x \\[5ex] \underline{\triangle EAD} \\[3ex] \hat{D_2} + 3x + x = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;a\;\;\triangle \\[3ex] \hat{D_2} + 4x = 180 \\[3ex] \hat{D_2} = 180 - 4x \\[5ex] \hat{B_2} + \hat{D_2} = 180^\circ \\[3ex] ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quad\;\;are\;\;supplementary \\[3ex] \implies \\[3ex] (180 - 5x) + (180 - 4x) = 180 \\[3ex] 180 - 5x + 180 - 4x = 180 \\[3ex] 360 - 9x = 180 \\[3ex] 360 - 180 = 9x \\[3ex] 180 = 9x \\[3ex] 9x = 180 \\[3ex] x = \dfrac{180}{9} \\[5ex] x = 20^\circ $

$ (a) \\[3ex] x = 55^\circ ... \angle\;\;in\;\;alternate\;\;segment = \angle\;\;between\;\;tangent\;XAY\;\;and\;\;chord\;AD \\[3ex] (b) \\[3ex] y + 65 = 180^\circ ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quadrilateral\;\;are\;\;supplementary \\[3ex] y = 180 - 65 \\[3ex] y = 115^\circ $

$ \angle WXZ = \angle WZX = p ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle WXZ \\[3ex] \angle VWZ = \angle WXZ + \angle WZX \\[3ex] ...exterior\;\;\angle\;\;of\;\;a\;\;\triangle = sum\;\;of\;\;two\;\;interior\;\;\angle s \\[3ex] 110 = p + p \\[3ex] 2p = 110 \\[3ex] p = \dfrac{110}{2} \\[5ex] p = 55 \\[3ex] \therefore \angle WXZ = 55^\circ $

$ \angle SRP + \angle SPR + \angle PSR = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle PSR \\[3ex] \angle SRP + 54 + 79 = 180 \\[3ex] \angle SRP + 133 = 180 \\[3ex] \angle SRP = 180 - 133 \\[3ex] \angle SRP = 47^\circ \\[3ex] (i) \\[3ex] \angle TPS = \angle SRP = 47^\circ ... \angle\;\;between\;\;tangent\;TPQ\;\;and\;\;chord\;PS = \angle\;\;in\;\;alternate\;\;segment \\[3ex] (ii) \\[3ex] \angle TPS + \angle SPR + \angle RPQ = 180^\circ ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] 47 + 54 + \angle RPQ = 180 \\[3ex] 101 + \angle RPQ = 180 \\[3ex] \angle RPQ = 180 - 101 \\[3ex] \angle RPQ = 79^\circ \\[3ex] \angle PRQ = 54^\circ ... alternate\;\;\angle s\;\;are\;\;equal \\[3ex] \angle PQR + \angle PRQ + \angle RPQ = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle PRQ \\[3ex] \angle PQR + 54 + 79 = 180 \\[3ex] \angle PQR + 133 = 180 \\[3ex] \angle PQR = 180 - 133 \\[3ex] \angle PQR = 47^\circ $

$ CP * PD = AP * PB ... Intersecting\;\;Chords\;\;Theorem \\[3ex] CP * 4 = 7 * 5 \\[3ex] CP = \dfrac{7 * 5}{4} \\[5ex] CP = 8.75 \\[3ex] |CP| = 8.75cm $

$ (i) \\[3ex] \angle B = \angle D ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal = Angle\;(A) \\[3ex] \angle A = \angle C ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal = Angle\;(A) \\[3ex] \therefore \triangle PAB \sim \triangle PCD ...AA\;\;Similarity\;\;Postulate \\[3ex] (ii) \\[3ex] \dfrac{|CD|}{|AB|} = \dfrac{|PC|}{|PA|} \\[5ex] \dfrac{|CD|}{9} = \dfrac{5}{7.5} \\[5ex] |CD| = \dfrac{9(5)}{7.5} \\[5ex] |CD| = 6\;cm \\[3ex] (iii) \\[3ex] Area\;\;of\;\;\triangle PAB \\[3ex] = \dfrac{1}{2} * |PA| * |AB| * \sin \angle A \\[5ex] = 0.5(7.5)(9)\sin \angle A \\[3ex] Area\;\;of\;\;\triangle PCD \\[3ex] = \dfrac{1}{2} * |PC| * |DC| * \sin \angle C \\[5ex] = 0.5(5)(6)\sin \angle C \\[3ex] Ratio\;\;of\;\;Areas \\[3ex] = \dfrac{Area\;\;of\;\;\triangle PAB}{Area\;\;of\;\;\triangle PCD} \\[5ex] = \dfrac{0.5(7.5)(9)\sin \angle A}{0.5(5)(6)\sin \angle C} \\[5ex] \angle A = \angle C \implies \sin \angle A = \sin \angle C \\[3ex] \implies \\[3ex] = \dfrac{22.5}{10} \\[5ex] = \dfrac{225}{100} \\[5ex] = \dfrac{9}{4} \\[5ex] = 9 : 4 $

$ \angle QSR = 48^\circ ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] \angle SQR = \angle QSR = 48^\circ ... base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle RSQ \\[3ex] \angle SRQ = \angle SRP + \angle PRQ ... diagram \\[3ex] \angle SRQ + \angle QSR + \angle SQR = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle RSQ \\[3ex] \implies \\[3ex] (\angle SRP + \angle PRQ) + \angle QSR + \angle SQR = 180 \\[3ex] 60 + \angle PRQ + 48 + 48 = 180 \\[3ex] \angle PRQ + 156 = 180 \\[3ex] \angle PRQ = 180 - 156 \\[3ex] \angle PRQ = 24^\circ $

$ \angle ONM = \angle OMN = 20^\circ ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle OMN \\[3ex] \angle NOM + \angle ONM + \angle OMN = 180^\circ ... sum\;\;of\;\;\angle s\;\;in\;\;\triangle NOM \\[3ex] \angle NOM + 20 + 20 = 180 \\[3ex] \angle NOM + 40 = 180 \\[3ex] \angle NOM = 180 - 40 \\[3ex] \angle NOM = 140^\circ \\[3ex] \angle NOM = 2 * \angle NLM ...\angle\;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] 140 = 2 * \angle NLM \\[3ex] 2 * \angle NLM = 140 \\[3ex] \angle NLM = \dfrac{140}{2} \\[5ex] \angle NLM = 70^\circ \\[5ex] \angle MTN = \angle LTN = 32^\circ ...diagram \\[3ex] \angle MNT = \angle MTN = 32^\circ ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle MNT \\[3ex] \angle LMN = \angle MNT + \angle MTN ...exterior\;\;\angle\;\;of\;\;a\;\;\triangle = sum\;\;of\;\;two\;\;interior\;\;opposite\;\;\angle s \\[3ex] \angle LMN = 32 + 32 = 64^\circ \\[5ex] \angle LNM + \angle NLM + \angle LMN = 180^\circ ... sum\;\;of\;\;\angle s\;\;in\;\;\triangle NLM \\[3ex] \angle LNM + 70 + 64 = 180 \\[3ex] \angle LNM + 134 = 180 \\[3ex] \angle LNM = 180 - 134 \\[3ex] \angle LNM = 46^\circ $

$ Reflex\;\;\angle AOC = 2 * \angle ADC ... \angle \;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] Reflex\;\;\angle AOC = 2 * 132 \\[3ex] Reflex\;\;\angle AOC = 264^\circ \\[3ex] x + Reflex\;\;\angle AOC = 360^\circ ...\angle s \;\;at\;\;a\;\;point \\[3ex] x + 264 = 360 \\[3ex] x = 360 - 264 \\[3ex] x = 96^\circ $

$ \angle OAC = 90^\circ ...radius\;OA \perp tangent\;AC\;\;at\;\;point\;\;of\;\;contact\;A \\[3ex] \angle OAC = \angle BAD = \angle OAD = \angle BAC = 90^\circ ... diagram \\[3ex] \angle AOC = 2 * \angle ABD ... \angle \;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] y = 2 * \angle ABD \\[3ex] 2 * \angle ABD = y \\[3ex] \angle ABD = \dfrac{y}{2} \\[5ex] \underline{\triangle ABD} \\[3ex] \angle ABD + \angle BAD + \angle ADB = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle ABD \\[3ex] \dfrac{y}{2} + 90 + 55 = 180 \\[5ex] \dfrac{y}{2} + 145 = 180 \\[5ex] \dfrac{y}{2} = 180 - 145 \\[5ex] \dfrac{y}{2} = 35 \\[5ex] y = 2(35) \\[3ex] y = 70^\circ \\[3ex] \underline{\triangle AOC} \\[3ex] \angle ACO + \angle AOC + \angle OAC = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle OAC \\[3ex] x + y + 90 = 180 \\[3ex] x + 70 + 90 = 180 \\[3ex] x + 160 = 180 \\[3ex] x = 180 - 160 \\[3ex] x = 20^\circ $

$ (i) \\[3ex] x = y ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] \angle TOP + z = 180^\circ ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] \angle TOP = 180 - z \\[3ex] \angle PTO = \angle TPO = x ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle TOP \\[3ex] \underline{\triangle TOP} \\[3ex] \angle TPO + \angle PTO + \angle TOP = 180^\circ ...sum\;\;of\;\;\angle s \;\;in\;\;a\;\;\triangle \\[3ex] x + x + (180 - z) = 180 \\[3ex] But\;\; x = y \\[3ex] And\;\;we\;\;need\;\;a\;\;relationship\;\;between\;\;x,\;y,\;z \\[3ex] x + y + 180 - z = 180 \\[3ex] x + y - z = 180 - 180 \\[3ex] x + y - z = 0 \\[3ex] x + y = 0 + z \\[3ex] x + y = z \\[3ex] (ii) \\[3ex] $ Construction: Draw the chord from point T to point R


$ \angle STR = \angle SRT = p ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle STR \\[3ex] \angle STR + \angle SRT + \angle TSR = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle STR \\[3ex] p + p + 118 = 180 \\[3ex] 2p = 180 - 118 \\[3ex] 2p = 62 \\[3ex] p = \dfrac{62}{2} \\[5ex] p = 31 \\[3ex] \therefore \angle STR = 31^\circ \\[3ex] \angle PTR = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] \angle STP = \angle STR + \angle PTR ...diagram \\[3ex] \angle STP = 31 + 90 \\[3ex] \angle STP = 121^\circ $

$ \angle BAQ = 30^\circ...given \\[3ex] \implies \\[3ex] \angle CAB = 30^\circ ... AB\;\;bisects\;\;\angle CAQ \\[3ex] \angle CDB = \angle CAB = 30^\circ ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] \angle BAQ = \angle ADB = 30^\circ ...\angle\;\;between\;\;tangent\;PAQ\;\;and\;\;chord\;AB = \angle\;\;in\;\;alternate\;\;segment \\[3ex] \angle ACB = \angle ADB = 30^\circ ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] \angle ABC + \angle ACB + \angle CAB = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle ABC \\[3ex] \angle ABC + 30 + 30 = 180 \\[3ex] \angle ABC + 60 = 180 \\[3ex] \angle ABC = 180 - 60 \\[3ex] \angle ABC = 120^\circ \\[3ex] \angle PAC = \angle ABC = 120^\circ ...\angle\;\;between\;\;tangent\;PAQ\;\;and\;\;chord\;AC = \angle\;\;in\;\;alternate\;\;segment \\[3ex] \angle PAC = \angle PAD + \angle DAC ... diagram \\[3ex] \angle PAD = \angle DAC = \dfrac{1}{2} * 120 = 60^\circ ... AD\;\;bisects\;\;\angle PAC \\[5ex] (i) \\[3ex] \underline{\triangle ABD} \\[3ex] \angle DAB = \angle DAC + \angle CAB ...diagram \\[3ex] \angle DAB = 60 + 30 \\[3ex] \angle DAB = 90^\circ \\[3ex] \implies \\[3ex] |BD| \;\;is\;\;a\;\;diameter \;\;because \angle DAB = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] (ii) \\[3ex] \underline{\triangle ABC} \\[3ex] \angle ACB = \angle CAB = 30^\circ \\[3ex] \angle ABC = 60^\circ \\[3ex] \implies \\[3ex] \triangle ABC \;\;is\;\;an\;\;isosceles\;\;\triangle $

$ \overline{BA} = \overline{CA} - \overline{CB} ...diagram \\[3ex] \overline{BA} = 12.5 - 4.5 \\[3ex] \overline{BA} = 8 \\[3ex] \overline{DA}^2 = \overline{BA} * \overline{CA} ...Intersecting\;\;Tangent-Secant\;\;Theorem \\[3ex] \overline{DA}^2 = 8 * 12.5 \\[3ex] \overline{DA}^2 = 100 \\[3ex] \overline{DA} = \sqrt{100} \\[3ex] \overline{DA} = 10 $

$ (a) \\[3ex] |PA| = |PB| = 12\;cm ...two\;\;tangents:\;\;PA\;\;and\;\;PB\;\;drawn\;\;from\;\;the\;\;same\;\;external\;\;point:\;P\;\;are\;\;equal\;\;in\;\;length \\[3ex] (b) \\[3ex] P\hat{A}O = 90^\circ ...radius\;OA \perp tangent\;PA\;\;at\;\;point\;\;of\;\;contact\;A \\[3ex] (c) \\[3ex] Quadrilateral\;PAOB = \triangle PAO + \triangle PBO \\[3ex] \underline{\triangle} \\[3ex] |OA| = |OB| = radius = perpendicular\;\;height = 4\;cm \\[3ex] |PB| = |PA| = base = 12\;cm \\[3ex] Area = \dfrac{1}{2} * base * perpendicular\;\;height \\[5ex] Area\;\;of\;\;Quadrilateral\;PAOB \\[3ex] = Area\;\;of\;\;\triangle PAO + Area\;\;of\;\;\triangle PBO \\[3ex] = \dfrac{1}{2} * 12 * 4 + \dfrac{1}{2} * 12 * 4 \\[5ex] = 24 + 24 \\[3ex] = 48\;cm^2 $

$ \angle ABC = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] (i) \\[3ex] \angle ACB + \angle BAC + \angle ABC = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle ABC \\[3ex] \angle ACB + 50 + 90 = 180 \\[3ex] \angle ACB + 140 = 180 \\[3ex] \angle ACB = 180 - 140 \\[3ex] \angle ACB = 40^\circ \\[3ex] (ii) \\[3ex] \angle EAC = \angle ACB = 40^\circ ...alternate\;\;\angle s\;\;are\;\;equal \\[3ex] \underline{Quadrilateral\;\;AEDC} \\[3ex] \angle D + \angle A = 180^\circ \\[3ex] ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quad\;\;are\;\;supplementary \\[3ex] \angle D = \angle EDC ...diagram \\[3ex] \angle A = \angle EAC = 40^\circ ... diagram \\[3ex] \implies \\[3ex] \angle EDC + \angle EAC = 180 \\[3ex] \angle EDC + 40 = 180 \\[3ex] \angle EDC = 180 - 40 \\[3ex] \angle EDC = 140^\circ \\[3ex] (iii) \\[3ex] $ Construction: Draw the chord from point B to point E and draw the chord from point E to point C


Because ∠EAB = 90° (40° + 50°) chord BE is a diameter...angle in a semicircle
This implies that chord BE (diameter) is perpendicular to chord AC (diameter)
Let K be the midpoint of chord BE

$ \angle KEC + \angle EKC + \angle ECK = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle KEC \\[3ex] \angle KEC + 90 + 50 = 180 \\[3ex] \angle KEC + 140 = 180 \\[3ex] \angle KEC = 180 - 140 \\[3ex] \angle KEC = 40^\circ \\[3ex] \angle BEC = \angle KEC = 40^\circ ... diagram $

$ (a) \\[3ex] (i) \\[3ex] \underline{Cyclic\;\;Quadrilateral\;\; EBCD} \\[3ex] \angle B + \angle D = 180^\circ ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quadrilateral\;\;are\;\;supplementary \\[3ex] \angle B + 110 = 180 \\[3ex] \angle B = 180 - 110 \\[3ex] \angle B = 70^\circ \\[3ex] \angle EBC = \angle B = 70^\circ ... diagram \\[3ex] \angle ABE = \angle FAE = 30^\circ ...\angle\;\;between\;\;tangent\;FAG\;\;and\;\;chord\;AE = \angle\;\;in\;\;alternate\;\;segment \\[3ex] \underline{Cyclic\;\;Quadrilateral\;\; ABCE} \\[3ex] \angle B = \angle ABE + \angle EBC ...diagram \\[3ex] \angle B = 30 + 70 \\[3ex] \angle B = 100^\circ \\[3ex] \angle E + \angle B = 180^\circ ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quadrilateral\;\;are\;\;supplementary \\[3ex] \angle E + 100 = 180 \\[3ex] \angle E = 180 - 100 \\[3ex] \angle E = 80^\circ \\[3ex] \angle AEC = \angle E = 80^\circ ... diagram \\[3ex] (ii) \\[3ex] $ Construction: Draw the radius from point O to point E


$ Reflex\;\; \angle EOC = 2 * \angle EDC ... \angle \;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] Reflex\;\; \angle EOC = 2 * 110 \\[3ex] Reflex\;\; \angle EOC = 220^\circ \\[3ex] Obtuse\;\;\angle EOC + Reflex\;\;\angle EOC = 360^\circ ...\angle s \;\;at\;\;a\;\;point \\[3ex] Obtuse\;\;\angle EOC + 220 = 360 \\[3ex] Obtuse\;\;\angle EOC = 360 - 220 \\[3ex] Obtuse\;\;\angle EOC = 140^\circ \\[3ex] \underline{\triangle EOC} \\[3ex] \angle OEC = \angle OCE = p ...base\;\;\angle s\;\;of\;\;isosceles\;\; \triangle EOC \\[3ex] \angle OEC + \angle OCE + Obtuse\;\;\angle EOC = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle EOC \\[3ex] p + p + 140 = 180 \\[3ex] 2p = 180 - 140 \\[3ex] 2p = 40 \\[3ex] p = \dfrac{40}{2} \\[5ex] p = 20 \\[3ex] \therefore \angle OEC = \angle OCE = 20^\circ \\[3ex] \underline{Cyclic\;\;Quadrilateral\;\; ABCE} \\[3ex] \angle C = \angle OCB + \angle OCE ... diagram \\[3ex] \angle C = 55 + 20 \\[3ex] \angle C = 75^\circ \\[3ex] \angle A + \angle C = 180^\circ ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quadrilateral\;\;are\;\;supplementary \\[3ex] \angle A + 75 = 180 \\[3ex] \angle A = 180 - 75 \\[3ex] \angle A = 105^\circ \\[3ex] \angle EAB = \angle A = 105^\circ ... diagram \\[3ex] \underline{\triangle AEB} \\[3ex] \angle AEB + \angle EAB + \angle ABE = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle ABE \\[3ex] \angle AEB + 105 + 30 = 180 \\[3ex] \angle AEB + 135 = 180 \\[3ex] \angle AEB = 180 - 135 \\[3ex] \angle AEB = 45^\circ \\[3ex] (b) \\[3ex] (i) \\[3ex] \underline{\triangle EAB} \\[3ex] Let\;\;|EB| = k \\[3ex] \dfrac{|EB|}{\sin \angle EAB} = \dfrac{|AB|}{\sin \angle AEB} ...Sine\;\;Law \\[5ex] \dfrac{k}{\sin 105} = \dfrac{5}{\sin 45} \\[5ex] k = \dfrac{5\sin 105}{\sin 45} \\[5ex] \underline{\triangle EBC} \\[3ex] Let\;\;|EC| = m \\[3ex] \dfrac{|EC|}{\sin \angle EBC} = \dfrac{|EB|}{\sin \angle BCE} ...Sine\;\;Law \\[5ex] \angle BCE = \angle OCB + \angle OCE ... diagram \\[3ex] \angle BCE = 55 + 20 \\[3ex] \angle BCE = 75^\circ \\[3ex] \implies \\[3ex] \dfrac{m}{\sin 70} = \dfrac{k}{\sin 75} \\[5ex] \dfrac{m}{\sin 70} = k \div \sin 75 \\[5ex] \dfrac{m}{\sin 70} = \dfrac{5\sin 105}{\sin 45} \div \sin 75 \\[5ex] \dfrac{m}{\sin 70} = \dfrac{5\sin 105}{\sin 45} * \dfrac{1}{\sin 75} \\[5ex] m = \dfrac{5 * \sin 105 * \sin 70}{\sin 45 * \sin 75} \\[5ex] m = \dfrac{4.538366856}{0.6830127019} \\[5ex] m = 6.644630244 \\[3ex] \underline{\triangle EOC} \\[3ex] Radius = |OE| = |OC| = r \\[3ex] |EC|^2 = |OE|^2 + |OC|^2 - 2(|OE|)(|OC|) \cos \angle EOC...Cosine\;\;Law \\[3ex] m^2 = r^2 + r^2 - 2(r)(r) * \cos 140 \\[3ex] m^2 = 2r^2 - 2r^2 \cos 140 \\[3ex] m^2 = 2r^2(1 - \cos 140) \\[3ex] 2r^2(1 - \cos 140) = m^2 \\[3ex] 2 * r^2 * (1 - \cos 140) = m^2 \\[3ex] r^2 = \dfrac{m^2}{2(1 - \cos 140)} \\[5ex] r^2 = \dfrac{6.644630244^2}{2(1 - (-0.7660444431))} \\[5ex] r^2 = \dfrac{44.15111108}{2(1 + 0.7660444431)} \\[5ex] r^2 = \dfrac{44.15111108}{2(1.7660444431)} \\[5ex] r^2 = \dfrac{44.15111108}{3.532088886} \\[5ex] r^2 = 12.5 \\[3ex] r = \sqrt{12.5} \\[3ex] r = 3.535533906 \\[3ex] \therefore radius \approx 3.5\;cm \\[3ex] (ii) \\[3ex] $ We can solve (ii) using at least two approaches.
Use any approach you prefer.
For KCSE students, use the 1st approach.

$ \underline{1st\;\;Approach:\;\;Intersecting\;\;Secant-Tangent\;\;Theorem} \\[3ex] |AF|^2 = |FE| * |FD| \\[3ex] But:\;\;|FD| = |FE| + |ED| ... diagram \\[3ex] |AF|^2 = 2.5(|FE| + |ED|) \\[3ex] |AF|^2 = 2.5(2.5 + 4.4) \\[3ex] |AF|^2 = 2.5(6.9) \\[3ex] |AF|^2 = 17.25 \\[3ex] |AF| = \sqrt{17.25} \\[3ex] |AF| = 4.153311931 \\[3ex] |AF| \approx 4.2\;cm \\[3ex] \underline{2nd\;\;Approach:\;\;Sine\;\;Law} \\[3ex] \underline{\triangle EDC} \\[3ex] \dfrac{\sin \angle ECD}{|ED|} = \dfrac{\sin \angle EDC}{|EC|} ...Sine\;\;Law \\[5ex] \dfrac{\sin \angle ECD}{4.4} = \dfrac{\sin 110}{m} \\[5ex] \dfrac{\sin \angle ECD}{4.4} = \dfrac{\sin 110}{6.644630244} \\[5ex] \sin \angle ECD = \dfrac{4.4 \sin 110}{6.644630244} \\[5ex] \sin \angle ECD = \dfrac{4.134647531}{6.644630244} \\[5ex] \sin \angle ECD = 0.6222539674 \\[3ex] \angle ECD = \sin^{-1}(0.6222539674) \\[3ex] \angle ECD = 38.48091859 \\[3ex] \underline{Cyclic\;\;Quadrilateral\;\; EBCD} \\[3ex] \angle C = \angle OCB + \angle OCE + \angle ECD ...diagram \\[3ex] \angle C = 55 + 20 + 38.48091859 \\[3ex] \angle C = 113.4809186^\circ \\[3ex] \angle BEF = \angle C = 113.4809186^\circ ... exterior\;\; \angle \:\:of\:\:a\:\:cyclic\:\:quadrilateral = interior\;\;opposite\;\;\angle \\[3ex] But: \\[3ex] \angle AEF + \angle AEB = \angle BEF ... diagram \\[3ex] \angle AEF + 45 = 113.4809186 \\[3ex] \angle AEF = 113.4809186 - 45 \\[3ex] \angle AEF = 68.48091859^\circ \\[3ex] \underline{\triangle AEF} \\[3ex] Let\;\;|AF| = n \\[3ex] \dfrac{|AF|}{\sin \angle AEF} = \dfrac{|FE|}{\sin \angle FAE} ...Sine\;\;Law \\[5ex] \dfrac{n}{\sin 68.48091859} = \dfrac{2.5}{\sin 30} \\[5ex] n = \dfrac{2.5 * \sin 68.48091859}{\sin 30} \\[5ex] n = \dfrac{2.325738648}{0.5} \\[5ex] n = 4.651477296 \\[3ex] \therefore |AF| \approx 4.7\;cm $

(a) There are at least two ways to prove that $\triangle ABC$ and $\triangle DCB$ are congruent
Use any approach you prefer

$ \underline{First\;\;Approach:\;\;HL} \\[3ex] \angle ABC = \angle DCB = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] \therefore \triangle ABC \;\;and\;\;\triangle DBC \;\;are\;\;right\;\;\triangle s \\[3ex] \implies \\[3ex] |AC| = |DB| = diameter = hypotenuse \;(H) \\[3ex] |BC| = |CB| ...diagram = leg\;(L) \\[3ex] \implies \\[3ex] HL\;\;Congruency \\[5ex] \underline{Second\;\;Approach:\;\;AAS} \\[3ex] \angle ABC = \angle DCB = 90^\circ ...\angle \;\;in\;\;a\;\;semicircle \\[3ex] ...angle\;(A) \\[3ex] \angle BAC = \angle CDB ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] ...angle\;(A) \\[3ex] |AC| = |DB| = diameter \\[3ex] ..side\;(S) \\[3ex] \implies \\[3ex] AAS\;\;Congruency $

Construction: Label the points on the circle


$ \angle OBC = 43^\circ ... vertical\;\;\angle s\;\;are\;\;equal \\[3ex] y = \angle OBC = \angle OCB = 43^\circ ...base\;\;\angle s \;\;of\;\;isosceles\;\;\triangle OBC $

Construction: Label the points on the circle


$ \angle BOC + \angle OBC + \angle OCB = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle OBC \\[3ex] \angle BOC + 43 + 43 = 180 \\[3ex] \angle BOC + 86 = 180 \\[3ex] \angle BOC = 180 - 86 \\[3ex] \angle BOC = 94^\circ \\[3ex] \angle BOC = 2 * x ...\angle\;\;at\;\;centre = 2 * \angle \;\;at\;\;circumference \\[3ex] 94 = 2x \\[3ex] 2x = 94 \\[3ex] x = \dfrac{94}{2} \\[5ex] x = 47^\circ $

Construction: Join the chord from Point D to Point B


$ \angle ABD = 71^\circ ...\angle\;\;between\;\;tangent\;TDV\;\;and\;\;chord\;AD = \angle\;\;in\;\;alternate\;\;segment \\[3ex] \angle ADB = \angle ABD = 71^\circ ...base\;\;\angle s\;\;of\;\;isosceles\;\; \triangle ABD \\[3ex] \angle DAB + \angle ADB + \angle ABD = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle ABD \\[3ex] \implies \\[3ex] \angle DAB + 71 + 71 = 180 \\[3ex] \angle DAB + 142 = 180 \\[3ex] \angle DAB = 180 - 142 \\[3ex] \angle DAB = 38^\circ \\[3ex] \angle BCD + \angle DAB = 180^\circ ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quadrilateral\;\;are\;\;supplementary \\[3ex] \angle BCD + 38 = 180 \\[3ex] \angle BCD = 180 - 38 \\[3ex] \angle BCD = 142^\circ $

$ |AD|\;\;is\;\;a\;\;diameter \\[3ex] \implies \\[3ex] \angle ACD = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] Sum\;\;of\;\;the\;\;interior\;\;\angle s\;\;of\;\;a\;\;regular\;\;polygon \\[3ex] = 180(n - 2) \\[3ex] For\;\;a\;\;regular\;\;hexagon:\;\;n = 6 \;\;(6\;\;sides) \\[3ex] Sum\;\;of\;\;the\;\;interior\;\;\angle s\;\;of\;\;a\;\;regular\;\;hexagon \\[3ex] = 180(6 - 2) \\[3ex] = 180(4) \\[3ex] = 720^\circ \\[3ex] Each\;\;interior\;\;\angle\;\;of\;\;a\;\;regular\;\;hexagon \\[3ex] = \dfrac{Sum}{n} \\[5ex] = \dfrac{720}{6} \\[5ex] = 120^\circ \\[3ex] \implies \\[3ex] \angle A = \angle B = \angle C = \angle D = \angle E = \angle F = 120^\circ \\[3ex] \underline{Cyclic\;\;Quadrilateral\;\;ABCD} \\[3ex] \angle ABC = \angle B = 120^\circ ...diagram \\[3ex] \angle ABC + \angle CDA = 180^\circ \\[3ex] ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quad\;\;are\;\;supplementary \\[3ex] 120 + \angle CDA = 180 \\[3ex] \angle CDA = 180 - 120 \\[3ex] \angle CDA = 60^\circ \\[3ex] \underline{\triangle ACD} \\[3ex] \angle CAD + \angle CDA + \angle ACD = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle ACD \\[3ex] \angle CAD + 60 + 90 = 180 \\[3ex] \angle CAD + 150 = 180 \\[3ex] \angle CAD = 180 - 150 \\[3ex] \angle CAD = 30^\circ $

Construction: Label the points on the circle


$ \angle CBD = \angle CDB = 35^\circ ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle CBD \\[3ex] m = \angle CBD = 35^\circ ... \angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal $

Construction: Label the points on the circle


$ \angle ABC = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] \angle ABC = n + 35 ...diagram \\[3ex] \implies \\[3ex] n + 35 = 90 \\[3ex] n = 90 - 35 \\[3ex] n = 55^\circ $

$ (i) \\[3ex] (a) \\[3ex] \angle HJM = 38^\circ ...alternate\;\;\angle s\;\;between\;\;parallel\;\;lines:|HJ| \;\;and\;\; |MK|\;\;are\;\;equal \\[3ex] (b) \\[3ex] \angle MJK = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle\;\;is\;\;a\;\;right\;\;\angle \\[3ex] (ii) \\[3ex] (a) \\[3ex] \underline{\triangle MJK} \\[3ex] \angle MKJ + \angle MJK + \angle JMK = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;a\;\;\triangle \\[3ex] \angle MKJ + 90 + 38 = 180 \\[3ex] \angle MKJ + 128 = 180 \\[3ex] \angle MKJ = 180 - 128 \\[3ex] \angle MKJ = 52^\circ \\[3ex] \angle MLJ = \angle MKJ = 52^\circ ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] (b) \\[3ex] $ We can solve (b) using at least two appraoches.
Use any approach you prefer.

$ \underline{First\;\;Approach} \\[3ex] \underline{\triangle MJL} \\[3ex] \angle LMJ = \angle MLJ = 52^\circ ...base\;\;\angle s\;\;of\;\;isosceles\;\; \triangle MJL \\[3ex] But:\;\; \angle LMK + \angle JMK = \angle LMJ ... diagram \\[3ex] \angle LMK + 38 = 52 \\[3ex] \angle LMK = 52 - 38 \\[3ex] \angle LMK = 14^\circ \\[3ex] \angle LJK = \angle LMK = 14^\circ ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] \underline{Second\;\;Approach} \\[3ex] \underline{\triangle MJL} \\[3ex] \angle LMJ = \angle MLJ = 52^\circ ...base\;\;\angle s\;\;of\;\;isosceles\;\; \triangle MJL \\[3ex] \angle MJL + \angle LMJ + \angle MLJ = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;a\;\;\triangle \\[3ex] \angle MJL + 52 + 52 = 180 \\[3ex] \angle MJL + 104 = 180 \\[3ex] \angle MJL = 180 - 104 \\[3ex] \angle MJL = 76^\circ \\[3ex] \angle MJL + \angle LJK = \angle MJK ...diagram \\[3ex] 76 + \angle LJK = 90 \\[3ex] \angle LJK = 90 - 76 \\[3ex] \angle LJK = 14^\circ \\[3ex] (c) \\[3ex] \underline{Cyclic\;\;Quadrilateral\;\;LMHJ} \\[3ex] \angle H + \angle L = 180^\circ ...sum\;\;of\;\;interior\;\;\angle s\;\;of\;\;Quadrilateral\;LMHJ \\[3ex] \angle L = \angle MLJ = 52^\circ ... diagram \\[3ex] \angle H = \angle JHM ... diagram \\[3ex] \implies \\[3ex] \angle JHM + 52 = 180 \\[3ex] \angle JHM = 180 - 52 \\[3ex] \angle JHM = 128^\circ $

$ \angle ABC = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] \angle AOD = 2 * \angle ABD ... \angle \;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] 98 = 2 * \angle ABD \\[3ex] 2 * \angle ABD = 98 \\[3ex] \angle ABD = \dfrac{98}{2} \\[5ex] \angle ABD = 49^\circ \\[3ex] \angle ABC = \angle ABD + \angle DBC ...diagram \\[3ex] 90 = 49 + \angle DBC \\[3ex] 49 + \angle DBC = 90 \\[3ex] \angle DBC = 90 - 49 \\[3ex] \angle DBC = 41^\circ $

$ \angle QST = \angle QRT = 42^\circ ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] \angle RST = 90^\circ ... \angle \;\;in\;\;a\;\;semicircle \\[3ex] \angle RSQ + \angle QST = \angle RST ...diagram \\[3ex] \angle RSQ + 42 = 90 \\[3ex] \angle RSQ = 90 - 42 \\[3ex] \angle RSQ = 48^\circ $

$ \angle SQR + \angle PQS = 180^\circ ... \angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] \angle SQR + 124 = 180 \\[3ex] \angle SQR = 180 - 124 \\[3ex] \angle SQR = 56^\circ \\[3ex] \angle STR = \angle SQR = 56^\circ ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal $

$ (a) \\[3ex] \angle BAD + \angle BCD = 180^\circ ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quadrilateral\;\;are\;\;supplementary \\[3ex] \angle BAD + 126 = 180 \\[3ex] \angle BAD = 180 - 126 \\[3ex] \angle BAD = 54^\circ \\[5ex] (b) \\[3ex] Reflex\;\;\angle BOD = 2 * \angle BCD ... \angle \;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] Reflex\;\;\angle BOD = 2 * 126 \\[3ex] Reflex\;\;\angle BOD = 252^\circ \\[3ex] Obtuse\;\;\angle BOD + Reflex\;\;\angle BOD = 360^\circ ...\angle s \;\;at\;\;a\;\;point \\[3ex] Obtuse\;\;\angle BOD + 252 = 360 \\[3ex] Obtuse\;\;\angle BOD = 360 - 252 \\[3ex] Obtuse\;\;\angle BOD = 108^\circ \\[3ex] \angle ODC + \angle BCD + \angle CBO + Obtuse\;\;\angle BOD = 360^\circ ...sum\;\;of\;\;interior\;\;\angle s\;\;of\;\;Quadrilateral\;CBOD \\[3ex] \angle ODC + 126 + 62 + 108 = 360 \\[3ex] \angle ODC + 296 = 360 \\[3ex] \angle ODC = 360 - 296 \\[3ex] \angle ODC = 64^\circ $

Construction: Join the chord from Point B to Point E


$ Sum\;\;of\;\;the\;\;interior\;\;\angle s\;\;of\;\;a\;\;regular\;\;polygon \\[3ex] = 180(n - 2) \\[3ex] For\;\;a\;\;regular\;\;hexagon:\;\;n = 6 \;\;(6\;\;sides) \\[3ex] Sum\;\;of\;\;the\;\;interior\;\;\angle s\;\;of\;\;a\;\;regular\;\;hexagon \\[3ex] = 180(6 - 2) \\[3ex] = 180(4) \\[3ex] = 720^\circ \\[3ex] Each\;\;interior\;\;\angle\;\;of\;\;a\;\;regular\;\;hexagon \\[3ex] = \dfrac{Sum}{n} \\[5ex] = \dfrac{720}{6} \\[5ex] = 120^\circ \\[3ex] \implies \\[3ex] \angle A = \angle B = \angle C = \angle D = \angle E = \angle F = 120^\circ \\[3ex] \underline{Cyclic\;\;Quadrilateral\;\;EBCD} \\[3ex] \angle BCD = \angle C = 120^\circ ...diagram \\[3ex] \angle BED + \angle BCD = 180^\circ \\[3ex] ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quad\;\;are\;\;supplementary \\[3ex] \angle BED + 120 = 180 \\[3ex] \angle BED = 180 - 120 \\[3ex] \angle BED = 60^\circ \\[3ex] \underline{\triangle OED} \\[3ex] \angle OED = \angle BED = 60^\circ ... diagram \\[3ex] \angle OED = \angle ODE = 60^\circ ...base\;\;\angle s\;\;of\;\;isosceles\;\; \triangle OED \\[3ex] \implies \\[3ex] |OE| = |OD| = 15\;cm ... isosceles\;\; \triangle OED \\[3ex] \angle EOD + \angle OED + \angle ODE = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle OED \\[3ex] \angle EOD + 60 + 60 = 180 \\[3ex] \angle EOD + 120 = 180 \\[3ex] \angle EOD = 180 - 120 \\[3ex] \angle EOD = 60^\circ \\[3ex] \implies \\[3ex] \triangle OED \;\;is\;\;an\;\;equilateral\;\;\triangle \\[3ex] \angle EOD = \angle OED = \angle ODE = 60^\circ \\[3ex] Similarly: \\[3ex] |OD| = |OE| = |ED| = 15\;cm ...equilateral\;\; \triangle OED \\[3ex] \implies \\[3ex] |AB| = |ED| = 15\;cm ...regular\;\;hexagon $

$ \angle RPQ = \angle QSR = y ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] \angle QRT = \angle RPQ ...\angle\;\;between\;\;tangent\;RT \;\;and\;\;chord\;RQ = \angle\;\;in\;\;alternate\;\;segment \\[3ex] 52 = y \\[3ex] y = 52^\circ $

$ \underline{\triangle RPQ} \\[3ex] \angle PRQ + \angle RPQ + \angle PQR = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;a\;\;\triangle \\[3ex] x + y + 70 = 180 \\[3ex] x + 52 + 70 = 180 \\[3ex] x + 122 = 180 \\[3ex] x = 180 - 122 \\[3ex] x = 58^\circ $

$ (a) \\[3ex] \angle OCE = 90^\circ ...radius\;OC \perp tangent\;ECF\;\;at\;\;point\;\;of\;\;contact\;C \\[3ex] (b) \\[3ex] \underline{\triangle OEC} \\[3ex] \triangle OEC \;\;is\;\;a\;\;right\;\;\triangle \\[3ex] SOHCAHTOA \\[3ex] |OC| = radius = r \\[3ex] \tan 27^\circ = \dfrac{opp}{adj} = \dfrac{|OC|}{|EC|} \\[5ex] \tan 27 = \dfrac{r}{12} \\[5ex] r = |OC| = 12\tan 27 \\[3ex] Area = \dfrac{|EC| * |OC|}{2} \\[5ex] Area = \dfrac{12 * 12 * \tan 27}{2} \\[5ex] Area = 72\tan 27 \\[3ex] Area = 72(0.5095254495) \\[3ex] Area = 36.68583236 \\[3ex] Area \approx 36.7cm^2 ...to\;\;3\;\;significant\;\;figures \\[3ex] (c) \\[3ex] \angle EOC + \angle OEC + \angle OCE = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle OEC \\[3ex] \angle EOC + 27 + 90 = 180 \\[3ex] \angle EOC + 117 = 180 \\[3ex] \angle EOC = 180 - 117 \\[3ex] \angle EOC = 63^\circ \\[3ex] Similarly: \\[3ex] \angle OAE = 90^\circ ...radius\;OA \perp tangent\;EA\;\;at\;\;point\;\;of\;\;contact\;A \\[3ex] \angle OEA = \angle OEC = 27^\circ \\[3ex] ... two\;\;tangents:\;AE\;\;and\;\;CE\;\;drawn\;\;from\;\;the\;\;same\;\;external\;\;point:\;E\;\;bisects\;\;the\;\;\angle\;\;formed\;\;by\;\;the\;\;tangents:\;\angle AEC \\[3ex] \implies \\[3ex] \angle EOA = 63^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle OEA \\[3ex] Obtuse\;\;\angle AOC = \angle EOA + \angle EOC ...diagram \\[3ex] Obtuse\;\;\angle AOC = 63 + 63 \\[3ex] Obtuse\;\;\angle AOC = 126^\circ \\[3ex] Obtuse\;\;\angle AOC = 2 * \angle ABC ... \angle \;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] 126 = 2 * \angle ABC \\[3ex] 2 * \angle ABC = 126 \\[3ex] \angle ABC = \dfrac{126}{2} \\[5ex] \angle ABC = 63^\circ \\[3ex] (d) \\[3ex] Obtuse\;\;\angle AOC + Reflex\;\;\angle AOC = 360^\circ ...\angle s \;\;at\;\;a\;\;point \\[3ex] 126 + Reflex\;\;\angle AOC = 360 \\[3ex] Reflex\;\;\angle AOC = 360 - 126 \\[3ex] Reflex\;\;\angle AOC = 234^\circ \\[3ex] Reflex\;\;\angle AOC = 2 * \angle ADC ... \angle \;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] 234 = 2 * \angle ADC \\[3ex] 2 * \angle ADC = 234 \\[3ex] \angle ADC = \dfrac{234}{2} \\[5ex] \angle ADC = 117^\circ \\[3ex] $ (e)
Let us do some construction to see the theorem we need to apply (Alternate Segment Theorem) before we find ∠BAO


$ \angle BAC = \angle BCF = 59^\circ ...\angle\;\;between\;\;tangent\;BCF\;\;and\;\;chord\;BC = \angle\;\;in\;\;alternate\;\;segment \\[3ex] \underline{\triangle AOC} \\[3ex] \angle OAC = \angle OCA = p ...base\;\;\angle s\;\;of\;\;isosceles\;\; \triangle AOC \\[3ex] \angle OAC + \angle OCA + Obtuse\;\;\angle AOC = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle AOC \\[3ex] p + p + 126 = 180 \\[3ex] 2p = 180 - 126 \\[3ex] 2p = 54 \\[3ex] p = \dfrac{54}{2} \\[5ex] p = 27 \\[3ex] \therefore \angle OAC = \angle OCA = 27^\circ \\[3ex] \angle BAC = \angle BAO + \angle OAC ...diagram \\[3ex] 59 = \angle BAO + 27 \\[3ex] \angle BAO + 27 = 59 \\[3ex] \angle BAO = 59 - 27 \\[3ex] \angle BAO = 32^\circ $

$ (i) \\[3ex] \angle QRS + \angle QPS = 180^\circ \\[3ex] ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quad\;\;are\;\;supplementary \\[3ex] \angle QRS + 73 = 180 \\[3ex] \angle QRS = 180 - 73 \\[3ex] \angle QRS = 107^\circ \\[3ex] (ii) \\[3ex] $ We can solve (ii) using at least two approaches.
Use any approach that you prefer.

$ \underline{First\;\;Approach} \\[3ex] \angle PQR + \angle PSR = 180^\circ \\[3ex] ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quad\;\;are\;\;supplementary \\[3ex] \angle PQR + 82 = 180 \\[3ex] \angle PQR = 180 - 82 \\[3ex] \angle PQR = 98^\circ \\[3ex] But:\;\; \angle RQS + \angle PQS = \angle PQR ... diagram \\[3ex] \angle RQS + 55 = 98 \\[3ex] \angle RQS = 98 - 55 \\[3ex] \angle RQS = 43^\circ \\[3ex] \underline{Second\;\;Approach} \\[3ex] \angle PSQ + \angle QPS + \angle PQS = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle PSQ \\[3ex] \angle PSQ + 73 + 55 = 180 \\[3ex] \angle PSQ + 128 = 180 \\[3ex] \angle PSQ = 180 - 128 \\[3ex] \angle PSQ = 52^\circ \\[3ex] But:\;\;\angle PSQ + \angle QSR = \angle PSR ...diagram \\[3ex] 52 + \angle QSR = 82 \\[3ex] \angle QSR = 82 - 52 \\[3ex] \angle QSR = 30^\circ \\[3ex] \angle RQS + \angle QSR + \angle QRS = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle SRQ \\[3ex] \angle RQS + 30 + 107 = 180 \\[3ex] \angle RQS + 137 = 180 \\[3ex] \angle RQS = 180 - 137 \\[3ex] \angle RQS = 43^\circ \\[3ex] (iii) \\[3ex] $ Construction: Draw the chord from point P to point R
Use a red color


$ \angle PRQ = \angle PSQ = 52^\circ ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] $

$ \angle ORP = \angle ORQ = \angle QRO = 20^\circ ... diagram \\[3ex] |OP| = |OQ| ...radius \\[3ex] \angle OPQ = \angle OQP ...base\;\;\angle s\;\;of\;\;isosceles\;\; \triangle OPQ \\[3ex] |QR| = |OP| ... given \\[3ex] \implies \\[3ex] |QR| = |OQ| ...transitive\;\;property \\[3ex] \implies \\[3ex] \angle QOR = \angle QRO = 20^\circ ...base\;\;\angle s\;\;of\;\;isosceles\;\; \triangle QOR \\[3ex] \angle OQP = \angle QOR + \angle QRO ...exterior\;\;\angle\;\;of\;\;a\;\;\triangle = sum\;\;of\;\;the\;\;two\;\;interior\;\;opposite\;\;\angle s \\[3ex] \angle OQP = 20 + 20 \\[3ex] \angle OQP = 40^\circ \\[3ex] \angle OPQ = \angle OQP = 40^\circ ...base\;\;\angle s\;\;of\;\;isosceles\;\; \triangle OPQ \\[3ex] \angle POQ + \angle OPQ + \angle OQP = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle OPQ \\[3ex] \angle POQ + 40 + 40 = 180 \\[3ex] \angle POQ + 80 = 180 \\[3ex] \angle POQ = 180 - 80 \\[3ex] \angle POQ = 100^\circ \\[3ex] x + \angle POQ + \angle QOR = 180^\circ ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] x + 100 + 20 = 180 \\[3ex] x + 120 = 180 \\[3ex] x = 180 - 120 \\[3ex] x = 60^\circ $

Construction: Draw the radius from Point O to POint A


$ |AC| = |CB| = \dfrac{|AB|}{2} \\[5ex] ...line\;OA\;\;\perp\;\;to\;\;chord\;AB;\;\;bisects\;\;the\;\;chord \\[3ex] |AC| = \dfrac{16}{2} = 8\;cm \\[5ex] |OA| = 10\;cm ...radius \\[3ex] \underline{Right\;\triangle\;\;OCA} \\[3ex] |OC|^2 + |AC|^2 = |OA|^2 ...Pythagorean\;\;theorem \\[3ex] |OC|^2 + 8^2 = 10^2 \\[3ex] |OC|^2 + 64 = 100 \\[3ex] |OC|^2 = 100 - 64 \\[3ex] |OC|^2 = 36 \\[3ex] |OC| = \sqrt{36} \\[3ex] |OC| = 6\;cm $

$ (a) \\[3ex] \angle CDB = 40^\circ ...alternate\;\;\angle s\;\;between\;\;parallel\;\;lines:|EC| \;\;and\;\; |AB|\;\;are\;\;equal \\[3ex] \angle DAE = 40^\circ ...\angle\;\;between\;\;tangent\;EA\;\;and\;\;chord\;DA = \angle\;\;in\;\;alternate\;\;segment \\[3ex] \angle EDA + \angle DAE + \angle AED = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle AED \\[3ex] \angle EDA + 40 + 45 = 180 \\[3ex] \angle EDA + 85 = 180 \\[3ex] \angle EDA = 180 - 85 \\[3ex] \angle EDA = 95^\circ \\[3ex] (i) \\[3ex] \angle EDA + \angle ADB + \angle CDB = 180^\circ ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] 95 + \angle ADB + 40 = 180 \\[3ex] \angle ADB + 135 = 180 \\[3ex] \angle ADB = 180 - 135 \\[3ex] \angle ADB = 45^\circ \\[3ex] (ii) \\[3ex] $ Construction: Draw the chord from point B to point C
This forms the cyclic quadrilateral ABCD


$ Obtuse\;\angle BOC = 2 * \angle CDB ... \angle \;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] \angle BOC = 2 * 40 \\[3ex] \angle BOC = 80^\circ \\[3ex] \underline{\triangle BOC} \\[3ex] \angle OCB = \angle OBC = p ...base\;\;\angle s\;\;of\;\;isosceles\;\; \triangle BOC \\[3ex] \angle OCB + \angle OBC + \angle BOC = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;a\;\;\triangle \\[3ex] p + p + 80 = 180 \\[3ex] 2p = 180 - 80 \\[3ex] 2p = 100 \\[3ex] p = \dfrac{100}{2} \\[5ex] p = 50 \\[3ex] \therefore \angle OCB = \angle OBC = 50^\circ \\[3ex] \underline{\triangle ADB} \\[3ex] \angle ADB + \angle DAB + \angle ABD = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;a\;\;\triangle \\[3ex] 45 + \angle DAB + 40 = 180 \\[3ex] \angle DAB + 85 = 180 \\[3ex] \angle DAB = 180 - 85 \\[3ex] \angle DAB = 95^\circ \\[3ex] \underline{Cyclic\;\;Quadrilateral\;ABCD} \\[3ex] \angle A + \angle C = 180^\circ ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quadrilateral\;\;are\;\;supplementary \\[3ex] \angle A = \angle DAB = 95^\circ \\[3ex] \angle C = \angle OCD + \angle OCB ... diagram \\[3ex] \angle C = \angle OCD + 50 \\[3ex] \implies \\[3ex] 95 + \angle OCD + 50 = 180 \\[3ex] \angle OCD + 145 = 180 \\[3ex] \angle OCD = 180 - 145 \\[3ex] \angle OCD = 35^\circ \\[3ex] (b) \\[3ex] $ Label: the lengths
Let the radius = r


We can solve (i) using at least two approaches.
Use any approach you prefer.
For KCSE students, use the 1st approach.

$ \underline{1st\;\;Approach:\;\;Intersecting\;\;Secant-Tangent\;\;Theorem} \\[3ex] |AE|^2 = |ED| * |EC| \\[3ex] But:\;\; |EC| = |ED| + |DC| ...diagram \\[3ex] \implies \\[3ex] |AE|^2 = 3.5(3.5 + 4.9) \\[3ex] |AE|^2 = 3.5(8.4) \\[3ex] |AE|^2 = 29.4 \\[3ex] |AE| = \sqrt{29.4} \\[3ex] |AE| = 5.422176685 \\[3ex] |AE| \approx 5.4\;cm \\[3ex] \underline{2nd\;\;Approach:\;\;Sine\;\;Law} \\[3ex] \underline{\triangle AED} \\[3ex] \dfrac{|AE|}{\sin \angle EDA} = \dfrac{|ED|}{\sin \angle DAE} ... Sine\;\;Law \\[5ex] \dfrac{|AE|}{\sin 95} = \dfrac{3.5}{\sin 40} \\[5ex] |AE| = \dfrac{3.5 * \sin 95}{\sin 40} \\[5ex] |AE| = \dfrac{3.486681443}{0.6427876097} \\[5ex] |AE| = 5.424313398 \\[3ex] |AE| \approx 5.4\;cm \\[3ex] (ii) \\[3ex] \underline{\triangle BDC} \\[3ex] \angle DBC + \angle BCD + \angle CDB = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;a\;\;\triangle \\[3ex] \angle BCD = \angle OCB + \angle OCD ... diagram \\[3ex] \angle BCD = 50 + 35 \\[3ex] \angle BCD = 85^\circ \\[3ex] \implies \\[3ex] \angle DBC + 85 + 40 = 180 \\[3ex] \angle DBC + 125 = 180 \\[3ex] \angle DBC = 180 - 125 \\[3ex] \angle DBC = 55^\circ \\[3ex] \dfrac{|BC|}{\sin \angle CDB} = \dfrac{|DC|}{\sin \angle DBC} ... Sine\;\;Law \\[5ex] \dfrac{|BC|}{\sin 40} = \dfrac{4.9}{\sin 55} \\[5ex] |BC| = \dfrac{4.9 * \sin 40}{\sin 55} ... eqn.(1) \\[5ex] \underline{\triangle BOC} \\[3ex] \dfrac{|BC|}{\sin \angle BOC} = \dfrac{|OC|}{\sin \angle OBC} ... Sine\;\;Law \\[5ex] |OC| = radius = r \\[3ex] \implies \\[3ex] \dfrac{|BC|}{\sin 80} = \dfrac{r}{\sin 50} \\[5ex] |BC| = \dfrac{r * \sin 80}{\sin 50} ... eqn.(2) \\[5ex] |BC| = |BC| \implies eqn.(1) = eqn.(2) \\[3ex] \dfrac{r \sin 80}{\sin 50} = \dfrac{4.9\sin 40}{\sin 55} \\[5ex] r = \dfrac{4.9 * \sin 40 * \sin 50}{\sin 80 * \sin 55} \\[5ex] r = \dfrac{2.412778995}{0.8067072841} \\[5ex] r = 2.990897743 \\[3ex] r \approx 3.0\;cm $

$ JM * TM = EM * RM ...Intersecting\;\;Chords\;\;Theorem \\[3ex] JM * TM = 8 * 15 \\[3ex] JM * TM = 120 \\[3ex] $ Test the options
12 * 9.5 = 114
14 * 8.5 = 119
16 * 7.5 = 120
18 * 6.5 = 117

$\overline{JM}$ and $\overline{TM}$ could be 16 and 7.5 because the product gives 120

$ \angle OPT = 90^\circ ... radius\;OP \perp tangent\;PT\:\:at\:\:point\:\:of\:\:contact\;P \\[3ex] \underline{\triangle QPT} \\[3ex] 30 + (x + 90) + (2x + x) = 180 ...sum\:\:of\:\:\angle s\:\:of\:\:a\:\:\triangle \\[3ex] 30 + x + 90 + 2x + x = 180 \\[3ex] 4x + 120 = 180 \\[3ex] 4x = 180 - 120 \\[3ex] 4x = 60 \\[3ex] x = \dfrac{60}{4} \\[5ex] x = 15^\circ \\[3ex] \angle PTO = 2x \\[3ex] \angle PTO = 2(15) \\[3ex] \angle PTO = 30^\circ $

(a) This is the minor segment (shaded blue)


OR
This is the major segment (shaded yellow)


(b) The straight line RS is a chord.

$ \angle OQP = \angle OPQ = 18^\circ ...base\;\;\angle s\;\;of\;\;isosceles\;\; \triangle OPQ \\[3ex] \angle OQT = 90^\circ \\[3ex] ...radius\;OQ \perp tangent\;QT\;\;at\;\;point\;\;of\;\;contact\;Q \\[3ex] \angle OQT = \angle OQP + \angle PQT ...diagram \\[3ex] 90 = 18 + \angle PQT \\[3ex] 18 + \angle PQT = 90 \\[3ex] \angle PQT = 90 - 18 \\[3ex] \angle PQT = 72^\circ $

For part (a), we can solve it using at least two approaches
Use any approach you prefer

$ (a) \\[3ex] \underline{First\;\;Approach} \\[3ex] \underline{Cyclic\;\;Quadrilateral\;\;PQRS} \\[3ex] \angle PSR + \angle PQR = 180^\circ ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quadrilateral\;\;are\;\;supplementary \\[3ex] \angle PSR + 105 = 180 \\[3ex] \angle PSR = 180 - 105 \\[3ex] \angle PSR = 75^\circ \\[3ex] \underline{Second\;\;Approach} \\[3ex] \angle QPR = \angle QRP = k ...base\;\;\angle s\;\;of\;\;isosceles\;\; \triangle PQR \\[3ex] \angle QPR + \angle QRP + \angle PQR = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle PQR \\[3ex] k + k + 105 = 180 \\[3ex] 2k = 180 - 105 \\[3ex] 2k = 75 \\[3ex] k = \dfrac{75}{2} \\[5ex] k = 37.5 \\[3ex] \therefore \angle QPR = \angle QRP = 37.5^\circ \\[3ex] \angle PSQ = \angle QRP = 37.5^\circ ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] \angle QSR = \angle QPR = 37.5^\circ ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] \angle PSR = \angle PSQ + \angle QSR ... diagram \\[3ex] \angle PSR = 37.5 + 37.5 \\[3ex] \angle PSR = 75^\circ \\[3ex] (b) \\[3ex] \angle SQR = \angle PSQ = 37.5^\circ ...alternate\;\;\angle s\;\;between\;\;parallel\;\;lines:|QR| \;\;and\;\; |PS|\;\;are\;\;equal \\[3ex] \angle PQS + \angle SQR = 105^\circ ... diagram \\[3ex] \angle PQS + 37.5 = 105 \\[3ex] \angle PQS = 105 - 37.5 \\[3ex] \angle PQS = 67.5^\circ $

$ \angle CFE = \dfrac{\overset{\huge\frown}{DB} + \overset{\huge\frown}{CE}}{2} \\[5ex] ... Angle\;\;of\;\;Intersecting\;\;Chords\;\;Theorem \\[3ex] \overset{\huge\frown}{CD} + \overset{\huge\frown}{CE} = 180^\circ ...arcs\;\;of\;\;semicircle \\[3ex] \overset{\huge\frown}{CE} = 180 - \overset{\huge\frown}{CD} \\[3ex] \overset{\huge\frown}{CE} = 180 - 46 \\[3ex] \overset{\huge\frown}{CE} = 134^\circ \\[3ex] \implies \\[3ex] \angle CFE = \dfrac{102 + 134}{2} \\[5ex] \angle CFE = \dfrac{236}{2} \\[5ex] \angle CFE = 118^\circ $

$ (a) \\[3ex] \angle CAB = \angle CBA ...base\;\;\angle s\;\;of\;\;isosceles\;\; \triangle ABC \;\;because\;\; AC = BC \\[3ex] \angle ACD = \angle CBA ...\angle\;\;between\;\;tangent\;CD\;\;and\;\;chord\;AC = \angle\;\;in\;\;alternate\;\;segment \\[3ex] \implies \\[3ex] \angle CAB = \angle ACD...equality\;\;with\;\;\angle CBA \\[3ex] \implies \\[3ex] \angle CAB = \angle ACD ...alternate\;\;\angle s\;\;between\;\;parallel\;\;lines:|AB| \;\;and\;\; |DC|\;\;are\;\;equal \\[3ex] \therefore |AB| || |DC| \\[3ex] (b) \\[3ex] \angle CAB = \angle CBA = \angle ACB = 60^\circ ... \triangle ABC \;\;is\;\;equilateral\;\triangle \\[3ex] \underline{Options} \\[3ex] |AB| || |DC|...Yes\;\;because\;\;alternate\;\;\angle s\;\;between\;\;parallel\;\;lines:|AB| \;\;and\;\; |DC|\;\;are\;\;equal \\[3ex] |AC|\;\;bisects\;\;\angle BCD ...Yes\;\;because \\[3ex] \angle ACB = \angle CAB\;\;and \\[3ex] \angle CAB = \angle ACD \\[3ex] \therefore \angle ACB = \angle ACD ...equality\;\;with\;\;\angle CAB \\[3ex] |AC|\;\;bisects\;\;\angle BAD ...No \\[3ex] $ Because there is no sufficient information to prove that $\angle CAD = \angle ACD$ or
Because there is no sufficient information to $\angle CAD = \angle CAB$

$ SOHCAHTOA \\[3ex] \tan 20 = \dfrac{\overline{OA}}{\overline{AP}} \\[5ex] 0.364 = \dfrac{\overline{OA}}{12} \\[5ex] \overline{OA} = 12 * 0.364 \\[3ex] \overline{OA} = 4.368 \\[3ex] \overline{OB} = \overline{OA} = 4.368\;cm ...radius \\[3ex] \cos 20 = \dfrac{\overline{AP}}{\overline{OP}} \\[5ex] 0.940 = \dfrac{12}{\overline{OP}} \\[5ex] 0.94 * \overline{OP} = 12 \\[3ex] \overline{OP} = \dfrac{12}{0.94} \\[5ex] \overline{OP} = 12.76595745 \\[3ex] \overline{OP} = \overline{OB} + \overline{BP} ...diagram \\[3ex] \overline{OB} + \overline{BP} = \overline{OP} \\[3ex] 4.368 + \overline{BP} = 12.76595745 \\[3ex] \overline{BP} = 12.76595745 - 4.368 \\[3ex] \overline{BP} = 8.397957447 \\[3ex] \overline{BP} \approx 8.4\;cm $

$ (a) \\[3ex] \underline{Cyclic\;\;Quadrilateral\;ABCD} \\[3ex] \angle BCD + \angle BAD = 180^\circ ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quadrilateral\;\;are\;\;supplementary \\[3ex] \angle BCD + 112 = 180 \\[3ex] \angle BCD = 180 - 112 \\[3ex] \angle BCD = 68^\circ \\[3ex] \angle BCO + \angle DCO = \angle BCD ... diagram \\[3ex] \angle BCO + 33 = 68 \\[3ex] \angle BCO = 68 - 33 \\[3ex] \angle BCO = 35^\circ \\[3ex] y = \angle BCO = 35^\circ ...base\;\;\angle s\;\;of\;\;isosceles\;\; \triangle COB \\[3ex] (b) \\[3ex] \underline{\triangle COB} \\[3ex] \angle COB + \angle BCO + y = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;a\;\;\triangle \\[3ex] \angle COB + 35 + 35 = 180 \\[3ex] \angle COB + 70 = 180 \\[3ex] \angle COB = 180 - 70 \\[3ex] \angle COB = 110^\circ \\[3ex] \angle COB = 2 * z ... \angle \;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] 110 = 2z \\[3ex] 2z = 110 \\[3ex] z = \dfrac{110}{2} \\[5ex] z = 55^\circ $

$ \angle AEC = \dfrac{\overset{\huge\frown}{AC} + \overset{\huge\frown}{DB}}{2} \\[5ex] ... Angle\;\;of\;\;Intersecting\;\;Chords\;\;Theorem \\[3ex] 58 = \dfrac{72 + \overset{\huge\frown}{DB}}{2} \\[5ex] 58(2) = 72 + \overset{\huge\frown}{DB} \\[3ex] 116 = 72 + \overset{\huge\frown}{DB} \\[3ex] 72 + \overset{\huge\frown}{DB} = 116 \\[3ex] \overset{\huge\frown}{DB} = 116 - 72 \\[3ex] \overset{\huge\frown}{DB} = 44^\circ $

$ \angle DAC = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle\;\;is\;\;a\;\;right\;\;\angle \\[3ex] \angle DCA + \angle DAC + \angle ADC = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle DAC \\[3ex] \angle DCA + 90 + 55 = 180 \\[3ex] \angle DCA + 145 = 180 \\[3ex] \angle DCA = 180 - 145 \\[3ex] \angle DCA = 35^\circ \\[3ex] \angle BAC = \angle DCA = 35^\circ ...alternate\;\;\angle s\;\;between\;\;parallel\;\;lines:|DC| \;\;and\;\; |AB|\;\;are\;\;equal \\[3ex] \underline{Cyclic\;\;Quadrilateral\;ABCD} \\[3ex] \angle D + \angle B = 180^\circ ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quadrilateral\;\;are\;\;supplementary \\[3ex] \angle D = \angle ADC = 55^\circ ... diagram \\[3ex] \angle B = \angle ABC ...diagram \\[3ex] \implies \\[3ex] 55 + \angle ABC = 180 \\[3ex] \angle ABC = 180 - 55 \\[3ex] \angle ABC = 125^\circ \\[3ex] \underline{\triangle ABC} \\[3ex] \angle ACB + \angle ABC + \angle BAC = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;a\;\;\triangle \\[3ex] \angle ACB + 125 + 35 = 180 \\[3ex] \angle ACB + 160 = 180 \\[3ex] \angle ACB = 180 - 160 \\[3ex] \angle ACB = 20^\circ $

Construction: Draw the radii from centre O to these points on the circumference:
point A
point B
point C
point D


$ radius = r = |OA| = |OB| = |OC| = |OD| = 13\;cm \\[3ex] ...given\;\;and\;\;diagram \\[3ex] |AN| = |NB| = \dfrac{|AB|}{2} = \dfrac{24}{2} = 12\;cm \\[3ex] ...line:\;ON\;\;\perp\;\;to\;\;chord:\;AB;\;\;bisects\;\;the\;\;chord \\[3ex] \underline{\triangle ONB} \\[3ex] |ON|^2 + |NB|^2 = |OB|^2 ... Pythagorean\;\;theorem \\[5ex] |ON|^2 + 12^2 = 13^2 \\[3ex] |ON|^2 + 144 = 169 \\[3ex] |ON|^2 = 169 - 144 \\[3ex] |ON|^2 = 25 \\[3ex] |ON| = \sqrt{25} \\[3ex] |ON| = 5 \\[5ex] |CM| = |MD| = \dfrac{|CD|}{2} = \dfrac{10}{2} = 5\;cm \\[3ex] ...line:\;OM\;\;\perp\;\;to\;\;chord:\;CD;\;\;bisects\;\;the\;\;chord \\[3ex] \underline{\triangle OMD} \\[3ex] |OM|^2 + |MD|^2 = |OD|^2 ... Pythagorean\;\;theorem \\[5ex] |OM|^2 + 5^2 = 13^2 \\[3ex] |OM|^2 + 25 = 169 \\[3ex] |OM|^2 = 169 - 25 \\[3ex] |OM|^2 = 144 \\[3ex] |OM| = \sqrt{144} \\[3ex] |OM| = 12 \\[5ex] \underline{Distance\;\;between\;\;chord\;|AB|\;\;and\;\;chord\;|CD|} \\[3ex] distance = |ON| + |OM| \\[3ex] distance = 5 + 12 \\[3ex] distance = 17\;cm $

$ x + 58 = 90^\circ ...radius\;OG \perp tangent\;JGH\;\;at\;\;point\;\;of\;\;contact\;G \\[3ex] x = 90 - 58 \\[3ex] x = 32^\circ \\[5ex] y = 58^\circ ... \angle\;\;in\;\;alternate\;\;segment = \angle\;\;between\;\;tangent\;JGH\;\;and\;\;chord\;GF $

$ RS * RT = RQ * RP ...Intersecting\;\;Secants\;\;Theorem \\[3ex] 6 * (6 + 4) = RQ * 15 \\[3ex] 6 * 10 = RQ * 15 \\[3ex] RQ * 15 = 60 \\[3ex] RQ = \dfrac{60}{15} \\[5ex] |RQ| = 4 $

Let us represent the information using a diagram.
Be reminded that the measure of an arc is the angle subtended at the center by the arc.


$ \angle ORA = \angle OAR = p ...base\;\;\angle s\;\;of\;\;isosceles\;\; \triangle ROA \\[3ex] \angle ORA + \angle OAR + \angle ROA = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle ROA \\[3ex] p + p + 100 = 180 \\[3ex] 2p = 180 - 100 \\[3ex] 2p = 80 \\[3ex] p = \dfrac{80}{2} \\[5ex] p = 40 \\[3ex] \therefore \angle ORA = \angle OAR = 40^\circ \\[3ex] \angle OAP = 90^\circ ...radius\;OA \perp tangent\;PA\;\;at\;\;point\;\;of\;\;contact\;A \\[3ex] \angle RAP = \angle OAR + \angle OAP ...diagram \\[3ex] \angle RAP = 40 + 90 = 130^\circ \\[3ex] \angle PRA = \angle ORA = 40^\circ ...diagram \\[3ex] \angle P = \angle RPA ...diagram \\[3ex] \angle RPA + \angle RAP + \angle PRA = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle RPA \\[3ex] \angle P + 130 + 40 = 180 \\[3ex] \angle P + 170 = 180 \\[3ex] \angle P = 180 - 170 \\[3ex] \angle P = 10^\circ $

$ \underline{\triangle BAX} \\[3ex] 30 + \angle BAX = 70 \\[3ex] ...exterior\;\;\angle\;\;of\;\;a\;\;\triangle = sum\;\;of\;\;two\;\;interior\;\;opposite\;\;\angle s \\[3ex] \angle BAX = 70 - 30 \\[3ex] \angle BAX = 40^\circ \\[3ex] \angle BAC = \angle BAX = 40^\circ ... diagram \\[3ex] \angle BOC = 2 * \angle BAC ... \angle \;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] \angle BOC = 2 * 40 \\[3ex] \angle BOC = 80^\circ \\[3ex] (a) \\[3ex] Acute\;\angle BOC + Reflex\;\angle BOC = 360^\circ ... \angle s\;\;at\;\;a\;\;point \\[3ex] 80 + Reflex\;\angle BOC = 360 \\[3ex] Reflex\;\angle BOC = 360 - 80 \\[3ex] Reflex\;\angle BOC = 280^\circ \\[3ex] (b) \\[3ex] \angle BOC + \angle COX = 180^\circ ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] 80 + \angle COX = 180 \\[3ex] \angle COX = 180 - 80 \\[3ex] \angle COX = 100^\circ \\[3ex] \angle CXO = \angle AXD = 70^\circ ...vertical\;\;\angle s\;\;are\;\;equal \\[3ex] \underline{\triangle XCO} \\[3ex] \angle XCO + \angle COX + \angle CXO = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;a\;\;\triangle \\[3ex] \angle XCO + 100 + 70 = 180 \\[3ex] \angle XCO + 170 = 180 \\[3ex] \angle XCO = 180 - 170 \\[3ex] \angle XCO = 10 \\[3ex] \angle ACO = \angle XCO = 10^\circ ... diagram $

(297.1) The exterior angle of a cyclic quadrilateral is equal to the interior opppsite angle.

$ (297.2.1) \\[3ex] M\hat{P}R = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle\;\;is\;\;a\;\;right\;\;\angle \\[3ex] (297.2.2) \\[3ex] \hat{P_3} = 11^\circ ...given \\[3ex] \hat{M_2} = \hat{P_3} = 11^\circ ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] \hat{L_4} = \hat{M_2} = 11^\circ ...alternate\;\;\angle s\;\;between\;\;parallel\;\;lines:|LN| \;\;and\;\; |RM|\;\;are\;\;equal \\[3ex] \hat{P_1} = \hat{L_4} = 11^\circ ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] $ (297.2.3)
LN || RM ...given
Because parallel lines are equidistant from each other:
LR = NM because each point on the line LN is the same distance as the corresponding point on the other line RM
Hence the point L on the line LN is equidistant from the point R on the line RM
Also, the point N on the line LN is equidistant from the point M on the line RM

$ (297.2.4) \\[3ex] \hat{P_3} + \hat{P_2} + \hat{P_1} = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle\;\;is\;\;a\;\;right\;\;\angle \\[3ex] 11 + \hat{P_2} + 11 = 90 \\[3ex] \hat{P_2} + 22 = 90 \\[3ex] \hat{P_2} = 90 - 22 \\[3ex] \hat{P_2} = 68^\circ \\[3ex] \hat{M_1} = \hat{P_2} = 68^\circ ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] (297.2.5) \\[3ex] \underline{Cyclic\;\;Quadrilateral\;LRPN} \\[3ex] \hat{L_1} = \hat{P_3} + \hat{P_2} ... exterior\;\; \angle \:\:of\:\:a\:\:cyclic\:\:quadrilateral = interior\;\;opposite\;\;\angle \\[3ex] \hat{L_1} = 11 + 68 \\[3ex] \hat{L_1} = 79^\circ $

Let us analyze each option

$ (1) \\[3ex] \angle ABC = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] \rightarrow \triangle ABC \;\;is\;\;a\;\;right\;\;\triangle \\[3ex] YES \\[3ex] (2) \\[3ex] \angle MAB = \angle MBA ...base\;\;\angle s\;\;of\;\;isosceles\;\; \triangle ABM \\[3ex] YES \\[3ex] (3) \\[3ex] \overset{\huge\frown}{BC} = \angle BMC...intercepted\;\;\overset{\huge\frown}{} = central\;\;\angle \\[3ex] YES \\[3ex] (4) \\[3ex] \overset{\huge\frown}{AB} \ne \angle ACB \\[3ex] NO $

$ (a) \\[3ex] \angle SPR = \angle RST = 50^\circ ... \angle\;\;in\;\;alternate\;\;segment = \angle\;\;between\;\;tangent\;UST\;\;and\;\;chord\;SR \\[3ex] \angle SOR = 2 * \angle SPR ... \angle \;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] \angle SOR = 2 * 50 \\[3ex] \angle SOR = 100^\circ \\[3ex] (i) \\[3ex] \angle ORS = \angle OSR = k ...base\;\;\angle s\;\;of\;\;isosceles\;\; \triangle SOR \\[3ex] \angle ORS + \angle OSR + \angle SOR = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle SOR \\[3ex] k + k + 100 = 180 \\[3ex] 2k = 180 - 100 \\[3ex] 2k = 80 \\[3ex] k = \dfrac{80}{2} \\[5ex] k = 40 \\[3ex] \therefore \angle ORS = \angle OSR = 40^\circ \\[3ex] (ii) \\[3ex] \angle RTS + \angle RTV = 180^\circ ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] \angle RTS + 150 = 180 \\[3ex] \angle RTS = 180 - 150 \\[3ex] \angle RTS = 30^\circ \\[3ex] \angle PRS = \angle RTS + \angle RST ...exterior\;\;\angle\;\;of\;\;a\;\;\triangle = sum\;\;of\;\;the\;\;two\;\;interior\;\;opposite\;\;\angle s \\[3ex] \angle PRS = 30 + 50 \\[3ex] \angle PRS = 80^\circ \\[3ex] \angle USP = \angle PRS = 80^\circ \\[3ex] (iii) \\[3ex] \underline{\triangle SPR} \\[3ex] \angle PSR + \angle PRS + \angle SPR = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;a\;\;\triangle \\[3ex] \angle PSR + 80 + 50 = 180 \\[3ex] \angle PSR + 130 = 180 \\[3ex] \angle PSR = 180 - 130 \\[3ex] \angle PSR = 50^\circ \\[3ex] \underline{Cyclic\;\;Quadrilateral\; PQRS} \\[3ex] \angle Q + \angle S = 180^\circ ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quadrilateral\;\;are\;\;supplementary \\[3ex] \angle PQR = \angle Q ... diagram \\[3ex] \angle S = \angle PSR = 50^\circ ... diagram \\[3ex] \implies \\[3ex] \angle PQR + 50 = 180 \\[3ex] \angle PQR = 180 - 50 \\[3ex] \angle PQR = 130^\circ \\[3ex] (b) \\[3ex] $ We can solve (b) using at least two approaches.
Use any approach you prefer.
For KCSE students, use the 1st approach.

$ (i) \\[3ex] \underline{1st\;\;Approach:\;\;Intersecting\;\;Secant-Tangent\;\;Theorem} \\[3ex] |RT| * |PT| = |ST|^2 ...Intersecting\;\;Secant-Tangent\;\;Theorem \\[3ex] |PT| = |PR| + |RT| ... diagram \\[3ex] \implies \\[3ex] |RT|(|PR| + |RT|) = |ST|^2 \\[3ex] 7(|PR| + 7) = 9^2 \\[3ex] 7(|PR| + 7) = 81 \\[3ex] |PR| + 7 = \dfrac{81}{7} \\[5ex] |PR| + 7 = 11.57142857 \\[3ex] |PR| = 11.57142857 - 7 \\[3ex] |PR| = 4.571428571 \\[3ex] |PR| \approx 4.57\;cm...to\;\;3\;\;significant\;\;figures \\[3ex] (ii) \\[3ex] \underline{2nd\;\;Approach:\;\;Cosine\;\;Law\;\;and\;\;Sine\;\;Law} \\[3ex] |SR|^2 = |RT|^2 + |ST|^2 - 2 * |RT| * |ST| * \cos \angle RTS ...Cosine\;\;Law \\[3ex] |SR|^2 = 7^2 + 9^2 - 2(7)(9) \cos 30 \\[3ex] = 49 + 81 - 126(0.8660254038) \\[3ex] = 130 - 109.1192009 \\[3ex] = 20.88079912 \\[3ex] |SR| = \sqrt{20.88079912} \\[3ex] |SR| = 4.569551304 \\[3ex] \underline{\triangle PSR} \\[3ex] \dfrac{||PR|}{\sin \angle PSR} = \dfrac{|SR|}{\sin \angle SPR}...Sine\;\;Law \\[5ex] \dfrac{|PR|}{\sin 50} = \dfrac{4.569551304}{\sin 50} \\[5ex] same\;\;denominator;\;\;equate\;\;the\;\;numerators \\[3ex] |PR| = 4.569551304 \\[3ex] |PR| \approx 4.57\;cm...to\;\;3\;\;significant\;\;figures \\[3ex] (ii) \\[3ex] |OR| = |OS| = radius = r \\[3ex] \underline{First\;\;Approach:\;\;Sine\;\;Law} \\[3ex] \dfrac{||OR|}{\sin \angle OSR} = \dfrac{|SR|}{\sin \angle SOR}...Sine\;\;Law \\[5ex] \dfrac{r}{\sin 40} = \dfrac{4.569551304}{\sin 100} \\[5ex] r = \dfrac{4.569551304 * \sin 40}{\sin 100} \\[5ex] r = \dfrac{2.93725096}{0.984807753} \\[5ex] r = 2.982562791 \\[3ex] r \approx 2.98\;cm ...to\;\;3\;\;significant\;\;figures \\[3ex] \underline{Second\;\;Approach:\;\;Cosine\;\;Law} \\[3ex] |SR|^2 = |OR|^2 + |OS|^2 - 2 * |OR| * |OS| * \cos \angle SOR ...Cosine\;\;Law \\[3ex] 20.88079912 = r^2 + r^2 - 2 * r * r * \cos 100 \\[3ex] 20.88079912 = 2r^2 - 2r^2 * -0.1736481777 \\[3ex] 20.88079912 = 2r^2 + 0.3472963553r^2 \\[3ex] 20.88079912 = 2.347296355r^2 \\[3ex] 3.347296355r^2 = 20.88079912 \\[3ex] r^2 = \dfrac{20.88079912}{2.347296355} \\[5ex] r^2 = 8.8956808 \\[3ex] r = \sqrt{8.8956808} \\[3ex] r = 2.982562791 \\[3ex] r \approx 2.98\;cm ...to\;\;3\;\;significant\;\;figures $

$ \underline{larger\;\;circle} \\[3ex] radius = |OB| \\[3ex] Equation:\;\; x^2 + y^2 = 144 \\[3ex] Compare:\;\;x^2 + y^2 = r^2 \\[3ex] r^2 = 144 \\[3ex] r = +\sqrt{144} \\[3ex] r = 12 \\[3ex] |OB| = 12 \\[3ex] \underline{smaller\;\;circle} \\[3ex] radius = |OA| \\[3ex] \underline{Ratio\;\;of\;\;Radii} \\[3ex] |OA| : |OB| = 4 : 5 \\[3ex] |OA| : 12 = 4 : 5 \\[3ex] \dfrac{|OA|}{12} = \dfrac{4}{5} \\[5ex] |OA| = \dfrac{12(4)}{5} \\[5ex] |OA| = 9.6 \\[3ex] \underline{\triangle AOB} \\[3ex] |AB|^2 = |OA|^2 + |OB|^2 - 2 * |OA| * |OB| \cos \angle AOB \\[3ex] ... Cosine\;\;Law \\[3ex] 20^2 = 9.6^2 + 12^2 - 2(9.6)(12) \cos \angle AOB \\[3ex] 400 = 92.16 + 144 - 230.4 * \cos \angle AOB \\[3ex] 400 = 236.16 - 230.4 * \cos \angle AOB \\[3ex] 230.4 * \cos \angle AOB = 236.16 - 400 \\[3ex] 230.4 * \cos \angle AOB = -163.84 \\[3ex] \cos \angle AOB = -\dfrac{163.84}{230.4} \\[5ex] \cos \angle AOB = -0.7111111111 \\[3ex] \angle AOB = \cos^{-1}{-0.7111111111} \\[3ex] \angle AOB = 135.3253904^\circ $

$ (301.1.1) \\[3ex] TRS\;\;is\;\;a\;\;diameter \\[3ex] \angle TQR = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle\;\;is\;\;a\;\;right\;\;\angle \\[3ex] \hat{M_1} = \angle TQR = 90^\circ ...corresponding\;\;\angle s\;\;between\;\;parallel\;\;lines:|MS| \;\;and\;\; |QR|\;\;are\;\;equal \\[3ex] \hat{M_1} + \hat{M_2} = 180^\circ ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] 90 + \hat{M_2} = 180 \\[3ex] \hat{M_2} = 180 - 90 \\[3ex] \hat{M_2} = 90^\circ \\[3ex] \implies \\[3ex] \hat{M_2} = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle\;\;is\;\;a\;\;right\;\;\angle \\[3ex] \therefore SQ \;\;is\;\;a\;\;diameter \\[3ex] (301.1.2) \\[3ex] \hat{Q_3} = \hat{T} ...\angle\;\;between\;\;tangent\;QP\;\;and\;\;chord\;QR = \angle\;\;in\;\;alternate\;\;segment \\[3ex] \implies \\[3ex] \sin \hat{Q_3} = \sin \hat{T} \\[3ex] \underline{\triangle QRP} \\[3ex] \sin \hat{Q_3} = \dfrac{RP}{RQ} ...SOHCAHTOA \\[5ex] \underline{\triangle QTR} \\[3ex] \sin \hat{T} = \dfrac{RQ}{RT} ...SOHCAHTOA \\[5ex] \implies \\[3ex] \dfrac{RP}{RQ} = \dfrac{RQ}{RT} \\[5ex] |RP| * |RT| = |RQ|^2 \\[3ex] |RT| = \dfrac{|RQ|^2}{|RP|} \\[5ex] radius\;\;of\;\;the\;\;larger\;\;circle = r = |RS| = |ST| = \dfrac{|RT|}{2} ...diagram \\[5ex] \underline{\triangle QTR} \\[3ex] |MS| = \dfrac{|QR|}{2} ...Converse\;\;of\;\;the\;\;Midpoint\;\;Theorem \\[5ex] 14 = \dfrac{|QR|}{2} \\[5ex] |QR| = 14(2) \\[3ex] |RQ| = |QR| = 28\;units \\[3ex] \underline{\triangle QRP} \\[3ex] |QR|^2 = |PQ|^2 + |RP|^2 ...Pythagorean\;\;Theorem \\[3ex] 28^2 = (\sqrt{640})^2 + |RP|^2 \\[3ex] 784 = 640 + |RP|^2 \\[3ex] 640 + |RP|^2 = 784 \\[3ex] |RP|^2 = 784 - 640 \\[3ex] |RP|^2 = 144 \\[3ex] |RP| = \sqrt{144} \\[3ex] |RP| = 12\;units \\[3ex] |RT| = \dfrac{|RQ|^2}{|RP|} \\[5ex] |RT| = \dfrac{28^2}{12} \\[5ex] |RT| = \dfrac{784}{12} \\[5ex] r = \dfrac{|RT|}{2} = \dfrac{784}{12} \div 2 \\[5ex] r = \dfrac{784}{12} * \dfrac{1}{2} \\[5ex] r = \dfrac{392}{12} \\[5ex] r = \dfrac{98}{3}\;units $

$ (x - 10) + 75 = 180^\circ \\[3ex] ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quad\;\;are\;\;supplementary \\[3ex] x - 10 + 75 = 180 \\[3ex] x + 65 = 180 \\[3ex] x = 180 - 65 \\[3ex] x = 115^\circ \\[3ex] Similarly \\[3ex] (y + 30) + 80 = 180^\circ \\[3ex] ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quad\;\;are\;\;supplementary \\[3ex] y + 30 + 80 = 180 \\[3ex] y + 110 = 180 \\[3ex] y = 180 - 110 \\[3ex] y = 70^\circ $

Let us represent the information in the circle
Construction:
(a.) Let O be the center of the circle.
(b.) Indicate the central angles corresponding to the minor arcs.
(c.) Indicate the congruent lines (chords)


$ \underline{\triangle AOE} \\[3ex] \angle OAE = \angle OEA = p ...base\;\;\angle s\;\;of\;\;isosceles\;\; \triangle AOE \\[3ex] \angle OAE + \angle OEA + \angle AOE = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle AOE \\[3ex] p + p + 106 = 180 \\[3ex] 2p = 180 - 106 \\[3ex] 2p = 74 \\[3ex] p = \dfrac{74}{2} \\[5ex] p = 37 \\[3ex] \therefore \angle OAE = \angle OEA = 37^\circ \\[3ex] \angle ACE = \dfrac{\overset{\huge\frown}{AE} - \overset{\huge\frown}{BD}}{2} ... Angle\;\;of\;\;Intersecting\;\;Chords\;\;Theorem \\[5ex] \implies \\[3ex] \angle ACE = \dfrac{\angle AOE - \angle BOD}{2} \\[5ex] \angle ACE = \dfrac{106 - 64}{2} \\[5ex] \angle ACE = \dfrac{42}{2} \\[5ex] \angle ACE = 21^\circ \\[3ex] \underline{\triangle ACE} \\[3ex] |AB| \cong |DE| ...given \\[3ex] \implies \\[3ex] \angle CAE = \angle CEA = k \\[3ex] \angle CAE + \angle CEA + \angle ACE = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle ACE \\[3ex] k + k + 21 = 180 \\[3ex] 2k = 180 - 21 \\[3ex] 2k = 159 \\[3ex] k = \dfrac{159}{2} \\[5ex] k = 79.5 \\[3ex] \therefore \angle CAE = \angle CEA = 79.5^\circ \\[3ex] \angle OEA + \angle OED = \angle CEA ...diagram \\[3ex] 37 + \angle OED = 79.5 \\[3ex] \angle OED = 79.5 - 37 \\[3ex] \angle OED = 42.5^\circ \\[3ex] \underline{\triangle EOD} \\[3ex] \angle OED = \angle ODE = 42.5^\circ ...base\;\;\angle s\;\;of\;\;isosceles\;\; \triangle EOD \\[3ex] \angle OED + \angle ODE + \angle EOD = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle EOD \\[3ex] 42.5 + 42.5 + \angle EOD = 180 \\[3ex] 85 + \angle EOD = 180 \\[3ex] \angle EOD = 180 - 85 \\[3ex] \angle EOD = 95^\circ \\[3ex] \implies \\[3ex] \overset{\huge\frown}{DE} = \angle EOD = 85^\circ $

$ (a) \\[3ex] (i) \\[3ex] \underline{Intersecting\;\;Secants\;\;Theorem} \\[3ex] PU * PT = PQ * PS \\[3ex] PT = PU + UT \\[3ex] PT = 50 + 40 \\[3ex] PT = 90\;cm \\[3ex] PS = PQ + QS ... diagram \\[3ex] \implies \\[3ex] PU * PT = PQ(PQ + QS) \\[3ex] 50 * 90 = 30(30 + QS) \\[3ex] 30(30 + QS) = 50 * 90 \\[3ex] 30 + QS = \dfrac{50 * 90}{30} \\[5ex] 30 + QS = 150 \\[3ex] QS = 150 - 30 \\[3ex] QS = 120\;cm \\[3ex] (ii) \\[3ex] QR = RS = \dfrac{QS}{2} = \dfrac{120}{2} = 60\;cm \\[5ex] ...line\;OR\;\;\perp\;\;to\;\;chord\;QS;\;\;bisects\;\;the\;\;chord \\[3ex] \underline{\triangle ORS} \\[3ex] |OR|^2 + |RS|^2 = |OS|^2 ...Pythagorean\;\;Theorem \\[3ex] |OR|^2 + 60^2 = 61^2 \\[3ex] |OR|^2 = 61^2 - 60^2 \\[3ex] = (61 + 60)(61 - 60) \\[3ex] = 121(1) \\[3ex] = 121 \\[3ex] |OR| = \sqrt{121} \\[3ex] |OR| = 11\;cm \\[3ex] (b) \\[3ex] let\;\;\angle ROS = \alpha \\[3ex] SOHCAHTOA \\[3ex] (i) \\[3ex] \sin \alpha = \dfrac{RS}{OS} \\[5ex] \sin \alpha = \dfrac{60}{61} \\[5ex] \sin \alpha = 0.9836065574 \\[3ex] \alpha = \sin^{-1}(0.9836065574) \\[3ex] \alpha = 79.61114218 \\[3ex] \therefore \angle ROS \approx 79.6^\circ ...to\;\;1\;\;decimal\;\;place \\[3ex] (ii) \\[3ex] QO = OS = r = 61\;cm...radii \\[3ex] \angle QOR = \angle ROS = \alpha \\[3ex] ...line\;OR\;\;\perp\;\;to\;\;chord\;QS;\;\;bisects\;\;its\;\;arc \\[3ex] let\;\;\angle QOS = \beta \\[3ex] \beta = 2 * \alpha \\[3ex] \beta = 2(79.61114218) \\[3ex] \beta = 159.2222844^\circ \\[3ex] \overset{\huge\frown}{QS} = \dfrac{\beta}{360} * 2\pi r \\[5ex] \overset{\huge\frown}{QS} = \dfrac{159.2222844}{360} * 2 * \dfrac{22}{7} * 61 \\[5ex] \overset{\huge\frown}{QS} = \dfrac{427352.6112}{2520} \\[5ex] \overset{\huge\frown}{QS} = 169.5843695 \\[3ex] \overset{\huge\frown}{QS} \approx 169.6\;cm $

$ (i) \\[3ex] \underline{\triangle RCP} \\[3ex] \angle CRP = \angle CPR = 25^\circ ...base\;\;\angle s\;\;of\;\;isosceles\;\; \triangle RCP \\[3ex] \angle RCP + \angle CRP + \angle CPR = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle RCP \\[3ex] x + 25 + 25 = 180 \\[3ex] x + 50 = 180 \\[3ex] x = 180 - 50 \\[3ex] x = 30^\circ \\[5ex] (ii) \\[3ex] \angle PRS = 25^\circ ...alternate\;\;\angle s\;\;between\;\;parallel\;\;lines:|CP| \;\;and\;\; |RS|\;\;are\;\;equal \\[3ex] \angle CRS = \angle CRP + \angle PRS ...diagram \\[3ex] \angle CRS = 25 + 25 \\[3ex] \angle CRS = 50^\circ \\[3ex] \underline{\triangle RCS} \\[3ex] \angle CSR = \angle CRS = 50^\circ ...base\;\;\angle s\;\;of\;\;isosceles\;\; \triangle RCS \\[3ex] \angle RCS + \angle CSR + \angle CRS = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle RCS \\[3ex] y + 50 + 50 = 180 \\[3ex] y + 100 = 180 \\[3ex] y = 180 - 100 \\[3ex] y = 80^\circ \\[5ex] $

$ \underline{Circle} \\[3ex] Equation:\;\; x^2 + y^2 = \dfrac{25}{4} \\[5ex] Compare:\;\; x^2 + y^2 = r^2 \\[3ex] \implies \\[3ex] r^2 = \dfrac{25}{4} \\[5ex] r = +\sqrt{\dfrac{25}{4}} \\[5ex] r = \dfrac{5}{2}\;cm \\[5ex] |OA| = |OB| = |OC| = |OD| = |OE| = |OF| = radius = r = \dfrac{5}{2}\;cm \\[5ex] \underline{Hexagon} \\[3ex] Sum\;\;of\;\;interior\;\;\angle s \\[3ex] = 180(n - 2) \\[3ex] = 180(6 - 2)...6-sided\;\;polygon \\[3ex] = 180(4) \\[3ex] Each\;\;interior\;\;\angle = \dfrac{Sum}{n} \\[5ex] = \dfrac{180(4)}{6} \\[5ex] = 30(4) \\[3ex] = 120^\circ \\[3ex] \implies \\[3ex] \angle FAB = \angle ABC = \angle BCD = \angle CDE = \angle DEF = \angle EFA = 120^\circ \\[3ex] \angle OFA = \angle OAF = \dfrac{120}{2} = 60^\circ...congruent\;\;\triangle s \\[5ex] The\;\;same\;\;applies\;\;to\;\;all\;\;other\;\;congruent\;\;\triangle s \\[3ex] \implies \\[3ex] each\;\;\triangle \;\;is\;\;an\;\;equilateral\;\;\triangle...each\;\;\angle\;\;of\;\;60^\circ \\[3ex] \implies \\[3ex] \underline{Sides\;\;of\;\;the\;\;Hexagon} \\[3ex] |AB| = |BC| = |CD| = |DE| = |EF| = |FA| = \dfrac{5}{2}\;cm \\[5ex] Perimeter\;\;of\;\;Hexagon = 6\left(\dfrac{5}{2}\right) ... 6\;\;sides \\[5ex] Perimeter = 3(5) \\[3ex] Perimeter = 15\;cm $

$ (a.) \\[3ex] \angle WXU = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] \angle XWU = \angle XUW = p ...base\;\;\angle s\;\;of\;\;isosceles\;\; \triangle WXU \\[3ex] \angle XWU + \angle XUW + \angle WXU = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle WXU \\[3ex] \implies \\[3ex] p + p + 90 = 180 \\[3ex] 2p = 180 - 90 \\[3ex] 2p = 90 \\[3ex] p = \dfrac{90}{2} \\[5ex] p = 45 \\[3ex] \therefore \angle XWU = \angle XUW = 45^\circ \\[3ex] \underline{First\;\;Reason} \\[3ex] z = \angle XWU = 45^\circ ...\angle\;\;between\;\;tangent\;TV\;\;and\;\;chord\;XU = \angle\;\;in\;\;alternate\;\;segment \\[5ex] \underline{Second\;\;Reason} \\[3ex] UW = diameter...given \\[3ex] Let\;\;C = center\;\;of\;\;the\;\;circle \\[3ex] \implies \\[3ex] WC = CU = radius \\[3ex] \angle WUV = 90^\circ ...radius\;CU \perp tangent\;TUV\;\;at\;\;point\;\;of\;\;contact\;U \\[3ex] \angle WUV = \angle XUW + z...diagram \\[3ex] \implies \\[3ex] \angle XUW + z = 90 \\[3ex] 45 + z = 90 \\[3ex] z = 90 - 45 \\[3ex] z = 45^\circ \\[5ex] \underline{Third\;\;Reason} \\[3ex] \angle WUT = 90^\circ ...radius\;CU \perp tangent\;TUV\;\;at\;\;point\;\;of\;\;contact\;U \\[3ex] \angle WUT + \angle XUW + z = 180^\circ ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] 90 + 45 + z = 180 \\[3ex] 135 + z = 180 \\[3ex] z = 180 - 135 \\[3ex] z = 45^\circ \\[5ex] (b.) \\[3ex] (i.) \\[3ex] x = 56^\circ ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] (ii.) \\[3ex] y = x + 19 ...exterior\;\;\angle\;\;of\;\;a\;\;\triangle = sum\;\;of\;\;the\;\;two\;\;interior\;\;opposite\;\;\angle s \\[3ex] y = 56 + 19 \\[3ex] y = 75^\circ $

$ \underline{Cyclic\;\;Quadrilateral\;PQRS} \\[3ex] \angle RQP = \angle TSP = 147^\circ \\[3ex] ... exterior\;\; \angle \:\:of\:\:a\:\:cyclic\:\:quad = interior\;\;opposite\;\;\angle \\[3ex] \underline{\triangle RQP} \\[3ex] \angle PRQ + \angle RQP + \angle RPQ = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;a\;\;\triangle \\[3ex] \angle PRQ + 147 + 21 = 180 \\[3ex] \angle PRQ + 168 = 180 \\[3ex] \angle PRQ = 180 - 168 \\[3ex] \angle PRQ = 12^\circ \\[3ex] \angle SRP = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle\;\;is\;\;a\;\;right\;\;\angle \\[3ex] \angle SRQ = \angle SRP + \angle PRQ ... diagram \\[3ex] \angle SRQ = 90 + 12 \\[3ex] \angle SRQ = 102^\circ $

(a.)
For us to use the one of the properties of angles between parallel lines:
Construction:
Extend Line YX to A
Extend Line ZW to B



$ (a.) \\[3ex] \angle XWB = \angle WXY = 84^\circ ...alternate\;\;\angle s\;\;are\;\;equal \\[3ex] \angle XYZ = \angle XWB = 84^\circ ... exterior\;\; \angle \:\:of\:\:a\:\:cyclic\:\:quadrilateral = interior\;\;opposite\;\;\angle \\[5ex] (b.) \\[3ex] \angle ODE = \angle BDE = \angle ADE = 35^\circ ...diagram \\[3ex] \angle OED = \angle ODE = 35^\circ ...base\;\;\angle s\;\;of\;\;isosceles\;\; \triangle EOD \\[3ex] \angle OEA = 90^\circ ...radius\;OE \perp tangent\;AE\;\;at\;\;point\;\;of\;\;contact\;E \\[3ex] \underline{\triangle ADE} \\[3ex] \angle AED = \angle OEA + \angle OED ...diagram \\[3ex] \angle AED = 90 + 35 = 125^\circ \\[3ex] z + \angle AED + \angle ADE = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;a\;\;\triangle \\[3ex] z + 125 + 35 = 180 \\[3ex] z + 160 = 180 \\[3ex] z = 180 - 160 \\[3ex] z = 20^\circ \\[5ex] y = \angle ODE + \angle OED ...exterior\;\;\angle\;\;of\;\;a\;\;\triangle = sum\;\;of\;\;the\;\;two\;\;interior\;\;opposite\;\;\angle s \\[3ex] y = 35 + 35 \\[3ex] y = 70^\circ \\[3ex] OR \\[3ex] \underline{\triangle AOE} \\[3ex] z + y + \angle OEA = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;a\;\;\triangle \\[3ex] 20 + y + 90 = 180 \[3ex] y + 110 = 180 \\[3ex] y = 180 - 110 \\[3ex] y = 70^\circ \\[5ex] y = 2x ... \angle \;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] 2x = y \\[3ex] 2x = 70 \\[3ex] x = \dfrac{70}{2} \\[5ex] x = 35^\circ \\[3ex] OR \\[3ex] x = 35^\circ ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal $

$ (i) \\[3ex] \angle BCD = \angle DAE = 70^\circ \\[3ex] ... exterior\;\; \angle \:\:of\:\;cyclic\:\:quad.\;ABCD = interior\;\;opposite\;\;\angle \\[3ex] (ii) \\[3ex] \angle BOD = 2 * \angle BCD ... \angle \;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] \angle BOD = 2 * 70 \\[3ex] \angle BOD = 140^\circ \\[3ex] (iii) \\[3ex] \angle OBD = \angle ODB = p ...base\;\;\angle s\;\;of\;\;isosceles\;\; \triangle OBD \\[3ex] \angle OBD + \angle ODB + \angle BOD = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle OBD \\[3ex] \implies \\[3ex] p + p + 140 = 180 \\[3ex] 2p = 180 - 140 \\[3ex] 2p = 40 \\[3ex] p = \dfrac{40}{2} \\[5ex] p = 20 \\[3ex] \therefore \angle OBD = \angle ODB = 20^\circ $

$ (a) \\[3ex] |OT| = |OR| = radius = r = 8\;cm \\[3ex] \underline{\triangle ROT} \\[3ex] |TR|^2 = |OT|^2 + |OR|^2 - 2 * |OT| * |OR| * \cos\angle ROT ...Cosine\;\;Law \\[3ex] 12^2 = 8^2 + 8^2 - 2(8)(8) \cos\angle ROT \\[3ex] 144 = 64 + 64 - 128\cos\angle ROT \\[3ex] 144 = 128 - 128\cos \angle ROT \\[3ex] 144 = 128(1 - \cos\angle ROT) \\[3ex] \dfrac{144}{128} = 1 - \cos\angle ROT \\[5ex] 1.125 = 1 - \cos \angle ROT \\[3ex] \cos \angle ROT = 1 - 1.125 \\[3ex] \cos \angle ROT = -0.125 \\[3ex] \angle ROT = \cos^{-1}(-0.125) \\[3ex] \angle ROT = 97.18075578^\circ \\[3ex] (b) \\[3ex] \angle OTR = \angle ORT = p ...base\;\;\angle s\;\;of\;\;isosceles\;\; \triangle ROT \\[3ex] \angle OTR + \angle ORT + \angle ROT = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle ROT \\[3ex] p + p + 97.18075578 = 180 \\[3ex] 2p = 180 - 97.18075578 \\[3ex] 2p = 82.81924422 \\[3ex] p = \dfrac{82.81924422}{2} \\[5ex] p = 41.40962211 \\[3ex] \therefore \angle OTR = \angle ORT = 41.40962211^\circ \\[3ex] \angle OTB = 90^\circ ...radius\;OT \perp tangent\;ATB\;\;at\;\;point\;\;of\;\;contact\;T \\[3ex] \angle OTR + \angle RTB = \angle OTB ... diagram \\[3ex] 41.40962211 + \angle RTB = 90 \\[3ex] \angle RTB = 90 - 41.40962211 \\[3ex] \angle RTB = 48.59037789^\circ \\[3ex] (c) \\[3ex] Length\;\;of\;\;arc\;RT = \dfrac{\angle ROT}{360} * 2\pi * r \\[5ex] = \dfrac{97.18075578}{360} * 2 * \dfrac{22}{7} * 8 \\[5ex] = \dfrac{34207.62603}{2520} \\[5ex] = 13.57445478\;cm $

$ x = 2y ...\angle s \:\:in\:\:the\:\:same\:\:segment\;\;are\;\;equal \\[3ex] 60 = 2y + 2x \\[3ex] exterior\;\;\angle\;\;of\;\;a\;\;\triangle = sum\;\;of\;\;two\;\;interior\;\;opposite\;\;\angle s \\[3ex] Substitute\;\;2y\;\;for\;\;x \\[3ex] 60 = 2y + 2(2y) \\[3ex] 60 = 2y + 4y \\[3ex] 60 = 6y \\[3ex] 6y = 60 \\[3ex] y = \dfrac{60}{6} \\[5ex] y = 10^\circ $

Construction: Draw the circle to represent the information.


$ |AT| + |TB| = |AB| ...diagram \\[3ex] |AT| + 3 = 15 \\[3ex] |AT| = 15 - 3 \\[3ex] |AT| = 12\;cm \\[3ex] \underline{Intersecting\;\;Chords\;\;Theorem} \\[3ex] |CT| * |TD| = |AT| * |TB| \\[3ex] |CT| * 4 = 12 * 3 \\[3ex] |CT| = \dfrac{12 * 3}{4} \\[5ex] |CT| = 9\;cm \\[3ex] Ratio\;\;in\;\;which\;\;T\;\;divides\;\;CD \\[3ex] = |CT| : |TD| \\[3ex] = 9 : 4 $

$ radius = |AO| = |OC| = 3.6\;cm ... diagram \\[3ex] diameter = |AC| = 2 * radius = 2(3.6) = 7.2\;cm \\[3ex] \underline{\triangle ABC} \\[3ex] \angle ABC = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle\;\;is\;\;a\;\;right\;\;\angle \\[3ex] \dfrac{\sin \angle BAC}{|BC|} = \dfrac{\sin \angle ABC}{|AC|} ... Sine\;\;Law \\[5ex] \dfrac{\sin \angle BAC}{4.1} = \dfrac{\sin 90}{7.2} \\[5ex] \sin \angle BAC = \dfrac{4.1 * \sin 90}{7.2} \\[5ex] \sin \angle BAC = 0.5694444444 \\[3ex] \angle BAC = \sin^{-1}(0.5694444444) \\[3ex] \angle BAC = 34.71149426^\circ \\[3ex] \angle BDC = \angle BAC = 34.71149426^\circ \\[3ex] ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] \underline{\triangle BCD} \\[3ex] \angle DBC + \angle BCD + \angle BDC = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;a\;\;\triangle \\[3ex] \angle DBC + 93 + 34.71149426 = 180 \\[3ex] \angle DBC + 127.7114943 = 180 \\[3ex] \angle DBC = 180 - 127.7114943 \\[3ex] \angle DBC = 52.28850574 \\[3ex] \angle DBC \approx 52.3^\circ ...to\;\;3\;\;significant\;\;figures $

$ (a.) \\[3ex] m\angle ACB = m\angle ACO + m\angle OCB ...diagram \\[3ex] 90 = m\angle ACO + 57 \\[3ex] m\angle + 57 = 90 \\[3ex] m\angle = 90 - 57 \\[3ex] m\angle = 33^\circ \\[3ex] (b.) \\[3ex] \angle AOB = 180^\circ ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] m\angle AOB = m\angle AOC + m\angle BOC ...diagram \\[3ex] 180 = m\angle AOC + 66 \\[3ex] m\angle AOC + 66 = 180 \\[3ex] m\angle AOC = 180 - 66 \\[3ex] m\angle AOC = 114^\circ \\[3ex] m\overset{\huge\frown}{AC} = m\angle AOC = 114^\circ ...intercepted\;\;arc = central\;\;\angle \\[3ex] (c.) \\[3ex] m\overset{\huge\frown}{CB} = 66^\circ ...intercepted\;\;arc = central\;\;\angle \\[3ex] (d.) \\[3ex] m\overset{\huge\frown}{AB} = 180^\circ ...intercepted\;\;arc = central\;\;\angle \\[3ex] (e.) \\[3ex] Obtuse\;\;\angle AOC + Reflex\;\;\angle AOC = 360^\circ ...\angle s\;\;at\;\;a\;\;point \\[3ex] 114 + Reflex\;\;\angle AOC = 360 \\[3ex] Reflex\;\;\angle AOC = 360 - 114 \\[3ex] Reflex\;\;\angle AOC = 246^\circ \\[3ex] m\overset{\huge\frown}{CBA} = Reflex\;\;\angle AOC = 246^\circ ...intercepted\;\;arc = central\;\;\angle $

Reason: Please review Number (7.) theorem
The measure of the angle at the center (central angle) is equal to the measure of it's intercepted arc and vice versa.

$ (a.)\;\; m\overset{\huge\frown}{BC} = 62^\circ \\[3ex] (b.)\;\; m\overset{\huge\frown}{CD} = 118^\circ \\[3ex] (c.)\;\; m\overset{\huge\frown}{AD} = 62^\circ \\[3ex] (d.)\;\; m\overset{\huge\frown}{AB} = 118^\circ \\[3ex] (e.)\;\; central\;\;\angle = 180^\circ ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] m\overset{\huge\frown}{ABC} = 180^\circ $

$ \angle PSR + \angle PQR = 180^\circ ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quadrilateral\;\;are\;\;supplementary \\[3ex] \angle PSR + 72 = 180 \\[3ex] \angle PSR = 180 - 72 \\[3ex] \angle PSR = 108^\circ \\[3ex] \angle SPR + \angle PSR + \angle PRS = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;a\;\;\triangle \\[3ex] y + 108 + 28 = 180 \\[3ex] y + 136 = 180 \\[3ex] y = 180 - 136 \\[3ex] y = 44^\circ $

$ \underline{Equation\;\;of\;a\;\;circle\;\;with\;\;center\;(h, k)} \\[3ex] (x - k)^2 + (y - k)^2 = r^2 \\[3ex] Compare\;\;with \\[3ex] (x - 2)^2 + y^2 = 1 \\[3ex] \implies \\[3ex] (x - 2)^2 + (y - 0)^2 = 1^2 \\[3ex] center = (h, k) = (2, 0) \\[3ex] radius = r = 1\;unit \\[3ex] r = |AB| = 1\;unit ...diagram \\[3ex] |OB| = distance\;\;between\;\;O(0, 0)\;\;and\;\;B(2, 0) \\[3ex] |OB| = \sqrt{(2 - 0)^2 + (0 - 0)^2}... Distance\;\;Formula \\[3ex] |OB| = \sqrt{2^2 + 0} \\[3ex] |OB| = \sqrt{4} \\[3ex] |OB| = 2\;units \\[3ex] \dfrac{\sin \angle AOB}{|AB|} = \dfrac{\sin \angle OAB}{|OB|}...Sine\;\;Law \\[5ex] \dfrac{\sin \angle AOB}{1} = \dfrac{\sin 90}{2} \\[5ex] \sin \angle AOB = \dfrac{1}{2} \\[5ex] \angle AOB = \sin^{-1}\left(\dfrac{1}{2}\right) \\[5ex] \angle AOB = 30^\circ $

$ (a) \\[3ex] \angle NOM = 2 * \angle NLM ... \angle \;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] \angle NOM = 2 * 27 \\[3ex] \angle NOM = 54^\circ \\[3ex] (b) \\[3ex] $ There are at least two approaches of doing (b)
Use any approach you prefer.

$ \underline{First\;\;Approach} \\[3ex] \angle NMQ = 27^\circ ...\angle\;\;between\;\;tangent\;QMT\;\;and\;\;chord\;NM = \angle\;\;in\;\;alternate\;\;segment \\[3ex] \underline{Second\;\;Approach} \\[3ex] \angle ONM = \angle OMN = p ...base\;\;\angle s\;\;of\;\;isosceles\;\; \triangle NOM \\[3ex] \angle ONM + \angle OMN + \angle NOM = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle NOM \\[3ex] p + p + 54 = 180 \\[3ex] 2p = 180 - 54 \\[3ex] 2p = 126 \\[3ex] p = \dfrac{126}{2} \\[5ex] p = 63 \\[3ex] \therefore \angle ONM = \angle OMN = 63^\circ \\[3ex] \angle OMQ = 90^\circ ...radius\;OM \perp tangent\;QMT\;\;at\;\;point\;\;of\;\;contact\;M \\[3ex] \angle NMQ + \angle OMN = \angle OMQ ...diagram \\[3ex] \angle NMQ + 63 = 90 \\[3ex] \angle NMQ = 90 - 63 \\[3ex] \angle NMQ = 27^\circ $

$ (i) \\[3ex] \underline{\triangle PRS} \\[3ex] \angle PSR = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] \implies \triangle PSR \;\;is\;\;a\;\;right-angled\;\;\triangle \\[3ex] \tan \angle PRS = \dfrac{PS}{RS} ...SOHCAHTOA \\[5ex] \tan a = \dfrac{16}{10} \\[5ex] \tan a = 1.6 \\[3ex] a = \tan^{-1}(1.6) \\[3ex] a = 57.99461679 \\[3ex] a \approx 58^\circ \\[5ex] (ii) \\[3ex] \angle PCQ = 2 * \angle PSQ ... \angle \;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] \angle PCQ = 2 * 68 \\[3ex] \angle PCQ = 136^\circ \\[3ex] \angle CQT = \angle CPT = 90^\circ \\[3ex] ...radius\;CQ \perp tangent\;QT\;\;at\;\;point\;\;of\;\;contact\;Q \\[3ex] ...radius\;CP \perp tangent\;PT\;\;at\;\;point\;\;of\;\;contact\;P \\[3ex] \angle PTQ + \angle CQT + \angle CPT + \angle PCQ = 360^\circ ...sum\;\;of\;\;interior\;\;\angle s\;\;of\;\;Quadrilateral\;CQTP \\[3ex] \implies \\[3ex] b + 90 + 90 + 136 = 360 \\[3ex] b + 316 = 360 \\[3ex] b = 360 - 316 \\[3ex] b = 44^\circ $

$ \underline{\triangle AOB} \\[3ex] \angle AOB + \angle OAB + \angle ABO = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;a\;\;\triangle \\[3ex] \angle AOB + 90 + 34 = 180 \\[3ex] \angle AOB + 124 = 180 \\[3ex] \angle AOB = 180 - 124 \\[3ex] \angle AOB = 56^\circ \\[3ex] $ (a.) Angle Measure Theorem: the side length facing an angle is opposite the angle measure as regards size.
small angle = ∠ABO = 34°
side length facing 34° = radius = |OA| = 6cm
Hence, the radius OA is the shortest distance from O to the tangent AB.

$ (b) \\[3ex] (i) \\[3ex] \dfrac{|OB|}{\sin 90} = \dfrac{|OA|}{\sin 34}...Sine\;\;Law \\[5ex] |OB| = \dfrac{|OA| * \sin 90}{\sin 34} \\[5ex] |OB| = \dfrac{6 * 1}{0.5591929035} \\[5ex] |OB| = 10.7297499\;cm \\[3ex] (ii) \\[3ex] |OC| = |OA| = 6\;cm \\[3ex] |OC| + |BC| = |OB| ... diagram \\[3ex] 6 + |BC| = 10.7297499 \\[3ex] |BC| = 10.7297499 - 6 \\[3ex] |BC| = 4.729749899\;cm $

$ (a) \\[3ex] \angle COB = 2 * \angle CAB ... \angle \;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] \angle COB = 2 * 32 \\[3ex] \angle COB = 64^\circ \\[3ex] (b) \\[3ex] \angle OCB = \angle OBC = p ...base\;\;\angle s\;\;of\;\;isosceles\;\; \triangle COB \\[3ex] \angle OCB + \angle OBC + \angle COB = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle COB \\[3ex] p + p + 64 = 180 \\[3ex] 2p = 180 - 64 \\[3ex] 2p = 116 \\[3ex] p = \dfrac{116}{2} \\[5ex] p = 58^\circ \\[3ex] \therefore \angle OCB = \angle OBC = 58^\circ \[3ex] \angle ACB = \angle CAB = 32^\circ ...base\;\;\angle s\;\;of\;\;isosceles\;\; \triangle CAB \\[3ex] \angle OCA + \angle ACB = \angle OCB ...diagram \\[3ex] \angle OCA + 32 = 58 \\[3ex] \angle OCA = 58 - 32 \\[3ex] \angle OCA = 26^\circ $

$ \angle CFG = \angle GFH = x ...diagram \\[3ex] \angle CGF = \angle CFG = x ...base\;\;\angle s\;\;of\;\;isosceles\;\; \triangle FGC \\[3ex] \angle GCH = \angle CFG + \angle CGF ...exterior\;\;\angle\;\;of\;\;a\;\;\triangle = sum\;\;of\;\;the\;\;two\;\;interior\;\;opposite\;\;\angle s \\[3ex] y = x + x \\[3ex] y = 2x $

There are at least four ways to prove that $\triangle ABC$ and $\triangle DCB$ are congruent.
Use any approach you prefer

$ \angle ADB = \angle ABD ...given \\[3ex] \implies \\[3ex] |AB| = |AD| ...isosceles\;\triangle \\[3ex] |BC| = |DC| ...Intersecting\;\;Tangents\;\;Theorem \\[3ex] \implies \\[3ex] \angle CBD = \angle CDB ...base\;\;\angle s\;\;of\;\;isosceles\;\; \triangle BCD \\[3ex] \angle ABC = \angle ABD + \angle CBD ...diagram \\[3ex] \angle ADC = \angle ADB + \angle CDB ...diagram \\[3ex] \implies \\[3ex] \angle ABC = \angle ADC \\[3ex] Also: \\[3ex] \angle BCA = \angle DCA \\[3ex] ... two\;\;tangents:\;BC\;\;and\;\;DC\;\;drawn\;\;from\;\;same\;\;external\;\;point:\;C\;\;bisects\;\;the\;\;\angle\;\;formed\;\;by\;\;the\;\;tangents:\;\angle BCD \\[3ex] |AC| = |AC| ...for\;\;\triangle ABC\;\;and\;\;\triangle ADC \\[3ex] \underline{First\;\;Approach:\;\;SSS} \\[3ex] Side:\;\; |AB| = |AD| \\[3ex] Side:\;\; |BC| = |DC| \\[3ex] Side:\;\; |AC| = |AC| \\[3ex] \underline{Second\;\;Approach:\;\;ASA} \\[3ex] Angle:\;\; \angle ABC = \angle ADC \\[3ex] Side:\;\; |BC| = |DC| \\[3ex] Angle:\;\; \angle BCA = \angle DCA \\[3ex] \underline{Third\;\;Approach:\;\;AAS} \\[3ex] Angle:\;\; \angle ABC = \angle ADC \\[3ex] Angle:\;\; \angle BCA = \angle DCA \\[3ex] Side:\;\; |AC| = |AC| \\[3ex] \underline{Fourth\;\;Approach:\;\;SAS} \\[3ex] Side:\;\; |BC| = |DC| \\[3ex] Angle:\;\; \angle BCA = \angle DCA \\[3ex] Side:\;\; |AC| = |AC| $

$ (i) \\[3ex] \angle EHG = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] \implies \triangle EHG \;\;is\;\;a\;\;right-angled\;\;\triangle \\[3ex] |EG|^2 = |EH|^2 + |GH|^2 ...Pythagorean\;\;Theorem \\[3ex] |EG|^2 = 8^2 + 12^2 \\[3ex] |EG|^2 = 64 + 144 \\[3ex] |EG|^2 = 208 \\[3ex] |EG| = \sqrt{208} \\[3ex] |EG| = 14.4222051 \\[3ex] |EG| \approx 14\;cm \\[5ex] (ii) \\[3ex] \underline{\triangle EHG} \\[3ex] \tan \angle EGH = \dfrac{|EH|}{|GH|} ...SOHCAHTOA \\[5ex] \tan \angle EGH = \dfrac{8}{12} \\[5ex] \tan \angle EGH = 0.6666666667 \\[3ex] \angle EGH = \tan^{-1}(0.6666666667) \\[3ex] \angle EGH = 33.69006753^\circ \\[3ex] \underline{\triangle EFH} \\[3ex] \angle EFH = \angle EGH = 33.69006753^\circ ...\angle s\;\;in\;\;the\;\;same\;\;segment\;\;are\;\;equal \\[3ex] \angle FEH + \angle EHF + \angle EFH = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;a\;\;\triangle \\[3ex] x + 75 + 33.69006753 = 180 \\[3ex] x + 108.6900675 = 180 \\[3ex] x = 180 - 108.6900675 \\[3ex] x = 71.30993247 \\[3ex] x \approx 71^\circ $

Construction:
(1.) Draw the diameter by joining the radius from |OD| to the circumference of the circle to G
This is to enable us to use the Intersecting Chords theorem so we can find the diameter.

(2.) Draw the diameter by joining the radius from |OE| to the circumference of the circle to H.
This is to enable us to use the Intersecting Secants theorem so we can find the value of x


$ |DG| = diameter = d \\[3ex] |DF| + |FG| = |DG| ...diagram \\[3ex] 2 + |FG| = d \\[3ex] |FG| = d - 2 \\[3ex] |DF| * |FG| = |AF| * |FB| ...Intersecting\;\;Chords\;\;Theorem \\[3ex] 2 * (d - 2) = 7 * 4 \\[3ex] d - 2 = \dfrac{7 * 4}{2} \\[5ex] d - 2 = 14 \\[3ex] d = 14 + 2 \\[3ex] d = 16\;cm \\[3ex] |HE| = diameter = d = 16\;cm \\[3ex] |AB| = |AF| + |FB| ... diagram \\[3ex] |AB| = 7 + 4 = 11\;cm \\[3ex] |EC| * (|HE| + |EC|) = |BC| * (|AB| + |BC|) ...Intersecting\;\;Secants\;\;Theorem \\[3ex] x * (16 + x) = 5(11 + 5) \\[3ex] 16x + x^2 = 5(16) \\[3ex] x^2 + 16x = 80 \\[3ex] x^2 + 16x - 80 = 0 \\[3ex] (x + 20)(x - 4) = 0 \\[3ex] x + 20 = 0 \;\;\;OR\;\;\; x - 4 = 0 \\[3ex] x = -20 \;\;\;OR\;\;\; x = 4 \\[3ex] $ Because the length cannot have a negative value:

$ x = 4\;cm $

Construction: Draw and label the radii:
from centre C to point P
from centre C to point Q


$ |CB| = |CP| = |CQ| = radius = r = 15\;cm ...given \\[3ex] |AB| = |AC| + r ...diagram \\[3ex] 27 = |AC| + 15 \\[3ex] |AC| + 15 = 27 \\[3ex] |AC| = 27 - 15 \\[3ex] |AC| = 12\;cm \\[3ex] \underline{\triangle CPA} \\[3ex] |CP|^2 = |PA|^2 + |AC|^2 ...Pythagorean\;\;theorem \\[3ex] 15^2 = |PA|^2 + 12^2 \\[3ex] |PA|^2 + 12^2 = 15^2 \\[3ex] |PA|^2 + 144 = 225 \\[3ex] |PA|^2 = 225 - 144 \\[3ex] |PA|^2 = 81 \\[3ex] |PA| = \sqrt{81} \\[3ex] |PA| = 9\;cm \\[3ex] |AQ| = |PA| = 9\;cm...A\;\;is\;\;the\;\;midpoint\;\;of\;\;chord\;PQ \\[3ex] \implies \\[3ex] |PQ| = |PA| + |AQ| \\[3ex] |PQ| = 9 + 9 \\[3ex] |PQ| = 18\;cm $

Construction: To use these theorems: Intersecting Tangents Theorem and Angle of Intersecting Tangents Theorem, we will need to draw the line from the centre of the circle to the intersection of the two tangents: Point B


$ \angle OCB = 90^\circ ...radius\;OC \perp tangent\;CB\;\;at\;\;point\;\;of\;\;contact\;C \\[3ex] \angle OAB = 90^\circ ...radius\;OA \perp tangent\;AB\;\;at\;\;point\;\;of\;\;contact\;A \\[3ex] \angle OBC = \angle OBA = \dfrac{\angle ABC}{2} = \dfrac{74}{2} = 37^\circ ... Angle\;\;of\;\;Intersecting\;\;Tangents\;\;Theorem \\[3ex] \angle COB + \angle OBC + \angle OCB = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle COB \\[3ex] \angle COB + 37 + 90 = 180 \\[3ex] \angle COB + 127 = 180 \\[3ex] \angle COB = 180 - 127 \\[3ex] \angle COB = 53^\circ \\[3ex] \angle AOB = \angle COB = 53^\circ ...Intersecting\;\;Tangents\;\;Theorem \\[3ex] \angle AOC = \angle AOB + \angle COB ...diagram \\[3ex] \angle AOC = 53 + 53 = 106^\circ \\[3ex] \angle AOC = 2 * \angle ADC ... \angle \;\;at\;\;centre = 2 * \angle\;\;at\;\;circumference \\[3ex] 106 = 2 * \angle ADC \\[3ex] 2 * \angle ADC = 106 \\[3ex] \angle ADC = \dfrac{106}{2} \\[5ex] \angle ADC = 53^\circ $

$ \angle OBC = 90^\circ \\[3ex] ...radius\;OB \perp tangent\;CBD\;\;at\;\;point\;\;of\;\;contact\;B \\[3ex] \angle ABC + \angle OBA = \angle OBC ...diagram \\[3ex] 38 + \angle OBA = 90 \\[3ex] \angle OBA = 90 - 38 \\[3ex] \angle OBA = 52^\circ \\[3ex] \angle OAB = \angle OBA = 52^\circ ...base\;\;\angle s\;\;of\;\;isosceles\;\; \triangle AOB \\[3ex] \angle AOB + \angle OAB + \angle OBA = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle AOB \\[3ex] \angle AOB + 52 + 52 = 180 \\[3ex] \angle AOB + 104 = 180 \\[3ex] \angle AOB = 180 - 104 \\[3ex] \angle AOB = 76^\circ \\[3ex] \angle CAB = \angle AOB + \angle OBA \\[3ex] ...exterior\;\;\angle\;\;of\;\;a\;\;\triangle = sum\;\;of\;\;the\;\;two\;\;interior\;\;opposite\;\;\angle s \\[3ex] \angle CAB = 76 + 52 \\[3ex] \angle CAB = 128^\circ $

$ |OD| = |OE| = radius = r = 6\;cm \\[3ex] Length\;\;of\;\;arc\;EFD = \dfrac{\angle EOD}{360} * 2 * \pi * r \\[5ex] \dfrac{7}{2}\pi = \dfrac{\angle EOD}{360} * 2 * \pi * 6 \\[5ex] \dfrac{7\pi}{2} = \dfrac{\angle EOD}{360} * 12\pi \\[5ex] \dfrac{7}{2} = \dfrac{\angle EOD}{30} \\[5ex] \dfrac{\angle EOD}{30} = \dfrac{7}{2} \\[5ex] \angle EOD = \dfrac{7 * 30}{2} \\[5ex] \angle EOD = 7(15) = 105^\circ \\[3ex] |ED|^2 = |OE|^2 + |OD|^2 - 2 * |OE| * |OD| * \cos \angle EOD...Cosine\;\;Law \\[3ex] |ED|^2 = 6^2 + 6^2 - 2(6)(6) * \cos 105^\circ \\[3ex] = 36 + 36 - 72(-0.2588190451) \\[3ex] = 72 + 18.63497125 \\[3ex] = 90.63497125 \\[3ex] |ED| = \sqrt{90.63497125} \\[3ex] |ED| = 9.520240083\;cm \\[3ex] |DC| * |EC| = |BC| * |AC| ...Intersecting\;\;Secants\;\;Theorem \\[3ex] |EC| = |ED| + |DC| ...diagram \\[3ex] |EC| = 9.520240083 + |DC| \\[3ex] |AC| = |AB| + |BC| ... diagram \\[3ex] |AC| = 7 + 9 = 16\;cm \\[3ex] Let\;\;|DC| = p...to\;\;make\;\;it\;\;simpler \\[3ex] \implies \\[3ex] |DC| * (9.520240083 + |DC|) = 9(16) \\[3ex] p(9.520240083 + p) = 144 \\[3ex] 9.520240083p + p^2 = 144 \\[3ex] p^2 + 9.520240083p - 144 = 0 \\[3ex] Compare\;\;to\;\;general\;\;form:\;\;ap^2 + bp + c = 0 \\[3ex] a = 1 \\[3ex] b = 9.520240083 \\[3ex] c = -144 \\[3ex] p = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} ...Quadratic\;\;Formula \\[5ex] p = \dfrac{-9.520240083 \pm \sqrt{9.520240083^2 - 4(1)(-144)}}{2(1)} \\[5ex] p = \dfrac{-9.520240083 \pm \sqrt{90.63497124 + 576}}{2} \\[5ex] p = \dfrac{-9.520240083 \pm \sqrt{666.6349712}}{2} \\[5ex] p = \dfrac{-9.520240083 \pm 25.81927519}{2} \\[5ex] p = \dfrac{-9.520240083 + 25.81927519}{2} \;\;\;OR\;\;\; p = \dfrac{-9.520240083 - 25.81927519}{2} \\[5ex] p = \dfrac{16.29903511}{2} \;\;\;OR\;\;\; p = \dfrac{-35.33951527}{2} \\[5ex] p = 8.149517553 \;\;\;OR\;\;\; p = -17.66975764 \\[3ex] $ Because the length cannot be negative, discard the negative value

$ p = 8.149517553 \\[3ex] p \approx 8.15\;cm...to\;\;3\;\;significant\;\;figures $

(a) Construction: To get a clearer picture of x and the remaining lengths, let us indicate the length of x and the radii.
Let:
|AP| = m
|CQ| = n


$ |AP| = |AR| = m ...two\;\;tangents:\;\;AP\;\;and\;\;AR\;\;drawn\;\;from\;\;the\;\;same\;\;external\;\;point:\;A\;\;are\;\;equal\;\;in\;\;length \\[3ex] |CR| = |CQ| = n ...two\;\;tangents:\;\;CR\;\;and\;\;CQ\;\;drawn\;\;from\;\;the\;\;same\;\;external\;\;point:\;C\;\;are\;\;equal\;\;in\;\;length \\[3ex] |AR| + |RC| = m + n = 8\;cm...diagram \\[3ex] |BQ| = |BP| = x ...two\;\;tangents:\;\;PB\;\;and\;\;QB\;\;drawn\;\;from\;\;the\;\;same\;\;external\;\;point:\;B\;\;are\;\;equal\;\;in\;\;length \\[3ex] |AB| + |CB| = 14 + 12 = 26\;cm \\[3ex] \implies \\[3ex] x + m + x + n = |AB| + |CB| \\[3ex] 2x + m + n = 26...eqn.(1) \\[3ex] But: \\[3ex] m + n = 8 ...eqn.(2) \\[3ex] Substitute\;\;eqn.(2)\;\;into\;\;eqn.(1) \\[3ex] 2x + 8 = 26 \\[3ex] 2x = 26 - 8 \\[3ex] 2x = 18 \\[3ex] x = \dfrac{18}{2} \\[5ex] x = 9\;cm \\[3ex] (b) \\[3ex] $ Let us determine angle B because we shall need it to answer (b)

$ \underline{\triangle ABC} \\[3ex] side = lower\;\;case\;\;letters \\[3ex] angle = upper\;\;case\;\;letters \\[3ex] Let: \\[3ex] |AB| = c = 14\;cm \\[3ex] |AC| = b = 8\;cm \\[3ex] |BC| = a = 12\;cm \\[3ex] b^2 = a^2 + c^2 - 2ac \cos B ...Cosine\;\;Law \\[3ex] 8^2 = 12^2 + 14^2 - 2(12)(14) \cos B \\[3ex] 64 = 144 + 196 - 336\cos B \\[3ex] 64 = 340 - 336\cos B \\[3ex] 336\cos B = 340 - 64 \\[3ex] \cos B = \dfrac{276}{336} \\[5ex] \cos B = 0.8214285714 \\[3ex] B = \cos^{-1}(0.8214285714) \\[3ex] B = 34.77194403^\circ \\[3ex] Area\;\;of\;\;\triangle ABC = \dfrac{1}{2}ac\sin B \\[5ex] = \dfrac{1}{2} * 12 * 14 * \sin 34.77194403^\circ \\[5ex] = 84(0.5703114079) \\[3ex] = 47.90615827\;cm^2 \\[3ex] $ We need to find the radius of the circle.
So, we can determine the area of the circle.
Construction: Draw a line from the centre O to point B
In other words, draw line |OB|


$ \angle OBP = \angle OBQ = \dfrac{\angle B}{2} ... Angle\;\;of\;\;Intersecting\;\;Tangents\;\;Theorem \\[5ex] \angle OBP = \dfrac{34.77194403}{2} \\[5ex] \angle OBP = 17.38597202^\circ \\[3ex] \underline{\triangle POB} \\[3ex] SOHCAHTOA \\[3ex] \tan \angle OBP = \dfrac{r}{x} \\[5ex] \tan 17.38597202 = \dfrac{r}{9} \\[5ex] 9\tan 17.38597202 = r \\[3ex] r = 9(0.3131121456) \\[3ex] r = 2.818009311\;cm \\[3ex] \underline{Circle\;RPQ} \\[3ex] Area\;\;of\;\;the\;\;circle = \pi r^2 \\[3ex] = \dfrac{22}{7} * (2.818009311)^2 \\[5ex] = 24.95798321\;cm^2 \\[3ex] $ What percent of the area of the triangle is the area of the circle?

$ What\;\;\%\;\;of\;\;47.90615827\;\;is\;\;24.95798321? \\[3ex] Let\;\;the\;\;variable\% = k \\[3ex] \dfrac{is}{of} = \dfrac{variable}{100} \\[5ex] \dfrac{24.95798321}{47.90615827} = \dfrac{k}{100} \\[5ex] k = \dfrac{24.95798321 * 100}{47.90615827} \\[5ex] k = 52.09765113\% \\[3ex] k \approx 52.1\%...to\;\;3\;\;significant\;\;figures \\[3ex] $ Approximately 52.1% of the total area of triangle is the area of the circle.