OR = OQ = Radius

∆ RQO is an isosceles triangle.

Isosceles triangle has the base angles the same.

∠ ORQ = ∠ OQR = 42°

∠ ROQ + 42° + 42° = 180° (sum of angles of a triangle)

∠ ROQ + 84 = 180

∠ ROQ = 180 - 84

∠ ROQ = 96°

QPR = $\frac{1}{2}$ x ∠ ROQ

QPR = $\frac{1}{2}$ x 96° = 48°

Draw an imaginary parallel line to cut t.

t = ∠ CGE + ∠ EGJ

∠ ACD = ∠ BCG = 35° (Vertically opposite angles)

∠ CGE = ∠ BCG = 35° (Alternate angles)

∠ HJG + ∠ IJG = 180° (Angles on a straight line)

∠ HJG = 140°

140° + ∠ IJG = 180°

∠ IJG = 180° - 140°

∠ IJG = 40°

∠ EGJ = IJG = 40° (Alternate angles)

t = ∠ CGE + ∠ EGJ

t = 35° + 40°

t = 75°

|OS| = |OQ| = Radius of the circle

∆ QOS is an isosceles triangle hence the base angles are the same.

∠ OSQ = ∠ OQS

∠ QOS + ∠ OSQ + ∠ OQS = 180°(Sum of angles of a triangle)

108° + ∠ OSQ + ∠ OQS = 180°

∠ OSQ + ∠ OQS = 180° - 108°

∠ OSQ + ∠ OQS = 72°

∠ OSQ = ∠ OQS = $\frac{\mathrm{72°}}{2}$ = 36°

From alternate segment theorem,

Draw an imaginary angle subtended by the arc QS

From the alternate segment theorem, ∠ QTS = ∠ SQR

∠ QTS = $\frac{\mathrm{\angle \; QOS}}{2}$ = $\frac{\mathrm{108°}}{2}$ = 54°

∠ QTS = ∠ SQR = 54°

From alternate segment theorem,

∠ MNP = ∠ QMP = 73°

∠ MNP + ∠ MPN + ∠ PMN = 180°(Sum of angles in a triangle)

73° + ∠ MPN + 58° = 180°

∠ MPN + 131° = 180°

∠ MPN = 180° - 131°

∠ MPN = 49°

PS = RS

∆ PSR is an isosceles triangle.

Isosceles triangles have the base angles the same.

∠ SPR = ∠ SRP

∠ PSR = 90°

∠ SPR + ∠ SRP + ∠ PSR = 180° (Sum of angles in a triangle)

∠ SPR + ∠ SRP + 90° = 180°

∠ SPR + ∠ SRP = 180° - 90°

∠ SPR + ∠ SRP = 90°

∠ SPR = ∠ SRP = $\frac{\mathrm{90°}}{2}$ = 45°

∠ PQS = ∠ SRP = 45°(Angles subtended by the same arc PS)