(a)

∠ QSP = 90°

∠ QSR + ∠ PSR = ∠ QSP = 90°

30° + ∠ PSR = 90°

∠ PSR = 90° - 30°

∠ PSR = 60°

From arc PR, ∠ RQP = ∠ PSR = 60°

∠ RQP + ∠ PQT = 180° (Sum of angles on a straight line)

60° + ∠ PQT = 180°

∠ PQT = 180° - 60°

∠ PQT = 120°

∠ PQT + ∠ RTP + ∠ QPS = 180°(Sum of angles in a triangle)

120° + 20° + ∠ QPS = 180°

140° + ∠ QPS = 180°

∠ QPS = 180° - 140°

∠ QPS = 40°

Angles subtended by the same arc are the same.

From arc QS, ∠ QRS = ∠ QPS = 40°

∠ QRS = 40°

(b)

∠ PQS + ∠ QPS + ∠ QSP = 180°(Sum of angles in a triangle)

∠ PQS + 40° + 90° = 180°

∠ PQS + 130° = 180°

∠ PQS = 180° - 130°

∠ PQS = 50°

(i)

∠ DCE + ∠ BCD = 180°(Sum of angles on a straight line)

51° + ∠ BCD = 180°

∠ BCD = 180° - 51°

∠ BCD = 129°

∠ BAD + ∠ BCD = 180°(Sum of opposite angles of cyclic quadrilateral)

∠ BAD + 129° = 180°

∠ BAD = 180° - 129°

∠ BAD = 51°

∠ CED + ∠ CEF = 180°(Sum of angles on a straight line)

∠ CED + 5t = 180°

∠ CED = 180° - 5t

∠ CED + ∠ ABC + ∠ BAD = 180°(Sum of angles in a triangle)

(180° - 5t) + 2t + 51° = 180°

180° - 5t + 2t + 51° = 180°

-3t + 231° = 180°

231° - 180° = 3t

51° = 3t

Divide both sides by 3

t = $\frac{\mathrm{51°}}{3}$ = 17°

(ii)

∠ ADC + ∠ ABC = 180°(Sum of opposite angles of cyclic quadrilateral)

∠ ADC + 2t = 180°

∠ ADC = 180° - 2t

But t = 17°

∠ ADC = 180° - 2 x 17°

∠ ADC = 180° - 34°

∠ ADC = 146°

(i)

∠ BAE = ∠ CEA = 40° (Alternate angles)

∠ AEB = ∠ BAE = 40° (Base angles equal)

∠ AEB + ∠ BAE + ∠ ABE = 180°(Sum of angles in a triangle)

40° + 40° + n° = 180°

80 + n = 180

n = 180 - 80

n = 100

(ii)

∠ ABE + ∠ DBE = 180° (Sum of angles on a straight line)

n° + ∠ DBE = 180°

But n = 100

100° + ∠ DBE = 180°

∠ DBE = 180° - 100°

∠ DBE = 80°

∠ DBE = ∠ BDE = 80°(Base angles equal)

∠ DBE + ∠ BDE + ∠ BED = 180°(Sum of angles in a triangle)

80° + 80° + m° = 180°

160° + m° = 180°

m° = 180° - 160°

m° = 20°

m = 20