(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

(a)

∠ DCE : ∠ CDE = 7:8

$\frac{\mathrm{\angle \; DCE}}{\mathrm{\angle \; CDE}}$ = $\frac{7}{8}$

But ∠ DCE = 56°

$\frac{\mathrm{56°}}{\mathrm{\angle \; CDE}}$ = $\frac{7}{8}$

Cross multiply

7 x ∠ CDE = 56° x 8

Divide both sides by 7

∠ CDE = $\frac{\mathrm{56°\; x\; 8}}{7}$ = 64°

∠ CDE + ∠ DCE + ∠ AEB = 180° (sum of interior angles of a triangle)

64° + 56° + ∠ AEB = 180°

120° + ∠ AEB = 180°

∠ AEB = 180° - 120°

∠ AEB = 60°

(b)

A quadrilateral is called cyclic if all the four vertices of it lie on a circle and the sum of its opposite angles is 180°

∠ ADC + ∠ CDE = 180°(sum of angles on a straight line)

But ∠ CDE = 64°

∠ ADC + 64° = 180°

∠ ADC = 180° - 64°

∠ ADC = 116°

∠ DCB + ∠ DCE = 180°(sum of angles on a straight line)

But ∠ DCE = 56°

∠ DCB + 56° = 180°

∠ DCB = 180° - 56°

∠ DCB = 124°

∠ ABE + ∠ ADC = 180° (sum of opposite angles of cyclic quadrilateral)

But ∠ ADC = 116°

(5x - 16)° + 116° = 180°

5x - 16 + 116 = 180

5x + 100 = 180

5x = 180 - 100

5x = 80

Divide both sides by 5

x = $\frac{80}{5}$ = 16

∠ BAE + ∠ DCB = 180° (sum of opposite angles of cyclic quadrilateral)

But ∠ DCB = 124°

$(\frac{3}{2}\mathrm{x}+\mathrm{4y})$^° + 124° = 180°

$\frac{3}{2}$x + 4y + 124 = 180

But x = 16

$\frac{3}{2}$ x 16 + 4y + 124 = 180

3 x 8 + 4y = 180 - 124

24 + 4y = 56

4y = 56 - 24

4y = 32

Divide both sides by 4

y = $\frac{32}{4}$ = 8

(i)

∠ ECF = ∠ CBD = 48°(corresponding angles)

∠ CBD + ∠ ABD = 180° (Sum of angles on a straight line)

2x + 34° + 48° = 180°

2x + 82° = 180°

2x = 180° - 82°

2x = 98°

Divide by 2

x = $\frac{\mathrm{98°}}{2}$ = 49°

(ii)

Let ∠ BDC = x

∠ BCD = $\frac{1}{3}$ ∠ BDC

∠ BCD = $\frac{1}{3}$ x x = $\frac{\mathrm{x}}{3}$

48° + x + $\frac{\mathrm{x}}{3}$ = 180° (sum of angles of a triangle)

Get rid of the fraction by multiplying both sides by the denominator, 3.

3 x 48° + 3 x x + 3 x $\frac{\mathrm{x}}{3}$ = 180° x 3

144° + 3x + x = 540°

4x = 540° - 144°

4x = 396°

Divide both sides by 4

x = $\frac{\mathrm{396°}}{4}$ = 99°

∴ ∠ BDC = 99°

(i)

∠ MNO = ∠ MNP + ∠ PNO

∠ MNO = 33° + 67° = 100°

(ii)

Using arc PM

∠ MOP = ∠ MNP = 33°

Sum of angles of a triangle is 180°

From ∆ MPO,

∠ PMO + ∠ MPO + ∠ MOP = 180°

67° + ∠ MPO + 33° = 180°

∠ MPO + 100° = 180°

∠ MPO = 180° - 100°

∠ MPO = 80°

(i)

CO and AO are radii hence both are equal.

CO = AO, ∆AOC is an isosceles triangle.

Isosceles triangles have base angles equare.

∠ BCA = ∠ OAC = 41°

∠ BCA + ∠ OAC + ∠ AOC = 180° (Sum of angles on a straight line)

41° + 41° + ∠ AOC = 180°

82° + ∠ AOC = 180°

∠ AOC = 180° - 82°

∠ AOC = 98°

∠ AOC + ∠ BOA = 180°

98° + ∠ BOA = 180°

∠ BOA = 180° - 98°

∠ BOA = 82°

Alternatively, ∠ BOA = 2 x ∠ BCA = 2 x 41° = 82°

Central angle is twice any inscribed angle subtended by the same arc.

(ii)

∠ BAC = 90° (Angle subtended by a diameter)

∠ BAO + ∠ OAC = ∠ BAC

∠ BCA = ∠ OAC = 41°

∠ BAO + 41° = 90°

∠ BAO = 90° - 41°

∠ BAO = 49°

(i)

∠ AEC = 2 x m

∠ AEC = 2m

∠ ADC + ∠ ABC = 180°

∠ ADC + m = 180°

∠ ADC = 180° - m

∠ AEC = ∠ ADC

2m = 180° - m

2m + m = 180°

3m = 180°

Divide both sides by 3

m = $\frac{\mathrm{180°}}{3}$ = 60°

(ii)

Method I

∠ AEC = 2m = 2 x 60° = 120°

∆ EAC is isosceles with base angles EAC and ECA

∠ EAC + ∠ ECA + ∠ AEC = 180°

∠ EAC + ∠ ECA + 120° = 180°

∠ EAC + ∠ ECA = 180° - 120°

∠ EAC + ∠ ECA = 60°

∠ EAC = ∠ ECA = $\frac{\mathrm{60°}}{2}$ = 30°

EAC = ECA = 30°

Note: the base angles of isosceles triangle are the same.

From ∆ ABC

m + ∠ BAC + ∠ BCA = 180° (Sum of angles of a triangle)

60° + (n + 30°) + (n + 30°) = 180°

60° + n + 30° + n + 30° = 180°

2n + 120° = 180°

2n = 180° - 120°

2n = 60°

Divide both sides by 2

n = $\frac{\mathrm{60°}}{2}$ = 30°

Method II

n + n + m + ∠ AEC top = 360° (Sum of angles of a quadrilateral)

Notes:

1. ABCE has four sides (Quadrilateral)

2. Sum of interior angles = (n - 2) x 180°

3. Sum of interior angles of a quadrilateral = (4 - 2) x 180° = 360°

m = 60°

AEC top + AEC down = 360°

AEC down = 2 x m = 2 x 60° = 120°

AEC top + 120° = 360°

AEC top = 360° - 120°

AEC top = 240°

n + n + 60° + 240° = 360°

2n + 300° = 360°

2n = 360° - 300°

2n = 60°

Divide both sides by 2

n = $\frac{\mathrm{60°}}{2}$ = 30°