In the figure, SHKP is a trapezium and triangles SPN and NSL are isosceles triangles. RG, RK and GK are straight lines. SP = SN = NL. Find

1. ∠x
2. ∠y

(a)
∠HSL = 180° - ∠x (Interior angles, SH//RK)
∠NLS = 180° - ∠x (Angles on a straight line)
∠NSL = 180° - ∠x (Isosceles triangle)

53° + 180° - ∠x + 180° - ∠x + ∠y + 20° = 180° (Angles on a straight line, GR)
53° + 180° + 180° + 20° - ∠x - ∠x + ∠y  = 180°
433° - 2∠x + ∠y = 180°
2∠x - ∠y = 433° - 180° 
2∠x - ∠y = 253°

∠y = 2∠x - 253° --- (1)

∠SNP
= ∠SPN
= 2 x (180° - ∠x)
= 360° - 2∠x (Exterior angle of a triangle)

∠y = 180° - (360° - 2∠x) - (360° - 2∠x)
∠y = 180° - 360° + 2∠x - 360° + 2∠x 
∠y = 180° - 360° - 360° + 2∠x + 2∠x
∠y = 4∠x - 540° (Angles sum of triangle) 

∠y = 4∠x - 540° --- (2)

(2) = (1)
4∠x - 540° = 2∠x - 253°
4∠x - 2∠x= 540° - 253°
2∠x = 287°

∠x
= 287° ÷ 2
= 143.5°

(b)
From (1)
∠y
=  2∠x - 253°
= 287° - 253° 
= 34°

Answer(s): (a) 143.5°; (b) 34°