In the figure, RGHN is a trapezium and triangles RNL and LRK are isosceles triangles. PF, PH and FH are straight lines. RN = RL = LK. Find

1. ∠q
2. ∠r

(a)
∠GRK = 180° - ∠q (Interior angles, RG//PH)
∠LKR = 180° - ∠q (Angles on a straight line)
∠LRK = 180° - ∠q (Isosceles triangle)

57° + 180° - ∠q + 180° - ∠q + ∠r + 20° = 180° (Angles on a straight line, FP)
57° + 180° + 180° + 20° - ∠q - ∠q + ∠r  = 180°
437° - 2∠q + ∠r = 180°
2∠q - ∠r = 437° - 180° 
2∠q - ∠r = 257°

∠r = 2∠q - 257° --- (1)

∠RLN
= ∠RNL
= 2 x (180° - ∠q)
= 360° - 2∠q (Exterior angle of a triangle)

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

∠r = 4∠q - 540° --- (2)

(2) = (1)
4∠q - 540° = 2∠q - 257°
4∠q - 2∠q= 540° - 257°
2∠q = 283°

∠q
= 283° ÷ 2
= 141.5°

(b)
From (1)
∠r
=  2∠q - 257°
= 283° - 257° 
= 26°

Answer(s): (a) 141.5°; (b) 26°