In the figure, HBCF is a trapezium and triangles HFE and EHD are isosceles triangles. GA, GC and AC are straight lines. HF = HE = ED. Find

1. ∠s
2. ∠t

(a)
∠BHD = 180° - ∠s (Interior angles, HB//GC)
∠EDH = 180° - ∠s (Angles on a straight line)
∠EHD = 180° - ∠s (Isosceles triangle)

57° + 180° - ∠s + 180° - ∠s + ∠t + 20° = 180° (Angles on a straight line, AG)
57° + 180° + 180° + 20° - ∠s - ∠s + ∠t  = 180°
437° - 2∠s + ∠t = 180°
2∠s - ∠t = 437° - 180° 
2∠s - ∠t = 257°

∠t = 2∠s - 257° --- (1)

∠HEF
= ∠HFE
= 2 x (180° - ∠s)
= 360° - 2∠s (Exterior angle of a triangle)

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

∠t = 4∠s - 540° --- (2)

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

∠s
= 283° ÷ 2
= 141.5°

(b)
From (1)
∠t
=  2∠s - 257°
= 283° - 257° 
= 26°

Answer(s): (a) 141.5°; (b) 26°