In the figure, ZSTX is a trapezium and triangles ZXV and VZU are isosceles triangles. YR, YT and RT are straight lines. ZX = ZV = VU. Find

1. ∠b
2. ∠c

(a)
∠SZU = 180° - ∠b (Interior angles, ZS//YT)
∠VUZ = 180° - ∠b (Angles on a straight line)
∠VZU = 180° - ∠b (Isosceles triangle)

56° + 180° - ∠b + 180° - ∠b + ∠c + 19° = 180° (Angles on a straight line, RY)
56° + 180° + 180° + 19° - ∠b - ∠b + ∠c  = 180°
435° - 2∠b + ∠c = 180°
2∠b - ∠c = 435° - 180° 
2∠b - ∠c = 255°

∠c = 2∠b - 255° --- (1)

∠ZVX
= ∠ZXV
= 2 x (180° - ∠b)
= 360° - 2∠b (Exterior angle of a triangle)

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

∠c = 4∠b - 540° --- (2)

(2) = (1)
4∠b - 540° = 2∠b - 255°
4∠b - 2∠b= 540° - 255°
2∠b = 285°

∠b
= 285° ÷ 2
= 142.5°

(b)
From (1)
∠c
=  2∠b - 255°
= 285° - 255° 
= 30°

Answer(s): (a) 142.5°; (b) 30°