In the figure, XPRU is a trapezium and triangles XUT and TXS are isosceles triangles. VN, VR and NR are straight lines. XU = XT = TS. Find

1. ∠s
2. ∠t

(a)
∠PXS = 180° - ∠s (Interior angles, XP//VR)
∠TSX = 180° - ∠s (Angles on a straight line)
∠TXS = 180° - ∠s (Isosceles triangle)

48° + 180° - ∠s + 180° - ∠s + ∠t + 21° = 180° (Angles on a straight line, NV)
48° + 180° + 180° + 21° - ∠s - ∠s + ∠t  = 180°
429° - 2∠s + ∠t = 180°
2∠s - ∠t = 429° - 180° 
2∠s - ∠t = 249°

∠t = 2∠s - 249° --- (1)

∠XTU
= ∠XUT
= 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 - 249°
4∠s - 2∠s= 540° - 249°
2∠s = 291°

∠s
= 291° ÷ 2
= 145.5°

(b)
From (1)
∠t
=  2∠s - 249°
= 291° - 249° 
= 42°

Answer(s): (a) 145.5°; (b) 42°