In the figure, TKLR is a trapezium and triangles TRP and PTN are isosceles triangles. SH, SL and HL are straight lines. TR = TP = PN. Find

1. ∠t
2. ∠v

(a)
∠KTN = 180° - ∠t (Interior angles, TK//SL)
∠PNT = 180° - ∠t (Angles on a straight line)
∠PTN = 180° - ∠t (Isosceles triangle)

52° + 180° - ∠t + 180° - ∠t + ∠v + 20° = 180° (Angles on a straight line, HS)
52° + 180° + 180° + 20° - ∠t - ∠t + ∠v  = 180°
432° - 2∠t + ∠v = 180°
2∠t - ∠v = 432° - 180° 
2∠t - ∠v = 252°

∠v = 2∠t - 252° --- (1)

∠TPR
= ∠TRP
= 2 x (180° - ∠t)
= 360° - 2∠t (Exterior angle of a triangle)

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

∠v = 4∠t - 540° --- (2)

(2) = (1)
4∠t - 540° = 2∠t - 252°
4∠t - 2∠t= 540° - 252°
2∠t = 288°

∠t
= 288° ÷ 2
= 144°

(b)
From (1)
∠v
=  2∠t - 252°
= 288° - 252° 
= 36°

Answer(s): (a) 144°; (b) 36°