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. ∠c
2. ∠d

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

54° + 180° - ∠c + 180° - ∠c + ∠d + 16° = 180° (Angles on a straight line, HS)
54° + 180° + 180° + 16° - ∠c - ∠c + ∠d  = 180°
430° - 2∠c + ∠d = 180°
2∠c - ∠d = 430° - 180° 
2∠c - ∠d = 250°

∠d = 2∠c - 250° --- (1)

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

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

∠d = 4∠c - 540° --- (2)

(2) = (1)
4∠c - 540° = 2∠c - 250°
4∠c - 2∠c= 540° - 250°
2∠c = 290°

∠c
= 290° ÷ 2
= 145°

(b)
From (1)
∠d
=  2∠c - 250°
= 290° - 250° 
= 40°

Answer(s): (a) 145°; (b) 40°