In the figure, NEFK is a trapezium and triangles NKH and HNG are isosceles triangles. LD, LF and DF are straight lines. NK = NH = HG. Find

1. ∠c
2. ∠d

(a)
∠ENG = 180° - ∠c (Interior angles, NE//LF)
∠HGN = 180° - ∠c (Angles on a straight line)
∠HNG = 180° - ∠c (Isosceles triangle)

54° + 180° - ∠c + 180° - ∠c + ∠d + 15° = 180° (Angles on a straight line, DL)
54° + 180° + 180° + 15° - ∠c - ∠c + ∠d  = 180°
429° - 2∠c + ∠d = 180°
2∠c - ∠d = 429° - 180° 
2∠c - ∠d = 249°

∠d = 2∠c - 249° --- (1)

∠NHK
= ∠NKH
= 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 - 249°
4∠c - 2∠c= 540° - 249°
2∠c = 291°

∠c
= 291° ÷ 2
= 145.5°

(b)
From (1)
∠d
=  2∠c - 249°
= 291° - 249° 
= 42°

Answer(s): (a) 145.5°; (b) 42°