In the figure, KCDG is a trapezium and triangles KGF and FKE are isosceles triangles. HB, HD and BD are straight lines. KG = KF = FE. Find

1. ∠w
2. ∠x

(a)
∠CKE = 180° - ∠w (Interior angles, KC//HD)
∠FEK = 180° - ∠w (Angles on a straight line)
∠FKE = 180° - ∠w (Isosceles triangle)

51° + 180° - ∠w + 180° - ∠w + ∠x + 21° = 180° (Angles on a straight line, BH)
51° + 180° + 180° + 21° - ∠w - ∠w + ∠x  = 180°
432° - 2∠w + ∠x = 180°
2∠w - ∠x = 432° - 180° 
2∠w - ∠x = 252°

∠x = 2∠w - 252° --- (1)

∠KFG
= ∠KGF
= 2 x (180° - ∠w)
= 360° - 2∠w (Exterior angle of a triangle)

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

∠x = 4∠w - 540° --- (2)

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

∠w
= 288° ÷ 2
= 144°

(b)
From (1)
∠x
=  2∠w - 252°
= 288° - 252° 
= 36°

Answer(s): (a) 144°; (b) 36°