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
- ∠w
- ∠x
(a)
∠CKE = 180° - ∠w (Interior angles, KC//HD)
∠FEK = 180° - ∠w (Angles on a straight line)
∠FKE = 180° - ∠w (Isosceles triangle)
52° + 180° - ∠w + 180° - ∠w + ∠x + 24° = 180° (Angles on a straight line, BH)
52° + 180° + 180° + 24° - ∠w - ∠w + ∠x = 180°
436° - 2∠w + ∠x = 180°
2∠w - ∠x = 436° - 180°
2∠w - ∠x = 256°
∠x = 2∠w - 256° --- (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 - 256°
4∠w - 2∠w= 540° - 256°
2∠w = 284°
∠w
= 284° ÷ 2
= 142°
(b)
From (1)
∠x
= 2∠w - 256°
= 284° - 256°
= 28°
Answer(s): (a) 142°; (b) 28°