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
- ∠x
- ∠y
(a)
∠CKE = 180° - ∠x (Interior angles, KC//HD)
∠FEK = 180° - ∠x (Angles on a straight line)
∠FKE = 180° - ∠x (Isosceles triangle)
51° + 180° - ∠x + 180° - ∠x + ∠y + 21° = 180° (Angles on a straight line, BH)
51° + 180° + 180° + 21° - ∠x - ∠x + ∠y = 180°
432° - 2∠x + ∠y = 180°
2∠x - ∠y = 432° - 180°
2∠x - ∠y = 252°
∠y = 2∠x - 252° --- (1)
∠KFG
= ∠KGF
= 2 x (180° - ∠x)
= 360° - 2∠x (Exterior angle of a triangle)
∠y = 180° - (360° - 2∠x) - (360° - 2∠x)
∠y = 180° - 360° + 2∠x - 360° + 2∠x
∠y = 180° - 360° - 360° + 2∠x + 2∠x
∠y = 4∠x - 540° (Angles sum of triangle)
∠y = 4∠x - 540° --- (2)
(2) = (1)
4∠x - 540° = 2∠x - 252°
4∠x - 2∠x= 540° - 252°
2∠x = 288°
∠x
= 288° ÷ 2
= 144°
(b)
From (1)
∠y
= 2∠x - 252°
= 288° - 252°
= 36°
Answer(s): (a) 144°; (b) 36°