In the figure, HBCF is a trapezium and triangles HFE and EHD are isosceles triangles. GA, GC and AC are straight lines. HF = HE = ED. Find
- ∠x
- ∠y
(a)
∠BHD = 180° - ∠x (Interior angles, HB//GC)
∠EDH = 180° - ∠x (Angles on a straight line)
∠EHD = 180° - ∠x (Isosceles triangle)
51° + 180° - ∠x + 180° - ∠x + ∠y + 22° = 180° (Angles on a straight line, AG)
51° + 180° + 180° + 22° - ∠x - ∠x + ∠y = 180°
433° - 2∠x + ∠y = 180°
2∠x - ∠y = 433° - 180°
2∠x - ∠y = 253°
∠y = 2∠x - 253° --- (1)
∠HEF
= ∠HFE
= 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 - 253°
4∠x - 2∠x= 540° - 253°
2∠x = 287°
∠x
= 287° ÷ 2
= 143.5°
(b)
From (1)
∠y
= 2∠x - 253°
= 287° - 253°
= 34°
Answer(s): (a) 143.5°; (b) 34°