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
- ∠y
- ∠z
(a)
∠BHD = 180° - ∠y (Interior angles, HB//GC)
∠EDH = 180° - ∠y (Angles on a straight line)
∠EHD = 180° - ∠y (Isosceles triangle)
56° + 180° - ∠y + 180° - ∠y + ∠z + 23° = 180° (Angles on a straight line, AG)
56° + 180° + 180° + 23° - ∠y - ∠y + ∠z = 180°
439° - 2∠y + ∠z = 180°
2∠y - ∠z = 439° - 180°
2∠y - ∠z = 259°
∠z = 2∠y - 259° --- (1)
∠HEF
= ∠HFE
= 2 x (180° - ∠y)
= 360° - 2∠y (Exterior angle of a triangle)
∠z = 180° - (360° - 2∠y) - (360° - 2∠y)
∠z = 180° - 360° + 2∠y - 360° + 2∠y
∠z = 180° - 360° - 360° + 2∠y + 2∠y
∠z = 4∠y - 540° (Angles sum of triangle)
∠z = 4∠y - 540° --- (2)
(2) = (1)
4∠y - 540° = 2∠y - 259°
4∠y - 2∠y= 540° - 259°
2∠y = 281°
∠y
= 281° ÷ 2
= 140.5°
(b)
From (1)
∠z
= 2∠y - 259°
= 281° - 259°
= 22°
Answer(s): (a) 140.5°; (b) 22°