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
- ∠h
- ∠i
(a)
∠BHD = 180° - ∠h (Interior angles, HB//GC)
∠EDH = 180° - ∠h (Angles on a straight line)
∠EHD = 180° - ∠h (Isosceles triangle)
51° + 180° - ∠h + 180° - ∠h + ∠i + 24° = 180° (Angles on a straight line, AG)
51° + 180° + 180° + 24° - ∠h - ∠h + ∠i = 180°
435° - 2∠h + ∠i = 180°
2∠h - ∠i = 435° - 180°
2∠h - ∠i = 255°
∠i = 2∠h - 255° --- (1)
∠HEF
= ∠HFE
= 2 x (180° - ∠h)
= 360° - 2∠h (Exterior angle of a triangle)
∠i = 180° - (360° - 2∠h) - (360° - 2∠h)
∠i = 180° - 360° + 2∠h - 360° + 2∠h
∠i = 180° - 360° - 360° + 2∠h + 2∠h
∠i = 4∠h - 540° (Angles sum of triangle)
∠i = 4∠h - 540° --- (2)
(2) = (1)
4∠h - 540° = 2∠h - 255°
4∠h - 2∠h= 540° - 255°
2∠h = 285°
∠h
= 285° ÷ 2
= 142.5°
(b)
From (1)
∠i
= 2∠h - 255°
= 285° - 255°
= 30°
Answer(s): (a) 142.5°; (b) 30°