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
- ∠f
- ∠g
(a)
∠BHD = 180° - ∠f (Interior angles, HB//GC)
∠EDH = 180° - ∠f (Angles on a straight line)
∠EHD = 180° - ∠f (Isosceles triangle)
49° + 180° - ∠f + 180° - ∠f + ∠g + 18° = 180° (Angles on a straight line, AG)
49° + 180° + 180° + 18° - ∠f - ∠f + ∠g = 180°
427° - 2∠f + ∠g = 180°
2∠f - ∠g = 427° - 180°
2∠f - ∠g = 247°
∠g = 2∠f - 247° --- (1)
∠HEF
= ∠HFE
= 2 x (180° - ∠f)
= 360° - 2∠f (Exterior angle of a triangle)
∠g = 180° - (360° - 2∠f) - (360° - 2∠f)
∠g = 180° - 360° + 2∠f - 360° + 2∠f
∠g = 180° - 360° - 360° + 2∠f + 2∠f
∠g = 4∠f - 540° (Angles sum of triangle)
∠g = 4∠f - 540° --- (2)
(2) = (1)
4∠f - 540° = 2∠f - 247°
4∠f - 2∠f= 540° - 247°
2∠f = 293°
∠f
= 293° ÷ 2
= 146.5°
(b)
From (1)
∠g
= 2∠f - 247°
= 293° - 247°
= 46°
Answer(s): (a) 146.5°; (b) 46°