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
- ∠b
- ∠c
(a)
∠BHD = 180° - ∠b (Interior angles, HB//GC)
∠EDH = 180° - ∠b (Angles on a straight line)
∠EHD = 180° - ∠b (Isosceles triangle)
51° + 180° - ∠b + 180° - ∠b + ∠c + 18° = 180° (Angles on a straight line, AG)
51° + 180° + 180° + 18° - ∠b - ∠b + ∠c = 180°
429° - 2∠b + ∠c = 180°
2∠b - ∠c = 429° - 180°
2∠b - ∠c = 249°
∠c = 2∠b - 249° --- (1)
∠HEF
= ∠HFE
= 2 x (180° - ∠b)
= 360° - 2∠b (Exterior angle of a triangle)
∠c = 180° - (360° - 2∠b) - (360° - 2∠b)
∠c = 180° - 360° + 2∠b - 360° + 2∠b
∠c = 180° - 360° - 360° + 2∠b + 2∠b
∠c = 4∠b - 540° (Angles sum of triangle)
∠c = 4∠b - 540° --- (2)
(2) = (1)
4∠b - 540° = 2∠b - 249°
4∠b - 2∠b= 540° - 249°
2∠b = 291°
∠b
= 291° ÷ 2
= 145.5°
(b)
From (1)
∠c
= 2∠b - 249°
= 291° - 249°
= 42°
Answer(s): (a) 145.5°; (b) 42°