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
- ∠c
- ∠d
(a)
∠BHD = 180° - ∠c (Interior angles, HB//GC)
∠EDH = 180° - ∠c (Angles on a straight line)
∠EHD = 180° - ∠c (Isosceles triangle)
51° + 180° - ∠c + 180° - ∠c + ∠d + 23° = 180° (Angles on a straight line, AG)
51° + 180° + 180° + 23° - ∠c - ∠c + ∠d = 180°
434° - 2∠c + ∠d = 180°
2∠c - ∠d = 434° - 180°
2∠c - ∠d = 254°
∠d = 2∠c - 254° --- (1)
∠HEF
= ∠HFE
= 2 x (180° - ∠c)
= 360° - 2∠c (Exterior angle of a triangle)
∠d = 180° - (360° - 2∠c) - (360° - 2∠c)
∠d = 180° - 360° + 2∠c - 360° + 2∠c
∠d = 180° - 360° - 360° + 2∠c + 2∠c
∠d = 4∠c - 540° (Angles sum of triangle)
∠d = 4∠c - 540° --- (2)
(2) = (1)
4∠c - 540° = 2∠c - 254°
4∠c - 2∠c= 540° - 254°
2∠c = 286°
∠c
= 286° ÷ 2
= 143°
(b)
From (1)
∠d
= 2∠c - 254°
= 286° - 254°
= 32°
Answer(s): (a) 143°; (b) 32°