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)
49° + 180° - ∠h + 180° - ∠h + ∠i + 25° = 180° (Angles on a straight line, AG)
49° + 180° + 180° + 25° - ∠h - ∠h + ∠i = 180°
434° - 2∠h + ∠i = 180°
2∠h - ∠i = 434° - 180°
2∠h - ∠i = 254°
∠i = 2∠h - 254° --- (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 - 254°
4∠h - 2∠h= 540° - 254°
2∠h = 286°
∠h
= 286° ÷ 2
= 143°
(b)
From (1)
∠i
= 2∠h - 254°
= 286° - 254°
= 32°
Answer(s): (a) 143°; (b) 32°