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
- ∠r
- ∠s
(a)
∠BHD = 180° - ∠r (Interior angles, HB//GC)
∠EDH = 180° - ∠r (Angles on a straight line)
∠EHD = 180° - ∠r (Isosceles triangle)
58° + 180° - ∠r + 180° - ∠r + ∠s + 23° = 180° (Angles on a straight line, AG)
58° + 180° + 180° + 23° - ∠r - ∠r + ∠s = 180°
441° - 2∠r + ∠s = 180°
2∠r - ∠s = 441° - 180°
2∠r - ∠s = 261°
∠s = 2∠r - 261° --- (1)
∠HEF
= ∠HFE
= 2 x (180° - ∠r)
= 360° - 2∠r (Exterior angle of a triangle)
∠s = 180° - (360° - 2∠r) - (360° - 2∠r)
∠s = 180° - 360° + 2∠r - 360° + 2∠r
∠s = 180° - 360° - 360° + 2∠r + 2∠r
∠s = 4∠r - 540° (Angles sum of triangle)
∠s = 4∠r - 540° --- (2)
(2) = (1)
4∠r - 540° = 2∠r - 261°
4∠r - 2∠r= 540° - 261°
2∠r = 279°
∠r
= 279° ÷ 2
= 139.5°
(b)
From (1)
∠s
= 2∠r - 261°
= 279° - 261°
= 18°
Answer(s): (a) 139.5°; (b) 18°