In the figure, BCD is parallel to EFG and ED cuts ∠CEF equally. Given that ∠BCE = 58° and ∠CHD = 113°, find
- ∠h
- ∠i
(a)
∠BCE = ∠CEF (Alternate angles)
∠h
= 58° ÷ 2
= 29°
(b)
∠CEH = ∠h = 29°
∠i
= 113° - 29°
= 84° (Exterior angle of a triangle)
Answer(s): (a) 29°; (b) 84°