In the figure, BCD is parallel to EFG and ED cuts ∠CEF equally. Given that ∠BCE = 42° and ∠CHD = 103°, find
- ∠x
- ∠y
(a)
∠BCE = ∠CEF (Alternate angles)
∠x
= 42° ÷ 2
= 21°
(b)
∠CEH = ∠x = 21°
∠y
= 103° - 21°
= 82° (Exterior angle of a triangle)
Answer(s): (a) 21°; (b) 82°