In the figure, BCD is parallel to EFG and ED cuts ∠CEF equally. Given that ∠BCE = 48° and ∠CHD = 118°, find
- ∠k
- ∠m
(a)
∠BCE = ∠CEF (Alternate angles)
∠k
= 48° ÷ 2
= 24°
(b)
∠CEH = ∠k = 24°
∠m
= 118° - 24°
= 94° (Exterior angle of a triangle)
Answer(s): (a) 24°; (b) 94°