In the figure, BCD is parallel to GHJ and the line GD cuts ∠CGJ into half. Given that DG and CH are straight lines, ∠BCG = 42°, ∠GHF = 38° and ∠CFD = 112°, find
- ∠k
- ∠n
- ∠m
(a)
∠GFH = ∠CFD = 112° (Vertically opposite angles)
∠k
= 180° - 112° - 38°
= 30° (Angles sum of triangle)
(b)
∠n
= 180° - 30°
= 150° (Interior angles)
(c)
∠m
= 180° - 30° - 30° - 38°
= 100° (Angles sum of triangle)
Answer(s): (a) 30°; (b) 150°; (c) 100°