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 = 41°, ∠GHF = 40° and ∠CFD = 112°, find
- ∠k
- ∠n
- ∠m
(a)
∠GFH = ∠CFD = 112° (Vertically opposite angles)
∠k
= 180° - 112° - 40°
= 28° (Angles sum of triangle)
(b)
∠n
= 180° - 28°
= 152° (Interior angles)
(c)
∠m
= 180° - 28° - 28° - 40°
= 99° (Angles sum of triangle)
Answer(s): (a) 28°; (b) 152°; (c) 99°