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 = 47°, ∠GHF = 35° and ∠CFD = 117°, find
- ∠e
- ∠g
- ∠f
(a)
∠GFH = ∠CFD = 117° (Vertically opposite angles)
∠e
= 180° - 117° - 35°
= 28° (Angles sum of triangle)
(b)
∠g
= 180° - 28°
= 152° (Interior angles)
(c)
∠f
= 180° - 28° - 28° - 35°
= 98° (Angles sum of triangle)
Answer(s): (a) 28°; (b) 152°; (c) 98°