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 = 50°, ∠GHF = 38° and ∠CFD = 113°, find
- ∠q
- ∠s
- ∠r
(a)
∠GFH = ∠CFD = 113° (Vertically opposite angles)
∠q
= 180° - 113° - 38°
= 29° (Angles sum of triangle)
(b)
∠s
= 180° - 29°
= 151° (Interior angles)
(c)
∠r
= 180° - 29° - 29° - 38°
= 92° (Angles sum of triangle)
Answer(s): (a) 29°; (b) 151°; (c) 92°