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 = 33° and ∠CFD = 119°, find
- ∠q
- ∠s
- ∠r
(a)
∠GFH = ∠CFD = 119° (Vertically opposite angles)
∠q
= 180° - 119° - 33°
= 28° (Angles sum of triangle)
(b)
∠s
= 180° - 28°
= 152° (Interior angles)
(c)
∠r
= 180° - 28° - 28° - 33°
= 97° (Angles sum of triangle)
Answer(s): (a) 28°; (b) 152°; (c) 97°