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 = 44°, ∠GHF = 37° and ∠CFD = 110°, find
- ∠q
- ∠s
- ∠r
(a)
∠GFH = ∠CFD = 110° (Vertically opposite angles)
∠q
= 180° - 110° - 37°
= 33° (Angles sum of triangle)
(b)
∠s
= 180° - 33°
= 147° (Interior angles)
(c)
∠r
= 180° - 33° - 33° - 37°
= 99° (Angles sum of triangle)
Answer(s): (a) 33°; (b) 147°; (c) 99°