In the figure, CDE is parallel to HJK and the line HE cuts ∠DHK into half. Given that EH and DJ are straight lines, ∠CDH = 41°, ∠HJG = 32° and ∠DGE = 116°, find
- ∠q
- ∠s
- ∠r
(a)
∠HGJ = ∠DGE = 116° (Vertically opposite angles)
∠q
= 180° - 116° - 32°
= 32° (Angles sum of triangle)
(b)
∠s
= 180° - 32°
= 148° (Interior angles)
(c)
∠r
= 180° - 32° - 32° - 32°
= 107° (Angles sum of triangle)
Answer(s): (a) 32°; (b) 148°; (c) 107°