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 = 45°, ∠HJG = 38° and ∠DGE = 117°, find
- ∠n
- ∠q
- ∠p
(a)
∠HGJ = ∠DGE = 117° (Vertically opposite angles)
∠n
= 180° - 117° - 38°
= 25° (Angles sum of triangle)
(b)
∠q
= 180° - 25°
= 155° (Interior angles)
(c)
∠p
= 180° - 25° - 25° - 38°
= 97° (Angles sum of triangle)
Answer(s): (a) 25°; (b) 155°; (c) 97°