In the figure, ABC is parallel to FGH and the line FC cuts ∠BFH into half. Given that CF and BG are straight lines, ∠ABF = 42°, ∠FGE = 38° and ∠BEC = 112°, find
- ∠e
- ∠g
- ∠f
(a)
∠FEG = ∠BEC = 112° (Vertically opposite angles)
∠e
= 180° - 112° - 38°
= 30° (Angles sum of triangle)
(b)
∠g
= 180° - 30°
= 150° (Interior angles)
(c)
∠f
= 180° - 30° - 30° - 38°
= 100° (Angles sum of triangle)
Answer(s): (a) 30°; (b) 150°; (c) 100°