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 = 33° and ∠BEC = 115°, find
- ∠e
- ∠g
- ∠f
(a)
∠FEG = ∠BEC = 115° (Vertically opposite angles)
∠e
= 180° - 115° - 33°
= 32° (Angles sum of triangle)
(b)
∠g
= 180° - 32°
= 148° (Interior angles)
(c)
∠f
= 180° - 32° - 32° - 33°
= 105° (Angles sum of triangle)
Answer(s): (a) 32°; (b) 148°; (c) 105°