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 = 45°, ∠FGE = 37° and ∠BEC = 117°, find
- ∠c
- ∠e
- ∠d
(a)
∠FEG = ∠BEC = 117° (Vertically opposite angles)
∠c
= 180° - 117° - 37°
= 26° (Angles sum of triangle)
(b)
∠e
= 180° - 26°
= 154° (Interior angles)
(c)
∠d
= 180° - 26° - 26° - 37°
= 98° (Angles sum of triangle)
Answer(s): (a) 26°; (b) 154°; (c) 98°