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 = 43°, ∠FGE = 41° and ∠BEC = 116°, find
- ∠b
- ∠d
- ∠c
(a)
∠FEG = ∠BEC = 116° (Vertically opposite angles)
∠b
= 180° - 116° - 41°
= 23° (Angles sum of triangle)
(b)
∠d
= 180° - 23°
= 157° (Interior angles)
(c)
∠c
= 180° - 23° - 23° - 41°
= 96° (Angles sum of triangle)
Answer(s): (a) 23°; (b) 157°; (c) 96°