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 = 39° and ∠BEC = 110°, find
- ∠d
- ∠f
- ∠e
(a)
∠FEG = ∠BEC = 110° (Vertically opposite angles)
∠d
= 180° - 110° - 39°
= 31° (Angles sum of triangle)
(b)
∠f
= 180° - 31°
= 149° (Interior angles)
(c)
∠e
= 180° - 31° - 31° - 39°
= 98° (Angles sum of triangle)
Answer(s): (a) 31°; (b) 149°; (c) 98°