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 = 47°, ∠FGE = 39° and ∠BEC = 115°, find
- ∠m
- ∠p
- ∠n
(a)
∠FEG = ∠BEC = 115° (Vertically opposite angles)
∠m
= 180° - 115° - 39°
= 26° (Angles sum of triangle)
(b)
∠p
= 180° - 26°
= 154° (Interior angles)
(c)
∠n
= 180° - 26° - 26° - 39°
= 94° (Angles sum of triangle)
Answer(s): (a) 26°; (b) 154°; (c) 94°