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 = 48°, ∠FGE = 37° and ∠BEC = 110°, find
- ∠k
- ∠n
- ∠m
(a)
∠FEG = ∠BEC = 110° (Vertically opposite angles)
∠k
= 180° - 110° - 37°
= 33° (Angles sum of triangle)
(b)
∠n
= 180° - 33°
= 147° (Interior angles)
(c)
∠m
= 180° - 33° - 33° - 37°
= 95° (Angles sum of triangle)
Answer(s): (a) 33°; (b) 147°; (c) 95°