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 = 50°, ∠FGE = 33° and ∠BEC = 116°, find
- ∠j
- ∠m
- ∠k
(a)
∠FEG = ∠BEC = 116° (Vertically opposite angles)
∠j
= 180° - 116° - 33°
= 31° (Angles sum of triangle)
(b)
∠m
= 180° - 31°
= 149° (Interior angles)
(c)
∠k
= 180° - 31° - 31° - 33°
= 97° (Angles sum of triangle)
Answer(s): (a) 31°; (b) 149°; (c) 97°