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 = 41°, ∠FGE = 33° and ∠BEC = 113°, find
- ∠j
- ∠m
- ∠k
(a)
∠FEG = ∠BEC = 113° (Vertically opposite angles)
∠j
= 180° - 113° - 33°
= 34° (Angles sum of triangle)
(b)
∠m
= 180° - 34°
= 146° (Interior angles)
(c)
∠k
= 180° - 34° - 34° - 33°
= 106° (Angles sum of triangle)
Answer(s): (a) 34°; (b) 146°; (c) 106°