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 = 46°, ∠FGE = 33° and ∠BEC = 117°, find
- ∠j
- ∠m
- ∠k
(a)
∠FEG = ∠BEC = 117° (Vertically opposite angles)
∠j
= 180° - 117° - 33°
= 30° (Angles sum of triangle)
(b)
∠m
= 180° - 30°
= 150° (Interior angles)
(c)
∠k
= 180° - 30° - 30° - 33°
= 101° (Angles sum of triangle)
Answer(s): (a) 30°; (b) 150°; (c) 101°