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 = 38° and ∠BEC = 120°, find
- ∠j
- ∠m
- ∠k
(a)
∠FEG = ∠BEC = 120° (Vertically opposite angles)
∠j
= 180° - 120° - 38°
= 22° (Angles sum of triangle)
(b)
∠m
= 180° - 22°
= 158° (Interior angles)
(c)
∠k
= 180° - 22° - 22° - 38°
= 101° (Angles sum of triangle)
Answer(s): (a) 22°; (b) 158°; (c) 101°