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 = 35° and ∠BEC = 115°, find
- ∠h
- ∠j
- ∠i
(a)
∠FEG = ∠BEC = 115° (Vertically opposite angles)
∠h
= 180° - 115° - 35°
= 30° (Angles sum of triangle)
(b)
∠j
= 180° - 30°
= 150° (Interior angles)
(c)
∠i
= 180° - 30° - 30° - 35°
= 104° (Angles sum of triangle)
Answer(s): (a) 30°; (b) 150°; (c) 104°