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 = 42° and ∠BEC = 117°, find
- ∠x
- ∠z
- ∠y
(a)
∠FEG = ∠BEC = 117° (Vertically opposite angles)
∠x
= 180° - 117° - 42°
= 21° (Angles sum of triangle)
(b)
∠z
= 180° - 21°
= 159° (Interior angles)
(c)
∠y
= 180° - 21° - 21° - 42°
= 92° (Angles sum of triangle)
Answer(s): (a) 21°; (b) 159°; (c) 92°