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 = 42°, ∠FGE = 38° and ∠BEC = 118°, find
- ∠v
- ∠x
- ∠w
(a)
∠FEG = ∠BEC = 118° (Vertically opposite angles)
∠v
= 180° - 118° - 38°
= 24° (Angles sum of triangle)
(b)
∠x
= 180° - 24°
= 156° (Interior angles)
(c)
∠w
= 180° - 24° - 24° - 38°
= 100° (Angles sum of triangle)
Answer(s): (a) 24°; (b) 156°; (c) 100°