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 = 45°, ∠FGE = 34° and ∠BEC = 115°, find
- ∠t
- ∠w
- ∠v
(a)
∠FEG = ∠BEC = 115° (Vertically opposite angles)
∠t
= 180° - 115° - 34°
= 31° (Angles sum of triangle)
(b)
∠w
= 180° - 31°
= 149° (Interior angles)
(c)
∠v
= 180° - 31° - 31° - 34°
= 101° (Angles sum of triangle)
Answer(s): (a) 31°; (b) 149°; (c) 101°