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 = 49°, ∠FGE = 41° and ∠BEC = 114°, find
- ∠a
- ∠c
- ∠b
(a)
∠FEG = ∠BEC = 114° (Vertically opposite angles)
∠a
= 180° - 114° - 41°
= 25° (Angles sum of triangle)
(b)
∠c
= 180° - 25°
= 155° (Interior angles)
(c)
∠b
= 180° - 25° - 25° - 41°
= 90° (Angles sum of triangle)
Answer(s): (a) 25°; (b) 155°; (c) 90°