In the figure, BC = BE, ∠CBE = 33° and ∠FDE = 18°. Find
- ∠e
- ∠f
(a)
∠BCE
= (180° - 33°) ÷ 2
= 73.5° (Isosceles triangle)
∠e
= 180° - 18° - 73.5°
= 88.5° (Angles sum of triangle)
(b)
∠BED
= 180° - 73.5°
= 106.5° (Angles on a straight line)
∠f
= 18° + 106.5°
= 124.5° (Exterior angle of a triangle)
Answer(s): (a) 88.5°; (b) 124.5°