In the figure, BC = BE, ∠CBE = 31° and ∠FDE = 15°. Find
- ∠d
- ∠e
(a)
∠BCE
= (180° - 31°) ÷ 2
= 74.5° (Isosceles triangle)
∠d
= 180° - 15° - 74.5°
= 90.5° (Angles sum of triangle)
(b)
∠BED
= 180° - 74.5°
= 105.5° (Angles on a straight line)
∠e
= 15° + 105.5°
= 120.5° (Exterior angle of a triangle)
Answer(s): (a) 90.5°; (b) 120.5°