In the figure, BC = BE, ∠CBE = 35° and ∠FDE = 18°. Find
- ∠t
- ∠v
(a)
∠BCE
= (180° - 35°) ÷ 2
= 72.5° (Isosceles triangle)
∠t
= 180° - 18° - 72.5°
= 89.5° (Angles sum of triangle)
(b)
∠BED
= 180° - 72.5°
= 107.5° (Angles on a straight line)
∠v
= 18° + 107.5°
= 125.5° (Exterior angle of a triangle)
Answer(s): (a) 89.5°; (b) 125.5°