In the figure, BC = BE, ∠CBE = 32° and ∠FDE = 14°. Find
- ∠t
- ∠v
(a)
∠BCE
= (180° - 32°) ÷ 2
= 74° (Isosceles triangle)
∠t
= 180° - 14° - 74°
= 92° (Angles sum of triangle)
(b)
∠BED
= 180° - 74°
= 106° (Angles on a straight line)
∠v
= 14° + 106°
= 120° (Exterior angle of a triangle)
Answer(s): (a) 92°; (b) 120°