In the figure, BC = BE, ∠CBE = 34° and ∠FDE = 16°. Find
- ∠p
- ∠q
(a)
∠BCE
= (180° - 34°) ÷ 2
= 73° (Isosceles triangle)
∠p
= 180° - 16° - 73°
= 91° (Angles sum of triangle)
(b)
∠BED
= 180° - 73°
= 107° (Angles on a straight line)
∠q
= 16° + 107°
= 123° (Exterior angle of a triangle)
Answer(s): (a) 91°; (b) 123°