In the figure, AB = AD, ∠BAD = 36° and ∠ECD = 17°. Find
- ∠g
- ∠h
(a)
∠ABD
= (180° - 36°) ÷ 2
= 72° (Isosceles triangle)
∠g
= 180° - 17° - 72°
= 91° (Angles sum of triangle)
(b)
∠ADC
= 180° - 72°
= 108° (Angles on a straight line)
∠h
= 17° + 108°
= 125° (Exterior angle of a triangle)
Answer(s): (a) 91°; (b) 125°