In the figure, AB = AD, ∠BAD = 35° and ∠ECD = 17°. Find
- ∠v
- ∠w
(a)
∠ABD
= (180° - 35°) ÷ 2
= 72.5° (Isosceles triangle)
∠v
= 180° - 17° - 72.5°
= 90.5° (Angles sum of triangle)
(b)
∠ADC
= 180° - 72.5°
= 107.5° (Angles on a straight line)
∠w
= 17° + 107.5°
= 124.5° (Exterior angle of a triangle)
Answer(s): (a) 90.5°; (b) 124.5°