In the figure, AB = AD, ∠BAD = 36° and ∠ECD = 14°. Find
- ∠j
- ∠k
(a)
∠ABD
= (180° - 36°) ÷ 2
= 72° (Isosceles triangle)
∠j
= 180° - 14° - 72°
= 94° (Angles sum of triangle)
(b)
∠ADC
= 180° - 72°
= 108° (Angles on a straight line)
∠k
= 14° + 108°
= 122° (Exterior angle of a triangle)
Answer(s): (a) 94°; (b) 122°