In the figure, HJ = HL, ∠JHL = 31° and ∠MKL = 16°. Find
- ∠h
- ∠i
(a)
∠HJL
= (180° - 31°) ÷ 2
= 74.5° (Isosceles triangle)
∠h
= 180° - 16° - 74.5°
= 89.5° (Angles sum of triangle)
(b)
∠HLK
= 180° - 74.5°
= 105.5° (Angles on a straight line)
∠i
= 16° + 105.5°
= 121.5° (Exterior angle of a triangle)
Answer(s): (a) 89.5°; (b) 121.5°