In the figure, HJ = HL, ∠JHL = 33° and ∠MKL = 18°. Find
- ∠y
- ∠z
(a)
∠HJL
= (180° - 33°) ÷ 2
= 73.5° (Isosceles triangle)
∠y
= 180° - 18° - 73.5°
= 88.5° (Angles sum of triangle)
(b)
∠HLK
= 180° - 73.5°
= 106.5° (Angles on a straight line)
∠z
= 18° + 106.5°
= 124.5° (Exterior angle of a triangle)
Answer(s): (a) 88.5°; (b) 124.5°