In the figure, HJ = HL, ∠JHL = 33° and ∠MKL = 15°. Find
- ∠p
- ∠q
(a)
∠HJL
= (180° - 33°) ÷ 2
= 73.5° (Isosceles triangle)
∠p
= 180° - 15° - 73.5°
= 91.5° (Angles sum of triangle)
(b)
∠HLK
= 180° - 73.5°
= 106.5° (Angles on a straight line)
∠q
= 15° + 106.5°
= 121.5° (Exterior angle of a triangle)
Answer(s): (a) 91.5°; (b) 121.5°