In the figure, JK = JM, ∠KJM = 31° and ∠NLM = 18°. Find
- ∠q
- ∠r
(a)
∠JKM
= (180° - 31°) ÷ 2
= 74.5° (Isosceles triangle)
∠q
= 180° - 18° - 74.5°
= 87.5° (Angles sum of triangle)
(b)
∠JML
= 180° - 74.5°
= 105.5° (Angles on a straight line)
∠r
= 18° + 105.5°
= 123.5° (Exterior angle of a triangle)
Answer(s): (a) 87.5°; (b) 123.5°