In the figure, PQ = PS, ∠QPS = 31° and ∠TRS = 12°. Find
- ∠g
- ∠h
(a)
∠PQS
= (180° - 31°) ÷ 2
= 74.5° (Isosceles triangle)
∠g
= 180° - 12° - 74.5°
= 93.5° (Angles sum of triangle)
(b)
∠PSR
= 180° - 74.5°
= 105.5° (Angles on a straight line)
∠h
= 12° + 105.5°
= 117.5° (Exterior angle of a triangle)
Answer(s): (a) 93.5°; (b) 117.5°