In the figure, PQ = PS, ∠QPS = 31° and ∠TRS = 15°. Find
- ∠n
- ∠p
(a)
∠PQS
= (180° - 31°) ÷ 2
= 74.5° (Isosceles triangle)
∠n
= 180° - 15° - 74.5°
= 90.5° (Angles sum of triangle)
(b)
∠PSR
= 180° - 74.5°
= 105.5° (Angles on a straight line)
∠p
= 15° + 105.5°
= 120.5° (Exterior angle of a triangle)
Answer(s): (a) 90.5°; (b) 120.5°