In the figure, PQ = PS, ∠QPS = 32° and ∠TRS = 16°. Find
- ∠t
- ∠v
(a)
∠PQS
= (180° - 32°) ÷ 2
= 74° (Isosceles triangle)
∠t
= 180° - 16° - 74°
= 90° (Angles sum of triangle)
(b)
∠PSR
= 180° - 74°
= 106° (Angles on a straight line)
∠v
= 16° + 106°
= 122° (Exterior angle of a triangle)
Answer(s): (a) 90°; (b) 122°