In the figure, PQRS is a parallelogram. O is the centre of the circle. Find
- ∠PSR
- ∠PQS
(a)
∠OQS = ∠QSO = 31° (Angle properties within a circle)
∠SOQ
= 180° - 31° - 31°
= 118° (Angles sum of triangle)
∠OPQ + ∠OQP = ∠QOS (Exterior angle of a triangle)
∠OPQ = ∠OQP (Isosceles triangle, OP = OQ)
∠OPQ
= 118° ÷ 2
= 59° (Exterior angle of a triangle)
∠PSR
= 180° - 59°
= 121° (Interior angles)
(b)
∠PQS
= 180° - 59° - 31°
= 90° (Angle sum of triangles)
Answer(s): (a) 121°; (b) 90°