In the figure, PQRS is a parallelogram. O is the centre of the circle. Find
- ∠PSR
- ∠PQS
(a)
∠OQS = ∠QSO = 45° (Angle properties within a circle)
∠SOQ
= 180° - 45° - 45°
= 90° (Angles sum of triangle)
∠OPQ + ∠OQP = ∠QOS (Exterior angle of a triangle)
∠OPQ = ∠OQP (Isosceles triangle, OP = OQ)
∠OPQ
= 90° ÷ 2
= 45° (Exterior angle of a triangle)
∠PSR
= 180° - 45°
= 135° (Interior angles)
(b)
∠PQS
= 180° - 45° - 45°
= 90° (Angle sum of triangles)
Answer(s): (a) 135°; (b) 90°
