In the figure, PQRS is a parallelogram. O is the centre of the circle. Find
- ∠PSR
- ∠PQS
(a)
∠OQS = ∠QSO = 40° (Angle properties within a circle)
∠SOQ
= 180° - 40° - 40°
= 100° (Angles sum of triangle)
∠OPQ + ∠OQP = ∠QOS (Exterior angle of a triangle)
∠OPQ = ∠OQP (Isosceles triangle, OP = OQ)
∠OPQ
= 100° ÷ 2
= 50° (Exterior angle of a triangle)
∠PSR
= 180° - 50°
= 130° (Interior angles)
(b)
∠PQS
= 180° - 50° - 40°
= 90° (Angle sum of triangles)
Answer(s): (a) 130°; (b) 90°