In the figure, PQRS is a parallelogram. O is the centre of the circle. Find
- ∠PSR
- ∠PQS
(a)
∠OQS = ∠QSO = 44° (Angle properties within a circle)
∠SOQ
= 180° - 44° - 44°
= 92° (Angles sum of triangle)
∠OPQ + ∠OQP = ∠QOS (Exterior angle of a triangle)
∠OPQ = ∠OQP (Isosceles triangle, OP = OQ)
∠OPQ
= 92° ÷ 2
= 46° (Exterior angle of a triangle)
∠PSR
= 180° - 46°
= 134° (Interior angles)
(b)
∠PQS
= 180° - 46° - 44°
= 90° (Angle sum of triangles)
Answer(s): (a) 134°; (b) 90°