Question

PQ is a tangent drawn from a point P to a circle with centre O and QOR is a diameter of the circle such that ∠POR = 120°. Find ∠OPQ.

Answer 1

∠POR + ∠POQ = 180° (straight angle = 180°)

∴ 120 +∠POQ = 180°

∠POQ = 60°

∠OQP = 90° (∵ radius is ⊥^r to r the tangent at the point of contact)

∴ ∠POQ + ∠PQO + ∠OPQ = 180° (∵ sum of the 3 angles of a triangle is 180°)

∴ 60 + 90 + ∠OPQ = 80°

∠OPQ = 180° – 150° = 30°