PQ is a tangent to a circle with centre 0 at the point P. If A Δ OPQ is an isosceles triangle, then ∠OQP is equal to

Question

PQ is a tangent to a circle with centre 0 at the point P. If A Δ OPQ is an isosceles triangle, then ∠ OQP is equal to

Philoid · Accepted Answer

Property 1: The tangent at a point on a circle is at right angles to the radius obtained by joining center and the point of tangency.

Property 2: Sum of all angles of a triangle = 180°.

By property 1, ∆POQ is right-angled at ∠OPQ (i.e., ∠OPQ = 90°).

∵ ∆POQ is an isosceles triangle

∴ ∠POQ = ∠OQP

By property 2,

∠POQ + ∠OQP + ∠QPO = 180°

⇒ ∠POQ + ∠OQP = 180° - ∠QPO

⇒ ∠POQ + ∠OQP = 180° - 90°

⇒ ∠POQ + ∠OQP = 90°

⇒ ∠OQP + ∠OQP = 90° [∵∠POQ = ∠OQP]

⇒ 2∠OQP = 90°

⇒ ∠OQP = 45°

Hence, ∠OQP = 45°