PQ is a tangent to a circle with centre O at point P. If ∆OPQ is an isosceles triangle, then find ∠OQP

PQ is a tangent to a circle with centre O at point P. If OPQ is an

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Question 9 (Choice - 2) PQ is a tangent to a circle with centre O at point P. If ∆OPQ is an isosceles triangle, then find ∠OQPPQ is the tangent and OQ is the radius We know that Tangent is perpendicular to radius ∴ ∠ OPQ = 90° Since Δ OPQ is isosceles OP = PQ ∴ ∠ OQP = ∠ POQ Thus, ∠ OQP = 1/2 × 90° = 45° (Isosceles triangle Property)

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