Q) In the given figure, O is the centre of the circle and QPR is a tangent to it at P. Prove that QAP + APR = 90°.

O is the centre of circles CBSE Board

Ans:

Since OA = OP (radii of same circle)

In Δ OAP,  ∠OPA = ∠ 0AP .. (i)

Since tangent is perpendicular to radius

∠ OPR = 900

OPA + APR = 90°

Substituting values from equation (i), we get

OAP + APR = 90°

∠ QAP + ∠ APR = 90°

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