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°.
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°