**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 = ∠ OAP .. (i)

Since tangent is perpendicular to radius

∠ OPR = 90^{0}

**∠** OPA + **∠** APR = 90°

Substituting values from equation (i), we get

**∠** OAP + **∠** APR = 90°

Since **∠** OAP = **∠** QAP (by line AO extended till point Q)

**∠ QAP + ****∠ APR = 90°**