Question

PQ is a tangent to a circle with centre O at the point P. If $\Delta$ OPQ is an isosceles triangle, then $\angle$ OQP is equal to

(a) ${30}^o$ (b) ${45}^o$ (c) ${60}^o$ (d) ${90}^o$

Answer 1

We found a similar question to the one that you posted.

In △OPQ, OP=PQ

Therefore POQ = ∠PQO

We know that

∠POQ + ∠PQO + ∠OPQ = 180

2∠PQO + 90 = 180

∠PQO= 90/2 = 45