Question

In fig. 10.85,PQ is a chord of a circle and PT is the tangent at p such that $\angle QPT={60}^O$ Then, find $\angle PQR$.

Answer 1

R.E.F image

As radius $\perp$ tangent

$\Rightarrow OPT={90}^{\circ }\to \angle OPQ={30}^{\circ }$

& OP = OQ = radius

$\Rightarrow \angle OPQ=\angle OQP={30}^{\circ }$

$\Rightarrow \angle POQ=\angle 180-30-30={120}^{\circ }$

As angle subtended by arc PQ on center, $\angle POQ$ is

double to that subtended at any other point

on circle $\left(R\right)\Rightarrow \angle PRQ=\frac{{120}^{\circ }}{2}={60}^{\circ }$