Question

In the following figure, PQ is a chord of a circle with centre O and PT is a tangent. If $\angle QPT={60}^o$, find $\angle PRQ$.

Answer 1

Given, $PQ$ is a chord of a circle with center $O$.

Also, $\angle QPT=60°$.

Let $x$ be the point on the tangent $PT$.

$\angle QPT+\angle OPT=90$

$\Rightarrow \angle OPT={30}^0-\angle QPT={90}^0-{60}^0={30}^0$

In $\Delta POQ$

$\angle POQ=180-(\angle OPQ+\angle PQO)=180-30-30={120}^0$

Minor arc $\angle POQ={120}^0$

Therefore Major arc $\angle POQ={360}^0-{120}^0={240}^0$

Angle subtended by an arc at centre is double the angle subtended by it on remaining part of circle

$\therefore \angle QRP=\frac{1}{2}\angle POQ={120}^0$