Question

$PQ$ is a tangent drawn from a point $P$ to a circle with centre $O$ and $QOR$ is a diameter of the circle such that $\angle POR={120}^o$, then $\angle OPQ$ is

• A. ${60}^o$
• B. ${45}^o$
• C. ${30}^o$
• D. ${90}^o$

Answer 1

The correct option is C ${30}^o$

Given- $PQ$ is a tangent to a circle with centre $O$ at $Q$. $QOR$ is a diameter of the given circle so that $\angle POR={120}^o$. To find out- $\angle OPQ=?$
Solution- $QOR$ is a diameter.
$\therefore OQ$ is a radius through the point of contact $Q$ of the tangent $PQ$. $\therefore \angle OQP={90}^o$ since the radius through the point of contact of a tangent to a circle is perpendicular to the tangent.$\therefore \angle OPQ+\angle OQP={120}^o$
(external angles of a triangle=sum of the internal opposite angles )
$\therefore \angle OPQ={120}^o-{90}^o={30}^o.$
Ans- Option C.