Question

In the following figure, $M$ is the centre of the circle. $PQ$ and $PR$ are tangents to the circle. If $\angle QPR={60}^o$, find $\angle QMR$ (in ${}^{\circ }$).

Answer 1

Line $PQ$ and $PR$ are tangents to the circle and seg $MQ$ and $MR$ are the radii through the point of contact.
$\angle MQP={90}^o$ and $\angle MRP={90}^o$
(Tangent is perpendicular to radius through the point of contact)
Also, $\angle QPR={60}^o$ (Given)
Now, $\square PQMR$
$\angle MQP+\angle QMR+\angle MRP+\angle QPR={360}^o$
(Sum of measures of angles of a quandrilateral is360${}^o$)
$\therefore {90}^o+\angle QMR+{90}^o+{60}^o={360}^o$
$\therefore {240}^o+\angle QMR={360}^o$
$\therefore \angle QMR={360}^o-{240}^o={120}^o$