Question

In the following figure, $Q$ is the centre of the circle and $PM$ and $PN$ are tangent segments to the circle. If $m\angle MPN={30}^o$, find $\angle MQN$.

Answer 1

In the given figure of the question $Q$ is the centre of the circle and $PM$ and $PN$ are tangents from external common point '$P$'.
$\angle MPN={30}^o$, to find $\angle MQN$
In $\square PMQN$,
$\angle PMQ=\angle PNQ={90}^o$
[Radius is $\perp$ to tangent at point of contact from the centre]
$\therefore \angle PMQ+\angle PNQ+\angle MPN+\angle MQN={360}^o$
[Sum of measures of interior angles of quadrilateral]
$\therefore {90}^o+{90}^o+{30}^o+\angle MQN={360}^o$
$\therefore \angle MQN={360}^o-({90}^o+{90}^o+{30}^o)$
$\therefore \angle MQN={150}^o$.