Question

In the given figure, $Q$ is the center of a circle and $PM$, $PN$ are tangent segments to the circle. If $\angle MPN={40}^o$, 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 an external common point '$P$'.
$\angle MPN={40}^o$, 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)
$\therefore {90}^o+{90}^o+{40}^o+\angle MQN={360}^o$
$\angle MQN={360}^o-({90}^o+{90}^o+{40}^o)$
$={360}^o-{220}^o$
$={140}^o$