Question

$PQT$ and $PR$ are tangents to the circle. If $\angle QPR={38}^o,\angle PRS={111}^o$, find $\angle TQS$.

Answer 1

PQ$=$PR[tangents]

$\Delta$PQR$\Rightarrow 2x+{38}^o={180}^o$

$\Rightarrow x={71}^o$

$\angle PRO={90}^o$, so $\angle ORQ={19}^o=\angle OQR$

So, $\angle QOR={62}^o$, Now $\angle PRS=111$

$\Rightarrow \angle ORS=111-{90}^o-{21}^o$

$\therefore \angle ROS={138}^o$

$\Rightarrow \angle QOS=360-138-62={160}^o$

$\therefore \angle OQS=\angle OSQ={10}^o$

$\therefore \angle SQT={90}^o-\angle OQS=90-10={80}^o$

$\therefore {80}^o$.