Question

In fig. $PQ$ is a tangent from an external point $P$ to a circle with centre $O$ and $OP$ cuts the circle at $T$ and $\angle QOR$ is a diameter. If $\angle PQR={130}^{\circ }$ and $S$ is a point on the circle, find $\angle 1+\angle 2$.

Answer 1

$\angle ROT=2\angle RST$ $\because$ (angle at centre = $2\times$ angle at circumference of the circle)

${130}^{\circ }=2\angle 2$

$\angle 2=130/2$

$={65}^{\circ }$

$\angle OQP={90}^{\circ }$ $\because$ (the point of contact of tangent and radius makes ${90}^{\circ })$

Side $QO$ of $△POQ$ is produced to R

$\angle QPO+\angle OQP=\angle ROT$ $\because$ (Ext. $\angle =$ sum of 2 opp. int $\angle s$ in a $△)$

$\angle 1+{90}^{\circ }={130}^{\circ }$

$\angle 1={130}^{\circ }-{90}^{\circ }$

$={40}^{\circ }$

$\angle 1+\angle 2={65}^{\circ }+{40}^{\circ }$

$={105}^{\circ }$

$\therefore \angle 1+\angle 2={105}^{\circ }$