Question

In the fig. 8.79, PQ is a tangent from an external point P to a circle with center O and OP cuts the circle at T and QOR is a diameter. If $\angle POR={130}^{\circ }$ and S is a point on the circle, find $\angle 1+\angle 2.$

Answer 1

$\angle ROT=2\angle RST$ since angle at centre $=2\times$ angle at circumference of the circle.

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

$\Rightarrow \angle 2=\frac{130}{2}={65}^{\circ }$

$\angle OPQ={90}^{\circ }$ since the point of contact of tangent and radius makes ${90}^{\circ }$

Side $OQ$ of $△POQ$ is produced to $R$

$\angle QPO+\angle OQP=\angle ROT$ since exterior angle$=$sum of two opposite interior angles in a triangle.

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

$\Rightarrow \angle 1={130}^{\circ }-{90}^{\circ }={40}^{\circ }$

$\angle 1+\angle 2={65}^{\circ }+{40}^{\circ }={105}^{\circ }$

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