Question

Find the angle between two radii at the centre of the circle as shown in the figure. Lines PA and PB are tangents to the circle at other ends of the radii and $\angle$APR $={130}^o$.

Answer 1

Given that $AP$ and $PB$ are tangents to given circle.

Point $O$ is center of the circle, so we have $\angle OAP=\angle OBP={90}^0$

Given $\angle APR={130}^0$

So we get $\angle APB={180}^0-\angle APR={180}^0-{130}^0={50}^0$

Now sum of angles of quadrilateral $OAPB$ mst be ${360}^0$

$\Rightarrow \angle AOB+{90}^0+{90}^0+{50}^0={360}^0$

$\Rightarrow \angle AOB={360}^0-{230}^0={130}^0$