Question

In the following figure, PQ and PR are tangents to the circle, with centre O. If $\angle QPR={60}^{\circ }$, find the value of $\angle OQR$.

Answer 1

Join QR

Now, PQ and PR are the tangents to the circle

PQ = PR and $\angle QPR={60}^{\circ }$

In quad $ORPQ,OQ\perp OP,OR\perp RP$

$\therefore \angle OQP={90}^{\circ },\angle ORP={90}^{\circ }and\angle QPR={60}^{\circ }$

$\angle QOR$$={360}^{\circ }-({90}^{\circ }+{90}^{\circ }+{60}^{\circ })={360}^{\circ }-{240}^{\circ }={120}^{\circ }$ (1 mark)

In$\Delta QOR,OQ=OR$ (radii of the same circle)

$\therefore \angle OQR=\angle QRO$ .....(i)

But $\angle OQR+\angle QRO+\angle QOR={180}^{\circ }$

$\angle OQR+\angle QEO+{120}^{\circ }={180}^{\circ }$

$\angle OQR+\angle QRO={180}^{\circ }-{120}^{\circ }={60}^{\circ }$

$\therefore \angle OQR+\angle OQR={60}^{\circ }$

[From (i)]

2 $\angle OQR={60}^{\circ }$

$\therefore \angle OQR={30}^{\circ }$ (2 marks)