Question

In the following figure, PQ and PR are tangents to the circle, with centre O. If $\angle QPR={60}^{\circ }$. Select the statements that are true,

• A. $\angle QOR={120}^{\circ }$
• B. $\angle OQR={30}^{\circ }$
• C. $\angle OQR={30}^{\circ }$
• D. $\angle QSR={60}^{\circ }$

Answer 1

Answer: (A), (B), (D)

The correct options are
A

$\angle QOR={120}^{\circ }$

B

$\angle OQR={30}^{\circ }$

D

$\angle QSR={60}^{\circ }$

Join QR

Now, PQ and PR the tangents to the circle

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

(i) 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 }$

(ii) 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 $\angle OQR={60}^{\circ }$

$\therefore \angle OQR={30}^{\circ }$

(ii) Arc QR subtends $\angle QOR$ at the centre and $\angle QSR$ at the remaining part of the circle.

$\therefore \angle QSR=\frac{1}{2}\angle QOR=\frac{1}{2}\times {120}^{\circ }={60}^{\circ }$