Question

In figure,two tangents RQ and RP are drawn from an external point R to the circle with centre O. If $\angle PRQ={120}^{\circ }$,then prove that,OR=PR+RQ.

Answer 1

O is the centre of the circle and $\angle PRQ={120}^{\circ }$

Construction: Join OP, OQ

To Prove: OP=PR+RQ


Proof: We know that,

Tangent to a circle is perpendicular to the radius at the point of contact i.e.,$OP\perp RP$ and $OQ\perp RQ.$

$\therefore \angle OPR=\angle OQR={90}^{\circ }$

Now, in $\Delta OPR$ and $\Delta OQR,$

OP=OQ [Radius of circle]

OR=OR [Common]

$\angle OPR=\angle OQR={90}^{\circ }\left[Each{90}^{\circ }\right]$

$\therefore \Delta OPR\cong \Delta OQR$
[By RHS congruence]

So, PR=QR [By c.p.c.t]

and $\angle ORP=\angle ORQ$= $\frac{{120}^{\circ }}{2}={60}^{\circ }$

Now, in $\Delta OPR$

$cos{60}^{\circ }=\frac{PR}{OR}[\because cos\theta =\frac{\text{Base}}{\text{Hypotenuse}}]$

$\frac{1}{2}=\frac{PR}{OR}$

OR=2PR

OR=PR+PR

OR=PR+RQ [$\because$ PR=RQ]

Hence, OR=PR+RQ. Hence Proved.