Two tangents RQ and RP are drawn from an external point R to the circle with centre O. If angle PRQ=120, then prove that OR=PR+RQ.

Answer :
 

We form our diagram , As :


And we know " A tangent to a circle is perpendicular to the radius at the point of tangency. "
So,
$\angle$ OPR  =  $\angle$ OQR  =  90$°$                                               ----- (  1 )

And In $∆$ OPR and $∆$ OQR

$\angle$ OPR  =  $\angle$ OQR  =  90$°$                                          ( From equation 1 )

OP  =  OQ                                                                                  (  Radii of same circle )

And

OR  =  OR                                                                                 ( Common side )

Hence
$∆$ OPR $\cong$ $∆$ OQR                                                     ( By RHS CONGRUENCY )

So,
RP  =  RQ                                        ---- ( 2 )                           (  BY CPCT )
And
$\angle$ ORP  =  $\angle$ ORQ                 ---- ( 3 )                         (  By CPCT ), So

$\angle$ PRQ   =  $\angle$ ORP +  $\angle$ ORQ  ,  Substitute $\angle$ PQR  =  120$°$ ( Given )  and from equation 3 , we get

$\angle$ ORP + $\angle$ ORP= 120$°$

2 $\angle$ ORP = 120$°$

$\angle$ ORP = 60$°$

And
we know Cos $\theta = \frac{Adjacent}{Hypotenuse}$
So,
In $∆$ OPR , we get

$$\begin{gathered} Cos \angle ORP = \frac{PR}{OR} \\ \Rightarrow Cos 60° = \frac{PR}{OR} \\ \Rightarrow \frac{1}{2} = \frac{PR}{OR} ( \mathrm{As} \mathrm{we} \mathrm{know} Cos 60° = \frac{1}{2} ) \\ \Rightarrow OR = 2 PR \\ \Rightarrow OR = PR + PR , \mathrm{Substitute} \mathrm{value} \mathrm{from} \mathrm{equation} 2 \mathrm{we} \mathrm{get} \\ \mathbf{\Rightarrow }\mathbf{ }\mathit{O}\mathit{R}\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{ }\mathit{P}\mathit{R}\mathbf{ }\mathbf{+}\mathbf{ }\mathbf{ }\mathit{R}\mathit{Q}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{(}\mathbf{ }\mathit{H}\mathit{e}\mathit{n}\mathit{c}\mathit{e}\mathbf{ }\mathit{p}\mathit{r}\mathit{o}\mathit{v}\mathit{e}\mathit{d}\mathbf{ }\mathbf{)} \end{gathered}$$