# 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.

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
So,
In $∆$ OPR , we get

• 176
Given in rd sharma. Mcqs
• -23
I got this through trigonometry
• -10
I didn;t know this...skipped.
• -46
Hint Use cos 60
• -25
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