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Dear Student,

Please find below the solution to the asked query:

We have our diagram , As :

Here we have join PQ .

We know from base angle theorem in a triangle if two sides are equal then their base angles also equal .

In APQ , PA =  PQ (  Radius of given circle )

So,

PAQ  =  PQA  =  30°                        (  As given PAQ  =   30° )

From angle sum property in triangle we know sum of all three internal angles of any triangle is 180° , So in APQ

PAQ +  PQA  +  APQ =  180°  , Substitute all values we get

30°  +  30°  + APQ =  180° 

APQ =  120° 

And

APQ +  RPQ +   RPB =  180°                          ( Linear pair angles )

Substitute all values we get

120°  + RPQ  +  30°  =  180°     

RPQ = 30° , So 

RPQ =  RPB = 30°                                                 --- ( 1 )

In PQR and  PBR 

PQ  =  PB                                                             ( Radius of circle )

RPQ =  RPB                                                ( From equation 1 )

And

PR  =  PR                                                             ( Common side )

Hence ,
PQR  PBR                                               (  By SAS rule )

So,

PQR =  PBR                                              ( By CPCT )

And

PBR  = 90°                                                    (  As we know QR is a tangent at " Q "  and we know a line from center to tangent is perpendicular , So  PQR  = 90°  ) 

Hence ,

BR is a tangent at " B " .                                       ( Hence proved )

Hope this information will clear your doubts about Circles.

If you have any more doubts just ask here on the forum and our experts will try to help you out as soon as possible.

Regards

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