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Dear Student,

Please find below the solution to the asked query:

We have our diagram , As :


Here we have join QT .


POT  +    TOQ  =  180° , ( Linear pair angles ) Substitute value from given diagram and get

140°  + TOQ =  180° ,

TOQ  =  40°                                                    --- ( 1 )

And  In TOQ  , OT = OQ ( Radius ) so from base angle theorem we get   

OQT =  OTQ                                             --- ( 2 )

And from angle sum property of triangle we get in TOQ  :

OQT  + OTQ  +  TOQ  =  180° , Substitute values from equation 1 and 2 we get

OQT  + OQT  + 40°  =  180° ,

2 OQT  =  140° ,

OQT  =  70°

And OQT  +    RQT =  180° , ( Linear pair angles ) Substitute above value and get

70°  + RQT =  180° ,

RQT  =  110°                                                      ( Ans ) 

And PQT  +    PST  =  180° , ( Opposite angles of cyclic quadrilateral PQTS ) Substitute value and get

70°  + PST =  180° ,      (  As we show OQT  =  70° and OQT  =  PQT same angles )

PST  =  110°, So

PSR  =  110°                                                   --- ( 3 )  ( Same angles )

And from angle sum property of triangle we get in PSR  :

PSR + SRP  +  RPS  =  180° , Substitute values from equation  3 and given values we get

110° + SRP  + 45°  =  180° ,

SRP + 155°  =  180° ,

SRP  =  25°

TRQ  =  25°                                                   --- ( 4 )  ( Same angles )

And from angle sum property of triangle we get in RTQ :

RTQ + RQT +  TRQ =  180° , Substitute values from equation  4 and as we solved , Then

RTQ + 110°  + 25°  =  180° ,

RTQ + 135°  =  180° ,

RTQ  =  45°                                                      ( Ans ) 


Hope this information will clear your doubts about topic.

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