# QP ll  TS  and angle QRS  =36 0 ,Calculate angle PQR Dear Student,

We have our diagram , As : Here we extend line TS that intersect line QR at " U "  , So  TU  | | QP (  As given QP | |TS and TU is part of TS )

So we take QR as transversal line , So

$\angle$ PQR =  $\angle$ QUS  =  x$°$ (  Alternate interior angles )

And

$\angle$ QUS + $\angle$ SUR =  180$°$   ( Linear pair angles )

x $°$ + $\angle$ SUR =  180$°$

$\angle$ SUR =  180$°$ - x $°$                                          ---- ( 1 )

And

$\angle$ TSR + $\angle$ USR =  180$°$   ( Linear pair angles )

2 x $°$ + $\angle$ USR =  180$°$

$\angle$ USR =  180$°$ - 2 x $°$                                        ---- ( 2 )

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

$\angle$ SRU + $\angle$ USR + $\angle$ SUR =  180$°$  , Substitute values from equation 1 and 2 and $\angle$ SRU = 36$°$ ( Given )  and get

36$°$  + 180$°$ - 2 x$°$ + 180$°$  - x $°$ =  180$°$

36$°$  + 180$°$ - 3 x$°$  =  0

3 x $°$ =  216$°$

x =  72$°$

Therefore,

$\angle$ PQR  = 72$°$                                                                 ( Ans )