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Dear Student,

Please find below the solution to the asked query:

Given , ABCD is a parallelogram and we know in parallelogram opposite angles are equal , So

$\angle$ BAD  =  $\angle$ BCD

$\angle$ BCD  = 75$°$                                             (  Form given diagram )

$\angle$ PCO =  75$°$                               --- ( 1 )  (  As $\angle$ BCD  =  $\angle$ PCO  same angles )

And

Also given PQRS is a parallelogram and we know in parallelogram adjacent angles re supplementry , So

$\angle$ PQR +   $\angle$ QPS = 180$°$

95$°$ + $\angle$ QPS = 180$°$                                             (  Form given diagram )

$\angle$ QPS = 85$°$  

$\angle$ CPO =  85$°$                               --- ( 2 )          (  As $\angle$ QPS  =  $\angle$ CPO  same angles )

From angle sum property in triangle POC we get

$\angle$ PCO  +  $\angle$ CPO  +  $\angle$ POC  =  180$°$  , Substitute values from equation 1 and 2 we get

75$°$ +  85$°$  + $\angle$ POC   =  180$°$ 

160$°$  + $\angle$ POC  =  180$°$ 

$\angle$ POC  =  20$°$                                --- ( 3 )

And


$\angle$ POC  + $\angle$ POD  =  180$°$                      ( Linear pair angles )

Now substitute value from equation 3 and from diagram we get

20$°$  +  x  =  180$°$ 

x  =  160$°$                                                    ( Ans )

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