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Please find below the solution to the asked query:

Given : ABCD is a rhombus , We know in rhombus diagonals intersect at right angle , So

$\angle$ AOB =  $\angle$ BOC  =  $\angle$ COD = $\angle$ DOA = 90$°$                                    --- ( 1 )

In $∆$ AOB from angle sum property of triangle we get :

$\angle$ OAB + $\angle$ OBA + $\angle$ AOB = 180$°$  , Now we substitute values from given diagram and equation 1 and get :

x + 53$°$ + 90$°$  = 180$°$ ,

x  + 143$°$ = 180$°$ ,

x  = 37$°$  

And in  $∆$ ABC we know :  AB =  BC (  As we know ABCD is a rhombus and sides of rhombus equal to each other ) 
So, from base angle theorem we get : 

$\angle$ BAC  =  $\angle$ BCA , Substitute values from given diagram we get : 

x  =  z  , And we calculate value of x  = 37$°$  , So

z  = 37$°$ 

Now in $∆$ BOC from angle sum property of triangle we get :

$\angle$ OBC + $\angle$ OCB+ $\angle$ BOC = 180$°$  , Now we substitute values from given diagram and equation 1 and get :

$\angle$ OBC  + z + 90$°$ = 180$°$ ,

$\angle$ OBC  + z = 90$°$ , As we calculate z  = 37$°$  , So

$\angle$ OBC + 37$°$ = 90$°$

$\angle$ OBC = 53$°$                                  --- ( 2 )

And

$\angle$ ABC  =  $\angle$ OBA + $\angle$ OBC , Now we substitute values from given diagram and equation 2 and get :

$\angle$ ABC  = 53$°$  + 53$°$  ,

$\angle$ ABC = 106$°$

And we know in rhombus opposite angles are equal to each other , So

$\angle$ CDA  =  $\angle$ ABC , Substitute value from above equation we get  :

$\angle$ CDA  = 106$°$  , Now we substitute value from given diagram and get :

y  = 106$°$

Therefore,

x  =  37$°$ , y  = 106$°$ and z  = 37$°$                                                         ( Ans )


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