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Answer:

 In the triangle ADP, ∠D = 52° and   ∠DAP  = 42°

Therefore y = ∠APQ = exterior angle of  ∠APD of triangle ADP = ∠D + ∠DAP = 52+42 = 94°

 Hence ∠RPQ = 180 - ∠APQ  =  180 - 94 = 86°(since APR is a straight line) ………………….A

 In triangle ARB, it is given that AR = BR.Therefore triangle ARB is isosceles and so

 ∠RBA =  ∠RAB = x. Now ABCD is a parallelogram and so AB is parallel to CD and so line segment PQ which is a part of CD is also parallel to AB. Therefore triangle PQR is similar to triangle ABR and so ∠RPQ=   ∠RAB  and ∠PQR = ∠RBA . But   ∠RBA =  ∠RAB = x.

Therefore  ∠RPQ = ∠PQR = x = 86° from “A”

In triangle PQR, ∠RPQ= ∠PQR = x = 86° and so 

∠PRQ = z =  180 -  (∠RPQ +  ∠PQR)  = 180 - (86+86)  = 8°

Thus  x = 86° , y = 94° and  z = 8°

 

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