Answer :

Given,

BP CQ


And,


AC BC


A = ABC (Since, AC = BC)


In


A + B + C = 180o


A + A + C = 180o


2A + C = 180o (i)


ACB + ACQ + QCD = 180o (Linear pair)


ACB + x = 110o (ii)


PBC + BCQ = 180o (Co. interior angle)


20o + A + ACB + x = 180o


A = 50o (iii)


Using (iii) in (i), we get


2 * 50o + ACB = 180o


ACB = 80o


Using value of ACB in (ii)l we get


80o + x = 110o


x = 30o

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