# In Fig. 9.53, if

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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