Given,

BP CQ

And,

AC ‖ BC

∠A = ∠ABC (Since, AC = BC)

In

∠A + ∠B + ∠C = 180^o

∠A + ∠A + ∠C = 180^o

2∠A + ∠C = 180^o (i)

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

∠ACB + x = 110^o (ii)

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

20^o + ∠A + ∠ACB + x = 180^o

∠A = 50^o (iii)

Using (iii) in (i), we get

2 * 50^o + ∠ACB = 180^o

∠ACB = 80^o

Using value of ∠ACB in (ii)l we get

80^o + x = 110^o

x = 30^o