Given,
BP ‖ CQ
And,
AC ‖ BC
∠A = ∠ABC (Since, AC = BC)
In ΔABC
∠A + ∠B + ∠C = 180°
∠A + ∠A + ∠C = 180°
2∠A + ∠C = 180°(i)
∠ACB + ∠ACQ + ∠QCD = 180°(Linear pair)
∠ACB + x = 110°(ii)
∠PBC + ∠BCQ = 180°(Co. interior angle)
20° + ∠A + ∠ACB + x = 180°
∠A = 50°(iii)
Using (iii) in (i), we get
2 x 50° + ∠ACB = 180°
∠ACB = 80°
Using value of ∠ACB in (ii) l we get
80° + x = 110°
x = 30°