It is given in the question that,

In cyclic quadrilateral ABCD, we have:

∠ A = (x + y + 10)^o

∠ B = (y + 20)^o

∠ C = (x + y – 30)^o

∠ D = (x + y)^o

As ABCD is a cyclic quadrilateral

∴ ∠ A + ∠ C = 180^o and ∠ B + ∠ D = 180^o

Now, ∠ A + ∠ C = 180^o

(x + y + 10)^o + (x + y – 30)^o = 180^o

2x + 2y – 20^o = 180^o

x + y = 100^o (i)

Also, ∠ B + ∠ D = 180^o

(y + 20)^o + (x + y)^o = 180^o

x + 2y + 20^o = 180^o

x + 2y = 160^o (ii)

On subtracting (i) from (ii), we get

y = (160 – 100)^o

y = 60^o

Putting the value of y in (i), we get

x + 60^o = 100^o

x = 100^o – 60^o

x = 40^o

∴ ∠ B = (y + 20)^o

∠ B = 60^o + 20^o = 80^o

Hence, option B is correct