(a) 40°
From the given figure TS ∥ BE and also BS is transversal line.
By the rule of alternate interior angles, ∠EBS = ∠BST = 40o
Then, ∠y = 90o … [∵diagonal bisect at 90o]
Consider triangle TSO,
By the rule of exterior angle property of triangle
∠STO + ∠TSO = ∠SOE
x + 40o = 90o
x = 90o – 40o
x = 50o
So, the value of y – x is = 90o – 40o = 50o