It is given that O is the centre of the circle and ∠DAB = 50o
We know that the radii of the circle are equal
OA = OB
From the figure we know that
∠OBA = ∠OAB = 50o
Consider △ OAB
Using the angle sum property
∠OAB + ∠OBA + ∠AOB = 180o
By substituting the values
50o + 50o + ∠AOB = 180o
On further calculation
∠AOB = 180o – 50o – 50o
By subtraction
∠AOB = 180o – 100o
So we get
∠AOB = 80o
From the figure we know that AOD is a straight line
It can be written as
x = 180o – ∠AOB
By substituting the values
x = 180o – 80o
By subtraction
x = 100o
We know that the opposite angles of a cyclic quadrilateral are supplementary
So we get
∠DAB + ∠BCD = 180o
By substituting the values
50o + ∠BCD = 180o
On further calculation
∠BCD = 180o – 50o
By subtraction
y = ∠BCD = 130o
Therefore, the value of x is 100o and y is 130o.