From the figure we know that ∠AOD and ∠OEC form right angles
So we get
∠AOD = ∠OEC = 90o
We know that OD || BC and OC is a transversal
From the figure we know that ∠AOD and ∠OEC are corresponding angles
∠AOD = ∠OEC
We know that ∠DOC and ∠OCE are alternate angles
∠DOC = ∠OCE = 30o
Angle subtended by an arc of a circle at its centre is twice the angle subtended by the same arc at a point on the circumference.
So we get
∠DOC = 2 ∠DBC
It can be written as
∠DBC = ½ ∠DOC
By substituting the values
∠DBC = 30/2
By division
y = ∠DBC = 15o
In the same way
∠ABD = ½ ∠AOD
By substituting the values
∠ABD = 90/2
By division
∠ABD = 45o
We know that
∠ABE = ∠ABC = ∠ABD + ∠DBC
So we get
∠ABE = ∠ABC = 45o + 15o
By addition
∠ABE = ∠ABC = 60o
Consider △ ABE
Using the angle sum property
∠BAE + ∠AEB + ∠ABE = 180o
By substituting the values
x + 90o + 60o = 180o
On further calculation
x = 180o – 90o – 60o
By subtraction
x = 180o – 150o
So we get
x = 30o
Therefore, the value of x is 30o and y is 15o.