Question

In the given figure, O is the centre of the circle and ∠BCO = 30°. Find x and y.

Answer 1

From the figure we know that ∠AOD and ∠OEC form right angles

So we get

∠AOD = ∠OEC = 90^o

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 = 30^o

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 = 15^o

In the same way

∠ABD = ½ ∠AOD

By substituting the values

∠ABD = 90/2

By division

∠ABD = 45^o

We know that

∠ABE = ∠ABC = ∠ABD + ∠DBC

So we get

∠ABE = ∠ABC = 45^o + 15^o

By addition

∠ABE = ∠ABC = 60^o

Consider △ ABE

Using the angle sum property

∠BAE + ∠AEB + ∠ABE = 180^o

By substituting the values

x + 90^o + 60^o = 180^o

On further calculation

x = 180^o – 90^o – 60^o

By subtraction

x = 180^o – 150^o

So we get

x = 30^o

Therefore, the value of x is 30^o and y is 15^o.