In the given figure, O is the centre of a circle, ∠AOB = 40° and �

1.	In the given figure, O is the centre of a circle, ∠AOB = 40° and ∠BDC = 100°, find ∠OBC.
Answer» We know that the angle subtended by an arc of a circle at the centre is double the angle subtended by the arc at any point on the circumference. So we get ∠AOB = 2 ∠ACB From the figure we know that ∠ACB = ∠DCB It can be written as ∠AOB = 2 ∠DCB We also know that ∠DCB = ½ ∠AOB By substituting the values ∠DCB = 40^o/2 By division ∠DCB = 20^o In △ DBC Using the angle sum property ∠BDC + ∠DCB + ∠DBC = 180^o By substituting the values we get 100^o + 20^o + ∠DBC = 180^o On further calculation ∠DBC = 180^o – 100^o – 20^o By subtraction ∠DBC = 180^o – 120^o So we get ∠DBC = 60^o From the figure we know that ∠OBC = ∠DBC = 60^o So we get ∠OBC = 60^o Therefore, ∠OBC = 60^o.

In the given figure, O is the centre of a circle, ∠AOB = 40° and ∠BDC = 100°, find ∠OBC.

Answer»

We know that the angle subtended by an arc of a circle at the centre is double the angle subtended by the arc at any point on the circumference.

So we get

∠AOB = 2 ∠ACB

From the figure we know that

∠ACB = ∠DCB

It can be written as

∠AOB = 2 ∠DCB

We also know that

∠DCB = ½ ∠AOB

By substituting the values

∠DCB = 40^o/2

By division

∠DCB = 20^o

In △ DBC

Using the angle sum property

∠BDC + ∠DCB + ∠DBC = 180^o

By substituting the values we get

100^o + 20^o + ∠DBC = 180^o

On further calculation

∠DBC = 180^o – 100^o – 20^o

By subtraction

∠DBC = 180^o – 120^o

So we get

∠DBC = 60^o

From the figure we know that

∠OBC = ∠DBC = 60^o

So we get

∠OBC = 60^o

Therefore, ∠OBC = 60^o.

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