Question

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

Answer 1

Step-by-step explanation:

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 = 40o/2

By division

∠DCB = 20o

In △ DBC Using the angle sum property

∠BDC + ∠DCB + ∠DBC = 180

By substituting the values we get

100o + 20o + ∠DBC = 180o

On further calculation

∠DBC = 180o – 100o – 20o

By subtraction

∠DBC = 180o – 120o

So we get

∠DBC = 60o

From the figure we know that

∠OBC = ∠DBC = 60o

So we get ∠OBC = 60o

Therefore, ∠OBC = 60o

Answer 2

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