In the given figure, O is the centre of a circle in which ∠OAB = 20�

1.	In the given figure, O is the centre of a circle in which ∠OAB = 20° and ∠OCB = 55°. Find(i) ∠BOC,(ii) ∠AOC.
Answer» (i) We know that OB = OC which is the radius The base angles of an isosceles triangle are equal So we get ∠OBC = ∠OCB = 55^o In △ BOC Using the angle sum property ∠BOC + ∠OCB + ∠OBC = 180^o By substituting the values ∠BOC + 55^o + 55^o = 180^o On further calculation ∠BOC = 180^o – 55^o – 55^o By subtraction ∠BOC = 180^o – 110^o So we get ∠BOC = 70^o (ii) We know that OA = OB which is the radius The base angles of an isosceles triangle are equal So we get ∠OBA= ∠OAB = 20^o In △ AOB Using the angle sum property ∠AOB + ∠OAB + ∠OBA = 180^o By substituting the values ∠AOB + 20^o + 20^o = 180^o On further calculation ∠AOB = 180^o – 20^o – 20^o By subtraction ∠AOB = 180^o – 40^o So we get ∠AOB = 140^o We know that ∠AOC = ∠AOB – ∠BOC By substituting the values ∠AOC = 140^o – 70^o So we get ∠AOC = 70^o

In the given figure, O is the centre of a circle in which ∠OAB = 20° and ∠OCB = 55°. Find(i) ∠BOC,(ii) ∠AOC.

Answer»

(i) We know that OB = OC which is the radius

The base angles of an isosceles triangle are equal

So we get

∠OBC = ∠OCB = 55^o

In △ BOC

Using the angle sum property

∠BOC + ∠OCB + ∠OBC = 180^o

By substituting the values

∠BOC + 55^o + 55^o = 180^o

On further calculation

∠BOC = 180^o – 55^o – 55^o

By subtraction

∠BOC = 180^o – 110^o

So we get

∠BOC = 70^o

(ii) We know that OA = OB which is the radius

The base angles of an isosceles triangle are equal

So we get

∠OBA= ∠OAB = 20^o

In △ AOB

Using the angle sum property

∠AOB + ∠OAB + ∠OBA = 180^o

By substituting the values

∠AOB + 20^o + 20^o = 180^o

On further calculation

∠AOB = 180^o – 20^o – 20^o

By subtraction

∠AOB = 180^o – 40^o

So we get

∠AOB = 140^o

We know that

∠AOC = ∠AOB – ∠BOC

By substituting the values

∠AOC = 140^o – 70^o

So we get

∠AOC = 70^o