Question

In the given figure, the lengths of arcs AB and BC are in the ratio 3:2. If ∠AOB = 96°, find: 

1. ∠BOC
2. ∠ABC

Answer 1

We know that for two arcs are in ratio 3: 2 then

∠AOB: ∠BOC = 3: 2

As give ∠AOB = 96°

So, 3x = 96

x = 32

There ∠BOC = 2 × 32 = 64°

The triangle thus formed, ΔAOB is an isosceles triangle with OA = OB as they are radii of the same circle.

Thus, ∠OBA = ∠BAO as they are opposite angles of equal sides of an isosceles triangle.

The sum of all the angles of a triangle is 180°

So, ∠AOB + ∠OBA + ∠BAO = 180°

2∠OBA + 96° = 180°          ...[as, ∠OBA = ∠BAO]

2∠OBA = 180° - 96° 

2∠OBA = 84°

∠OBA = 42°

as, ∠OBA = ∠BAO So,

∠OBA = ∠BAO = 42°

The triangle thus formed, ΔBOC is an isosceles triangle with OB = OC as they are radii of the same circle.

Thus, ∠OBC = ∠OCB as they are opposite angles of equal sides of an isosceles triangle.

The sum of all the angles of a triangle is 180°

So, ∠BOC + ∠OBC + ∠OCB = 180°

2∠OBC + 64° = 180°          ...[as, ∠OBC = ∠OCB] 

2∠OBC = 180° - 64° 

2∠OBC =  116° 

∠OBC = 58°

As ∠OBC = ∠OCB So,

∠OBC = ∠OCB = 58° 

∠ABC = ∠BOA + ∠OBC

= 42° + 58°

= 100°