Question

In the given figure, arc AB and arc BC are equal in length. If ∠AOB = 48°, find:
(i) ∠BOC
(ii) ∠OBC
(iii) ∠AOC
(iv) ∠OAC

Answer 1

We know that the arc of equal lengths subtends equal angles at the center.

hence ∠AOB = ∠BOC = 48°
Then ∠AOC = ∠AOB + ∠BOC = 48° + 48° = 96°
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 + 48° = 180°              as ∠OBC = ∠OCB
 2∠OBC = 180° - 48°
 2∠OBC = 132°
∠OBC = 66°
as ∠OBC = ∠OCB 
So, ∠OBC = ∠OCB = 66°

The triangle thus formed, ΔAOC is an isosceles triangle with OA = OC as they are radii of the same circle.
Thus ∠OAC = ∠OCA as they are opposite angles of equal sides of an isosceles triangle.

The sum of all the angles of a triangle is 180°.
So, ∠COA + ∠OAC + ∠OCA = 180°
2∠OAC + 96° = 180°                 as, ∠OAC = ∠OCA 
2∠OAC = 180° - 96°
2∠OAC = 84° 
∠OAC = 42°
as ∠OCA = ∠OAC
So, ∠OCA = ∠OAC = 42°.