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

(i) We know that,

If two arcs are of a circle are equal, they subtend equal angles at the center.

∴ ∠BOC = ∠AOB = 48°.

Hence, ∠BOC = 48°.

(ii) Join BC.

In △ OBC,

⇒ OC = OB (Radius of same circle)

⇒ ∠OBC = ∠OCB = y (let) [Angle opposite to equal sides are equal]

By angle sum property of triangle,

⇒ ∠OBC + ∠OCB + ∠BOC = 180°

⇒ y + y + 48° = 180°

⇒ 2y + 48° = 180°

⇒ 2y = 180° - 48°

⇒ 2y = 132°

⇒ y = $\dfrac{132°}{2}$

⇒ y = 66°

⇒ ∠OBC = 66°.

Hence, ∠OBC = 66°.

(iii) From figure,

⇒ ∠AOC = ∠AOB + ∠BOC = 48° + 48° = 96°.

Hence, ∠AOC = 96°.

(iv) Join AC.

In △ AOC,

⇒ OC = OA (Radius of same circle)

⇒ ∠OAC = ∠OCA = x (let) [Angle opposite to equal sides are equal]

By angle sum property of triangle,

⇒ ∠OAC + ∠OCA + ∠AOC = 180°

⇒ x + x + 96° = 180°

⇒ 2x + 96° = 180°

⇒ 2x = 180° - 96°

⇒ 2x = 84°

⇒ x = $\dfrac{84°}{2}$

⇒ x = 42°

⇒ ∠OAC = 42°.

Hence, ∠OAC = 42°.