In the given figure, the lengths of arcs AB and BC are in the ratio 3 : 2.

If ∠AOB = 96°, find :

(i) ∠BOC

(ii) ∠ABC

(i) We know that,

Ratio of the angles subtended by the chords on the center is equal to the ratio of the chords.

$\Rightarrow \dfrac{∠AOB}{∠BOC} = \dfrac{AB}{BC} \\[1em] \Rightarrow \dfrac{96°}{∠BOC} = \dfrac{3}{2} \\[1em] \Rightarrow ∠BOC = \dfrac{2}{3} \times 96° = 2 \times 32° = 64°.$

Hence, ∠BOC = 64°.

(ii) In △ AOB,

⇒ OA = OB (Radius of same circle)

⇒ ∠OBA = ∠OAB = x (let) [Angle opposite to equal sides are equal]

By angle sum property of triangle,

⇒ ∠OAB + ∠OBA + ∠AOB = 180°

⇒ x + x + 96° = 180°

⇒ 2x + 96° = 180°

⇒ 2x = 180° - 96°

⇒ 2x = 84°

⇒ x = $\dfrac{84°}{2}$

⇒ x = 42°

⇒ ∠OBA = 42°.

In △ BOC,

⇒ OB = OC (Radius of same circle)

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

By angle sum property of triangle,

⇒ ∠OCB + ∠OBC + ∠BOC = 180°

⇒ y + y + 64° = 180°

⇒ 2y + 64° = 180°

⇒ 2y = 180° - 64°

⇒ 2y = 116°

⇒ y = $\dfrac{116°}{2}$

⇒ y = 58°

⇒ ∠OBC = 58°.

From figure,

⇒ ∠ABC = ∠OBA + ∠OBC = 42° + 58° = 100°.

Hence, ∠ABC = 100°.