In the given figure, O is the centre of the circle and the length of arc AB is twice the length of arc BC.

If ∠AOB = 100°, find :

(i) ∠BOC

(ii) ∠OAC

(i) We know that,

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

$\Rightarrow \dfrac{∠AOB}{∠BOC} = \dfrac{AB}{BC} \\[1em] \Rightarrow \dfrac{100°}{∠BOC} = \dfrac{2}{1} \\[1em] \Rightarrow ∠BOC = \dfrac{1}{2} \times 100° = 50°.$

Hence, ∠BOC = 50°.

(ii) Join AC.

From figure,

⇒ ∠AOC = ∠AOB + ∠BOC = 100° + 50° = 150°.

In △ OAC,

⇒ OA = OC (Radius of same circle)

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

By angle sum property of triangle,

⇒ ∠OAC + ∠OCA + ∠AOC = 180°

⇒ x + x + 150° = 180°

⇒ 2x + 150° = 180°

⇒ 2x = 180° - 150°

⇒ 2x = 30°

⇒ x = $\dfrac{30°}{2}$

⇒ x = 15°

⇒ ∠OAC = 15°.

Hence, ∠OAC = 15°.