In the given figure, O is the center of the circle. ∠OAB and ∠OCB are 30° and 40° respectively. ∠AOC is equal to :

1. 70°

2. 80°

3. 150°

4. 140°

Join AC.

In △AOC,

Since,

OA = OC (Radius of same circle)

∴ ∠OAC = ∠OCA = x (let)

By angle sum property of triangle,

⇒ ∠OAC + ∠OCA + ∠AOC = 180°

⇒ x + x + ∠AOC = 180°

⇒ ∠AOC = 180° - 2x

In △AOC,

By angle sum property of triangle,

⇒ ∠BAC + ∠ACB + ∠CBA = 180°

⇒ (30° + x) + (40° + x) + ∠CBA = 180°

⇒ ∠CBA + 70° + 2x = 180°

⇒ ∠CBA = 180° - 70° - 2x

⇒ ∠CBA = 110° - 2x

We know that,

The angle which an arc subtends at the center is double that which it subtends at any point on the remaining part of the circumference.

∴ ∠AOC = 2∠CBA

⇒ 180° - 2x = 2(110° - 2x)

⇒ 180° - 2x = 220° - 4x

⇒ 4x - 2x = 220° - 180°

⇒ 2x = 40°

⇒ x = $\dfrac{40°}{2}$ = 20°.

⇒ ∠AOC = 180° - 2x

⇒ ∠AOC = 180° - 2(20°) = 180° - 40° = 140°.

Hence, Option 4 is the correct option.