In the given figure, O is the centre of the circle in which ∠OBA = 30° and ∠OCA = 40°. Then, ∠BOC = ?

1. 70°

2. 100°

3. 120°

4. 140°

Join BC.

In △BOC,

Since,

OB = OC (Radius of same circle)

∴ ∠OBC = ∠OCB = x (let)

By angle sum property of triangle,

⇒ ∠OBC + ∠OCB + ∠BOC = 180°

⇒ x + x + ∠BOC = 180°

⇒ ∠BOC = 180° - 2x

In △AOC,

By angle sum property of triangle,

⇒ ∠BAC + ∠CBA + ∠ACB = 180°

⇒ ∠BAC + (30° + x) + (40° + x) = 180°

⇒ ∠BAC + 70° + 2x = 180°

⇒ ∠BAC = 180° - 70° - 2x

⇒ ∠BAC = 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.

∴ ∠BOC = 2∠BAC

⇒ 180° - 2x = 2(110° - 2x)

⇒ 180° - 2x = 220° - 4x

⇒ 4x - 2x = 220° - 180°

⇒ 2x = 40°

⇒ x = $\dfrac{40^{\circ}}{2}$ = 20°

⇒ ∠BOC = 180° - 2x

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

Hence, option 4 is the correct option.