In the given figure, O is the centre of a circle, ∠OAB = 30° and ∠OCB = 40°. Calculate ∠AOC.

Join AC.

As, OA = OC = radius of circle.

Let ∠OAC = ∠OCA = x (As angles opposite to equal sides are equal)

We know that,

Sum of angles of triangle = 180°

∴ ∠OAC + ∠OCA + ∠AOC = 180°

⇒ x + x + ∠AOC = 180°

⇒ ∠AOC = 180° - 2x

From figure,

⇒ ∠BAC = ∠BAO + OAC = 30° + x

⇒ ∠BCA = ∠BCO + OCA = 40° + x

Now, in ∆ABC

⇒ ∠ABC = 180° - ∠BAC - ∠BCA [Angle sum property of a triangle]

= 180° - (30° + x) - (40° + x)

= 180° - 30° - x - 40° - x

= 180° - 70° - 2x

= 110° - 2x.

We know that,

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∠ABC

⇒ 180° - 2x = 2(110° - 2x)

⇒ 180° - 2x = 220° - 4x

⇒ -2x + 4x = 220° - 180°

⇒ 2x = 40°

⇒ x = 20°.

Thus, ∠AOC = 180° - 2x = 180° - 2(20°)

= 180° - 40° = 140°.

Hence, ∠AOC = 140°.