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

Options
70°
80°
150°
140°
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Solution
140°
Explanation:
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 = `(40°)/2` = 20°
⇒ ∠AOC = 180° − 2x
⇒ ∠AOC = 180° − 2(20°)
= 180° − 40°
= 140°
