In the given figure, O is the centre of a circle and ∠AOB = 140°. Then, ∠ACB = ?
(a) 70°
(b) 80°
(c) 110°
(d) 40°

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Solution

(c) 110°
Join AB.
Then chord AB subtends ∠AOB at the centre and ∠ADB at a point D of the remaining parts of a circle.

∴∠AOB = 2∠ADB
$\Rightarrow ∠ ADB = \frac{1}{2} ∠ AOB = (\frac{1}{2} \times 140 °) = 70 °$
In the cyclic quadrilateral, we have:
∠ADB + ∠ACB = 180°
⇒ 70° + ∠ACB = 180°
∴∠ACB = (180° - 70°) = 110°

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