In the given figure, O is the centre of a circle and ∠AOB = 140°.

Question:

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

(b) 80°
(c) 110°
(d) 40°

 

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

$\Rightarrow \angle \mathrm{ADB}=\frac{1}{2} \angle \mathrm{AOB}=\left(\frac{1}{2} \times 140^{\circ}\right)=70^{\circ}$

In the cyclic quadrilateral, we have:
∠ADB + ∠ACB = 180°

⇒ 70° + ∠ACB = 180°
∠ACB = (180° - 70°) = 110°

 

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