In the given figure, O is the centre of a circle. If ∠OAB = 40°, then ∠ACB = ?
(a) 40°
(b) 50°
(c) 60°
(d) 70°

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Solution

(b) 50°

OA = OB
⇒ ∠OBA = ∠OAB = 40°
In ΔAOB, we have:
∠AOB + ∠OAB + ∠OBA = 180°
⇒ ∠AOB + 40° + 40° = 180°
⇒ ∠AOB = (180° - 80°) = 100°
∴ $∠ ACB = \frac{1}{2} ∠ AOB = (\frac{1}{2} \times 100 °) = 50 °$

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In the given figure, O is the centre of a circle and $∠ A O B = 140^{\circ} . T h e n ∠ A C B = ?$

(a) $70^{\circ}$

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In the given figure, O is the centre of a circle. If $∠ O A B = 40^{\circ}$ and C is a point on the circle then $∠ A C B = ?$

(a) $40^{\circ}$

(b) $50^{\circ}$

(d) $100^{\circ}$

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∠ A C B

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