Question

# In the given figure, O is the centre of a circle. If ∠OAB = 40° and C is a point on the circle, then ∠ACB = ? (a) 40° (b) 50° (c) 80° (d) 100°

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Solution

## (b) 50° OA = OB ⇒ ∠OBA = ∠OAB = 40° Now, ∠AOB = 180° - (40° + 40°) = 100° ∴ $\angle \mathrm{ACB}=\frac{1}{2}\angle \mathrm{AOB}=\left(\frac{1}{2}×100\right)°=50°$

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