50°. Solution: In $\triangle\text{QAB, OA} = \text{OB}$ [both are the radius of a circle] $\angle\text{OAB} = \angle\text{OBA}\Rightarrow \angle\text{OBA} = 40^\circ$ [angles opposite to equal sides are equal] Also, $\angle\text{AOB} = \angle\text{OBA}\Rightarrow \angle\text{BAO} = 180^\circ$ [by angle sum property of a triangle] $\angle\text{AOB} + 40^\circ + 40^\circ = 180^\circ$ $\Rightarrow\ \angle\text{AOB} = 180^\circ – 80^\circ = 100^\circ$ We know that, in a circle, the angle subtended by an arc at the centre is twice the angle subtended by it at the remaining part of the circle. $\angle\text{AOB} = 2 \angle\text{ACB} \Rightarrow 100^\circ =2 \angle\text{ACB}$ $\angle\text{ACB} = \frac{100}{2} = 50^\circ$
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