In the figure,if angle OAB = 40^{circ}, then | vedclass

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(D) In $\Delta OAB$,$OA = OB$ (Radii of the same circle).Therefore,$\angle OAB = \angle OBA = 40^{\circ}$ (Angles opposite to equal sides are equal).I

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$50$

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(D) In $\Delta OAB$,$OA = OB$ (Radii of the same circle).Therefore,$\angle OAB = \angle OBA = 40^{\circ}$ (Angles opposite to equal sides are equal).In $\Delta OAB$,the sum of angles is $180^{\circ}$.So,$\angle AOB = 180^{\circ} - (40^{\circ} + 40^{\circ}) = 180^{\circ} - 80^{\circ} = 100^{\circ}$.According to the theorem,the angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle.Therefore,$\angle ACB = \frac{1}{2} \angle AOB = \frac{1}{2} 	imes 100^{\circ} = 50^{\circ}$.

In the figure,if $\angle OAB = 40^{\circ},$ then $\angle ACB$ is equal to: (in $^{\circ}$)

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