In the figure,A, B and C are three points on | vedclass

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(D) Given,a circle with centre $O$ such that $\angle AOB = 60^{\circ}$ and $\angle BOC = 30^{\circ}$.Since $\angle AOC = \angle AOB + \angle BOC$,we h

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$45$

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(D) Given,a circle with centre $O$ such that $\angle AOB = 60^{\circ}$ and $\angle BOC = 30^{\circ}$.Since $\angle AOC = \angle AOB + \angle BOC$,we have $\angle AOC = 60^{\circ} + 30^{\circ} = 90^{\circ}$.The arc $ABC$ subtends an angle $\angle AOC = 90^{\circ}$ at the centre of the circle.According to the theorem,the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.Therefore,$\angle ADC = \frac{1}{2} \angle AOC$.$\angle ADC = \frac{1}{2} (90^{\circ}) = 45^{\circ}$.

In the figure,$A, B$ and $C$ are three points on a circle with centre $O$ such that $\angle BOC = 30^{\circ}$ and $\angle AOB = 60^{\circ}$. If $D$ is a point on the circle other than the arc $ABC$,find $\angle ADC$. (in $^{\circ}$)

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