In the figure,BC is a diameter of the circle | vedclass

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(C) In $\Delta OAB$,we have:$OA = OB$ [Radii of the same circle]Therefore,$\angle ABO = \angle BAO$ [Angles opposite to equal sides are equal].Given $

Vedclass · Accepted Answer

$60$

Vedclass · Answer

(C) In $\Delta OAB$,we have:$OA = OB$ [Radii of the same circle]Therefore,$\angle ABO = \angle BAO$ [Angles opposite to equal sides are equal].Given $\angle BAO = 60^{\circ}$,so $\angle ABO = 60^{\circ}$.In $\Delta OAB$,the sum of angles is $180^{\circ}$,so $\angle AOB = 180^{\circ} - (60^{\circ} + 60^{\circ}) = 60^{\circ}$.Since $BC$ is a diameter,$\angle AOC = 180^{\circ} - \angle AOB = 180^{\circ} - 60^{\circ} = 120^{\circ}$.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.Thus,$\angle ADC = \frac{1}{2} \angle AOC = \frac{1}{2} 	imes 120^{\circ} = 60^{\circ}$.

In the figure,$BC$ is a diameter of the circle and $\angle BAO = 60^{\circ}$. Then $\angle ADC$ is equal to: (in $^{\circ}$)

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