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(B) Given that $ABCD$ is a cyclic quadrilateral and $AB$ is the diameter of the circle.Since $ABCD$ is a cyclic quadrilateral,the sum of opposite angl

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$50$

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(B) Given that $ABCD$ is a cyclic quadrilateral and $AB$ is the diameter of the circle.Since $ABCD$ is a cyclic quadrilateral,the sum of opposite angles is $180^{\circ}$.$\angle ADC + \angle ABC = 180^{\circ}$Given $\angle ADC = 140^{\circ}$,we have:$140^{\circ} + \angle ABC = 180^{\circ}$$\angle ABC = 180^{\circ} - 140^{\circ} = 40^{\circ}$Since $AB$ is the diameter,the angle subtended by the diameter at the circumference is $90^{\circ}$.Therefore,$\angle ACB = 90^{\circ}$.In $\Delta ABC$,the sum of angles is $180^{\circ}$:$\angle BAC + \angle ABC + \angle ACB = 180^{\circ}$$\angle BAC + 40^{\circ} + 90^{\circ} = 180^{\circ}$$\angle BAC + 130^{\circ} = 180^{\circ}$$\angle BAC = 180^{\circ} - 130^{\circ} = 50^{\circ}$

$ABCD$ is a cyclic quadrilateral such that $AB$ is a diameter of the circle circumscribing it and $\angle ADC = 140^{\circ}$,then $\angle BAC$ is equal to (in $^{\circ}$)

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