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(A) $1$. In a cyclic quadrilateral $ABCD$,the sum of opposite angles is $180^{\circ}$.$2$. Therefore,$\angle ABC + \angle ADC = 180^{\circ}$.$3$. Give

Vedclass · Accepted Answer

$60$

Vedclass · Answer

(A) $1$. In a cyclic quadrilateral $ABCD$,the sum of opposite angles is $180^{\circ}$.$2$. Therefore,$\angle ABC + \angle ADC = 180^{\circ}$.$3$. Given $\angle ADC = 150^{\circ}$,we have $\angle ABC = 180^{\circ} - 150^{\circ} = 30^{\circ}$.$4$. Since $AB$ is the diameter,the angle subtended by the diameter at the circumference is $90^{\circ}$. Thus,$\angle ACB = 90^{\circ}$.$5$. In $	riangle ABC$,the sum of angles is $180^{\circ}$.$6$. $\angle BAC + \angle ABC + \angle ACB = 180^{\circ}$.$7$. $\angle BAC + 30^{\circ} + 90^{\circ} = 180^{\circ}$.$8$. $\angle BAC = 180^{\circ} - 120^{\circ} = 60^{\circ}$.

In a circle with centre $P$,$AB$ is a diameter and $ABCD$ is a cyclic quadrilateral. If $\angle ADC = 150^{\circ}$,then find $\angle BAC$. (in $^{\circ}$)

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