In a cyclic quadrilateral ABCD,diagonals AC a | vedclass

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(D) In a cyclic quadrilateral,angles subtended by the same arc at the circumference are equal.Since $ABCD$ is a cyclic quadrilateral,the angles subten

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$50$

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(D) In a cyclic quadrilateral,angles subtended by the same arc at the circumference are equal.Since $ABCD$ is a cyclic quadrilateral,the angles subtended by the same arc $AB$ at the circumference are $\angle ACB$ and $\angle ADB$.Therefore,$\angle ACB = \angle ADB = 78^{\circ}$.Now,consider the triangle $	riangle ABC$. The sum of the interior angles of a triangle is $180^{\circ}$.Thus,$\angle ABC + \angle BAC + \angle ACB = 180^{\circ}$.Substituting the given values: $\angle ABC + 52^{\circ} + 78^{\circ} = 180^{\circ}$.$\angle ABC + 130^{\circ} = 180^{\circ}$.$\angle ABC = 180^{\circ} - 130^{\circ} = 50^{\circ}$.

In a cyclic quadrilateral $ABCD$,diagonals $AC$ and $BD$ intersect at $P$. If $\angle BAC = 52^{\circ}$ and $\angle ADB = 78^{\circ}$,then find $\angle ABC$. (in $^{\circ}$)

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