In the given figure,ABCD is a cyclic quadrila | vedclass

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(B) We know that angles in the same segment of a circle are equal.Therefore,$\angle BDC = \angle BAC$.Given $\angle BAC = 50^{\circ}$,so $\angle BDC =

Vedclass · Accepted Answer

$85$

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(B) We know that angles in the same segment of a circle are equal.Therefore,$\angle BDC = \angle BAC$.Given $\angle BAC = 50^{\circ}$,so $\angle BDC = 50^{\circ}$.Now,$\angle ADC = \angle ADB + \angle BDC$.Substituting the values,$\angle ADC = 45^{\circ} + 50^{\circ} = 95^{\circ}$.Since $ABCD$ is a cyclic quadrilateral,the sum of opposite angles is $180^{\circ}$.Therefore,$\angle ABC + \angle ADC = 180^{\circ}$.$\angle ABC + 95^{\circ} = 180^{\circ}$.$\angle ABC = 180^{\circ} - 95^{\circ} = 85^{\circ}$.

In the given figure,$ABCD$ is a cyclic quadrilateral in which diagonals $AC$ and $BD$ intersect at $M.$ If $\angle BAC = 50^{\circ}$ and $\angle ADB = 45^{\circ}$,then find $\angle ABC$. (in $^{\circ}$)

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