ABCD is a cyclic quadrilateral in which angle | vedclass

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

Vedclass · Accepted Answer

$80$

Vedclass · Answer

(C) 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 arc $AB$ at the circumference are $\angle ADB$ and $\angle ACB$.Therefore,$\angle ACB = \angle ADB = 55^{\circ}$.Now,consider the triangle $	riangle ABC$. The sum of the angles in a triangle is $180^{\circ}$.$\angle BAC + \angle ABC + \angle ACB = 180^{\circ}$.Substituting the given values: $45^{\circ} + \angle ABC + 55^{\circ} = 180^{\circ}$.$100^{\circ} + \angle ABC = 180^{\circ}$.$\angle ABC = 180^{\circ} - 100^{\circ} = 80^{\circ}$.

$ABCD$ is a cyclic quadrilateral in which $\angle BAC = 45^{\circ}$ and $\angle ADB = 55^{\circ},$ then $\angle ABC = \dots$ (in $^{\circ}$)

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