In the figure,ABCD is a cyclic quadrilateral | vedclass

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(A) We know that angles in the same segment of a circle are equal.Therefore,$\angle CAD = \angle DBC = 55^{\circ}$.Now,the total angle $\angle DAB = \

Vedclass · Accepted Answer

$80$

Vedclass · Answer

(A) We know that angles in the same segment of a circle are equal.Therefore,$\angle CAD = \angle DBC = 55^{\circ}$.Now,the total angle $\angle DAB = \angle CAD + \angle BAC$.Substituting the values,$\angle DAB = 55^{\circ} + 45^{\circ} = 100^{\circ}$.Since $ABCD$ is a cyclic quadrilateral,the sum of its opposite angles is $180^{\circ}$.Thus,$\angle DAB + \angle BCD = 180^{\circ}$.Substituting the value of $\angle DAB$,we get $100^{\circ} + \angle BCD = 180^{\circ}$.Therefore,$\angle BCD = 180^{\circ} - 100^{\circ} = 80^{\circ}$.

In the figure,$ABCD$ is a cyclic quadrilateral in which $AC$ and $BD$ are its diagonals. If $\angle DBC = 55^{\circ}$ and $\angle BAC = 45^{\circ}$,find $\angle BCD$. (in $^{\circ}$)

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