In a cyclic quadrilateral ABCD,AB parallel CD | vedclass

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(D) $1$. In a cyclic quadrilateral,the sum of opposite angles is $180^{\circ}$. Thus,$\angle A + \angle C = 180^{\circ}$ and $\angle B + \angle D = 18

Vedclass · Accepted Answer

$\angle A = 80^{\circ}, \angle C = 100^{\circ}, \angle D = 100^{\circ}$

Vedclass · Answer

(D) $1$. In a cyclic quadrilateral,the sum of opposite angles is $180^{\circ}$. Thus,$\angle A + \angle C = 180^{\circ}$ and $\angle B + \angle D = 180^{\circ}$.$2$. Given $\angle B = 80^{\circ}$,we have $\angle D = 180^{\circ} - 80^{\circ} = 100^{\circ}$.$3$. Since $AB \parallel CD$,the sum of interior angles on the same side of the transversal is $180^{\circ}$. Thus,$\angle A + \angle D = 180^{\circ}$.$4$. Substituting $\angle D = 100^{\circ}$,we get $\angle A + 100^{\circ} = 180^{\circ}$,which implies $\angle A = 80^{\circ}$.$5$. Finally,using $\angle A + \angle C = 180^{\circ}$,we get $80^{\circ} + \angle C = 180^{\circ}$,so $\angle C = 100^{\circ}$.$6$. Therefore,the angles are $\angle A = 80^{\circ}, \angle C = 100^{\circ}, \angle D = 100^{\circ}$.

In a cyclic quadrilateral $ABCD$,$AB \parallel CD$. If $\angle B = 80^{\circ}$,then find the remaining angles of $ABCD$.

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