In parallelogram ABCD,angle A = angle B + 50^ | vedclass

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(A) In a parallelogram,adjacent angles are supplementary,so $\angle A + \angle B = 180^{\circ}$.Given that $\angle A = \angle B + 50^{\circ}$.Substitu

Vedclass · Accepted Answer

$\angle A = 115^{\circ}, \angle B = 65^{\circ}, \angle C = 115^{\circ}, \angle D = 65^{\circ}$

Vedclass · Answer

(A) In a parallelogram,adjacent angles are supplementary,so $\angle A + \angle B = 180^{\circ}$.Given that $\angle A = \angle B + 50^{\circ}$.Substituting the value of $\angle A$ in the first equation: $(\angle B + 50^{\circ}) + \angle B = 180^{\circ}$.$2 \angle B + 50^{\circ} = 180^{\circ}$.$2 \angle B = 130^{\circ}$,which gives $\angle B = 65^{\circ}$.Then,$\angle A = 65^{\circ} + 50^{\circ} = 115^{\circ}$.Since opposite angles of a parallelogram are equal,$\angle C = \angle A = 115^{\circ}$ and $\angle D = \angle B = 65^{\circ}$.Thus,the angles are $\angle A = 115^{\circ}, \angle B = 65^{\circ}, \angle C = 115^{\circ}, \angle D = 65^{\circ}$.

In parallelogram $ABCD$,$\angle A = \angle B + 50^{\circ}$. Find the measure of each angle of the quadrilateral $ABCD$.

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