ABCD is a trapezium such that AB || CD. If an | vedclass

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(A) In a trapezium $ABCD$ with $AB || CD$,the sum of consecutive interior angles between parallel lines is $180^{\circ}$.Therefore,$\angle A + \angle

Vedclass · Accepted Answer

$\angle A = 110^{\circ}, \angle B = 100^{\circ}, \angle C = 80^{\circ}, \angle D = 70^{\circ}$

Vedclass · Answer

(A) In a trapezium $ABCD$ with $AB || CD$,the sum of consecutive interior angles between parallel lines is $180^{\circ}$.Therefore,$\angle A + \angle D = 180^{\circ}$ and $\angle B + \angle C = 180^{\circ}$.Substituting the given values:$(y + 60^{\circ}) + (3y - 80^{\circ}) = 180^{\circ} \implies 4y - 20^{\circ} = 180^{\circ} \implies 4y = 200^{\circ} \implies y = 50^{\circ}$.$(x + 60^{\circ}) + (3x - 40^{\circ}) = 180^{\circ} \implies 4x + 20^{\circ} = 180^{\circ} \implies 4x = 160^{\circ} \implies x = 40^{\circ}$.Now,calculating the angles:$\angle A = 50^{\circ} + 60^{\circ} = 110^{\circ}$.$\angle B = 40^{\circ} + 60^{\circ} = 100^{\circ}$.$\angle C = 3(40^{\circ}) - 40^{\circ} = 120^{\circ} - 40^{\circ} = 80^{\circ}$.$\angle D = 3(50^{\circ}) - 80^{\circ} = 150^{\circ} - 80^{\circ} = 70^{\circ}$.

$ABCD$ is a trapezium such that $AB || CD$. If $\angle A = y + 60^{\circ}$,$\angle B = x + 60^{\circ}$,$\angle C = 3x - 40^{\circ}$ and $\angle D = 3y - 80^{\circ}$,then find the measure of each angle of $ABCD$.

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