In parallelogram ABCD,angle A is an obtuse an | vedclass

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(A) In quadrilateral $AMCN$,the sum of angles is $360^{\circ}$.Given $AM \perp BC$ and $AN \perp CD$,we have $\angle AMC = 90^{\circ}$ and $\angle ANC

Vedclass · Accepted Answer

$120^{\circ}, 60^{\circ}, 120^{\circ}, 60^{\circ}$

Vedclass · Answer

(A) In quadrilateral $AMCN$,the sum of angles is $360^{\circ}$.Given $AM \perp BC$ and $AN \perp CD$,we have $\angle AMC = 90^{\circ}$ and $\angle ANC = 90^{\circ}$.In quadrilateral $AMCN$,$\angle MCN + \angle AMC + \angle ANC + \angle MAN = 360^{\circ}$.$\angle C + 90^{\circ} + 90^{\circ} + 60^{\circ} = 360^{\circ}$.$\angle C + 240^{\circ} = 360^{\circ}$,so $\angle C = 120^{\circ}$.In a parallelogram,opposite angles are equal,so $\angle A = \angle C = 120^{\circ}$.Adjacent angles are supplementary,so $\angle B = 180^{\circ} - 120^{\circ} = 60^{\circ}$ and $\angle D = 180^{\circ} - 120^{\circ} = 60^{\circ}$.Thus,the angles are $120^{\circ}, 60^{\circ}, 120^{\circ}, 60^{\circ}$.

In parallelogram $ABCD$,$\angle A$ is an obtuse angle,$AM \perp BC$,and $AN \perp CD$. If $\angle MAN = 60^{\circ}$,find each of the angles of quadrilateral $ABCD$.

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