In a parallelogram ABCD,angle A = 5 angle B,t | vedclass

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(A) In a parallelogram $ABCD$,the sum of adjacent angles is $180^o$. Therefore,$\angle A + \angle B = 180^o$.Given that $\angle A = 5 \angle B$,we sub

Vedclass · Accepted Answer

$150$

Vedclass · Answer

(A) In a parallelogram $ABCD$,the sum of adjacent angles is $180^o$. Therefore,$\angle A + \angle B = 180^o$.Given that $\angle A = 5 \angle B$,we substitute this into the equation:$5 \angle B + \angle B = 180^o$$6 \angle B = 180^o$$\angle B = 30^o$.Since $\angle A = 5 \angle B$,then $\angle A = 5 	imes 30^o = 150^o$.In a parallelogram,opposite angles are equal,so $\angle C = \angle A$.Therefore,$\angle C = 150^o$.

In a parallelogram $ABCD$,$\angle A = 5 \angle B$,then $\angle C = \ldots$ (in $^o$)

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