In the figure,if AB parallel CD,CD parallel E | vedclass

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(B) Given: $AB \parallel CD$ and $CD \parallel EF$.Since lines parallel to the same line are parallel to each other,we have $AB \parallel EF$.Let the

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$126$

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(B) Given: $AB \parallel CD$ and $CD \parallel EF$.Since lines parallel to the same line are parallel to each other,we have $AB \parallel EF$.Let the transversal intersect the lines $AB$,$CD$,and $EF$ at points $P$,$Q$,and $R$ respectively.Since $AB \parallel EF$,the interior alternate angles are equal. Therefore,$\angle x = \angle z$  .......... $(1)$Again,$AB \parallel CD$,so the sum of interior angles on the same side of the transversal is $180^o$.Thus,$y + z = 180^o$.Given $y: z = 3: 7$,let $y = 3k$ and $z = 7k$.Then,$3k + 7k = 180^o \Rightarrow 10k = 180^o \Rightarrow k = 18^o$.So,$z = 7 	imes 18^o = 126^o$.From $(1)$,$x = z = 126^o$.

In the figure,if $AB \parallel CD$,$CD \parallel EF$ and $y: z = 3: 7$,find $x$. (in $^o$)

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