In the given figure,if PQ || RS,RS || MN and | vedclass

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(D) Given that $PQ || RS$ and $RS || MN$. Therefore,$PQ || MN$.From the figure,the angle $x$ and $z$ are alternate interior angles if we consider the

Vedclass · Accepted Answer

$110$

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(D) Given that $PQ || RS$ and $RS || MN$. Therefore,$PQ || MN$.From the figure,the angle $x$ and $z$ are alternate interior angles if we consider the transversal line,but looking at the positions,$x$ and $z$ are interior angles on the same side of the transversal for lines $PQ$ and $MN$,so $x + z = 180^o$.Also,$y$ and $z$ are consecutive interior angles between lines $RS$ and $MN$,so $y + z = 180^o$.This implies $y = x$ (as both are supplementary to $z$).Given $y: z = 7: 11$,let $y = 7k$ and $z = 11k$.Since $y + z = 180^o$,we have $7k + 11k = 180^o$,which gives $18k = 180^o$,so $k = 10^o$.Thus,$y = 7 	imes 10^o = 70^o$ and $z = 11 	imes 10^o = 110^o$.Since $x + z = 180^o$ (consecutive interior angles for $PQ || MN$),we have $x + 110^o = 180^o$.Therefore,$x = 180^o - 110^o = 70^o$.Wait,checking the figure again: $x$ and $y$ are consecutive interior angles between $PQ$ and $RS$,so $x + y = 180^o$. Since $y = 70^o$,$x = 180^o - 70^o = 110^o$.

In the given figure,if $PQ || RS$,$RS || MN$ and $y: z = 7: 11$,then find $x$. (in $^o$)

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