In the figure,if PQ parallel RS,angle MXQ = 1 | vedclass

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(B) To solve this,we draw a line $AB$ passing through point $M$ such that $AB \parallel PQ$. Since $PQ \parallel RS$ and $AB \parallel PQ$,it follows

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$85$

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(B) To solve this,we draw a line $AB$ passing through point $M$ such that $AB \parallel PQ$. Since $PQ \parallel RS$ and $AB \parallel PQ$,it follows that $AB \parallel RS$.Now,consider the transversal $XM$ intersecting parallel lines $PQ$ and $AB$. The interior angles on the same side of the transversal are supplementary.Therefore,$\angle QXM + \angle XMB = 180^o$.Given $\angle QXM = 135^o$,we have $135^o + \angle XMB = 180^o$,which gives $\angle XMB = 180^o - 135^o = 45^o$ (Equation $1$).Next,consider the transversal $YM$ intersecting parallel lines $RS$ and $AB$. The alternate interior angles are equal.Therefore,$\angle BMY = \angle MYR$.Given $\angle MYR = 40^o$,we have $\angle BMY = 40^o$ (Equation $2$).Finally,$\angle XMY = \angle XMB + \angle BMY$.Substituting the values from Equation $1$ and Equation $2$,we get $\angle XMY = 45^o + 40^o = 85^o$.

In the figure,if $PQ \parallel RS$,$\angle MXQ = 135^o$ and $\angle MYR = 40^o$,find $\angle XMY$. (in $^o$)

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