In the figure,if AB parallel CD,angle BMX = 1 | vedclass

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(A) Draw a line $PQ \parallel AB$,passing through point $X$ as shown in the figure.Now,$PQ \parallel AB$ and $AB \parallel CD$.Therefore,$CD \parallel

Vedclass · Accepted Answer

$110$

Vedclass · Answer

(A) Draw a line $PQ \parallel AB$,passing through point $X$ as shown in the figure.Now,$PQ \parallel AB$ and $AB \parallel CD$.Therefore,$CD \parallel PQ$ (since $AB \parallel CD$ and $AB \parallel PQ$).Consider the transversal $MX$ intersecting parallel lines $AB$ and $PQ$. The interior angles on the same side of the transversal are supplementary.$	herefore \angle BMX + \angle MXQ = 180^{\circ}$Given $\angle BMX = 125^{\circ}$,we have:$125^{\circ} + \angle MXQ = 180^{\circ}$$	herefore \angle MXQ = 180^{\circ} - 125^{\circ} = 55^{\circ}$ $(1)$Now,consider the transversal $NX$ intersecting parallel lines $CD$ and $PQ$. The alternate interior angles are equal.$	herefore \angle NXQ = \angle XNC = 55^{\circ}$ $(2)$Finally,$\angle MXN = \angle MXQ + \angle NXQ$.Substituting the values from $(1)$ and $(2)$:$\angle MXN = 55^{\circ} + 55^{\circ} = 110^{\circ}$.

In the figure,if $AB \parallel CD$,$\angle BMX = 125^{\circ}$ and $\angle CNX = 55^{\circ}$,find $\angle MXN$. (in $^{\circ}$)

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