Question

In figure, if $PQ\parallel RS,\angle MXQ = {135^0}$ and $\angle MYR = {40^0}$, find $\angle XMY$.

Answer 1

Hint: In this question, we will use some basic points of lines and angles. We will use the theorem which states that the lines which are parallel to the same line are parallel to each other.

Complete step by step answer:
Given that,
$PQ\parallel RS,\angle MXQ = {135^0}$ and $\angle MYR = {40^0}$.
We have to find $\angle XMY$.
First, we need to draw a line AB which is parallel to the line PQ, through point M, i.e. $AB\parallel PQ$. As shown in the figure.

Now, we know that the lines which are parallel to the same line are parallel to each other.
So,
$AB\parallel PQ$ and $PQ\parallel RS$
Thus,
$AB\parallel RS$.
Now,
We know that the sum of interior angles on the same side of the transversal are supplementary.
Since,
$AB\parallel PQ$ and MX is the traversal.
Therefore,
$
   \Rightarrow \angle XMB + \angle QXM = {180^0} \\
   \Rightarrow \angle XMB + {135^0} = {180^0} \\
   \Rightarrow \angle XMB = {180^0} - {135^0} \\
   \Rightarrow \angle XMB = {45^0} \\
$
Similarly, we have
$AB\parallel RS$ and MY is the traversal.
$\angle BMY = \angle MYR$ [ alternate interior angles ]
So,
$\angle BMY = {40^0}$.
From the figure, we can see that,
$\angle XMB + \angle BMY = \angle XMY$. ………. (i)
We have,
$\angle BMY = {40^0}$ and $\angle XMB = {45^0}$.
Putting these values in equation (i), we will get
$
   \Rightarrow \angle XMY = {40^0} + {45^0} \\
   \Rightarrow \angle XMY = {85^0} \\
$

Hence, we have found $\angle XMY = {85^0}$.

Note: Whenever we ask this type of question, we should always remember some basic theorems and properties of lines and angles. A line that passes through two lines at two distinct points in the same plane is called a transversal line and if two lines crossed are parallel, then the alternate angles are equal. Using these basic concepts, we can easily find out the answer of our question.