Question

In the figure, if $ PQ\parallel ST,\angle PQR = {110^0} $ and $ \angle RST = {130^0}, $ find $ \angle QRS. $

Answer 1

Hint: Draw a flat parallel line passing through point $ R. $ Other construction can also be done to obtain the answers as per the requirements of the solution, and then use the properties of lines and angles.

Complete step-by-step answer:

Observe the diagram
We have done a construction for the simplicity of the question.
Let us assume, XY is a line, passing through R in such a way that it is parallel to PQ and ST. i.e.
 $ PQ\parallel ST\parallel XY $
Now, since, parts of parallel lines are parallel, we can say that,
 $ PQ\parallel XR $
And, using the transverse line rule for the transverse QR, we can say that
$\Rightarrow \angle PQR + \angle QRX = {180^0} $ (since, the sum of the transverse angles between two parallel lines is equal to $ {180^0} $ )
It is given to us that,
 $ \angle PQR = {110^0} $
By substituting this value in the above equation, we get
$\Rightarrow {110^0} + \angle QRX = {180^0} $
By rearranging it, we get
$\Rightarrow \angle QRX = {180^0} - {110^0} $
 $ \therefore \angle QRX = 70^\circ $
Again, by using the property that, the parts of parallel lines are parallel, we can say that,
$ ST\parallel RY $
And, using the transverse line rule for the transverse RS, we can say that
$\Rightarrow \angle RST + \angle SRY = {180^0} $ (since, the sum of the transverse angles between two parallel lines is equal to $ {180^0} $)
It is given to us that,
$ \angle RST = {130^0} $
By substituting this value in the above equation, we get
$\Rightarrow {130^0} + \angle SRY = {180^0} $
By rearranging it, we get
$ \angle SRY = {180^0} - {130^0} $
$ \therefore \angle SRY = 50 $
Now, we know that, a straight line always makes and angle of $ {180^0} $
$ \therefore \angle XRY = {180^0} $
But, $ \angle XRY = \angle QRS + \angle QRX + \angle SRY $
Therefore, we can write
$ \angle QRS + \angle QRX + \angle SRY = {180^0} $
By substituting the values of $ \angle QRX $ and $ \angle SRY $ , we get
$ \angle QRS + {70^0} + {50^0} = {180^0} $
By re-arranging it, we get
$ \angle QRS = {180^0} - {70^0} - {50^0} $
$ \therefore \angle QRS = 60^\circ $

Note: Sometimes, you need to understand that a construction is necessary to simplify the question. Like in this question, without the construction that we did, there would have been to ground to find a relationship between the given information and $ \angle QRS $ . Drawing one line parallel to the given lines made the question very easy for us.