Question

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

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
$\Rightarrow \mathrm{\angle }PQR+\mathrm{\angle }QRX={180}^0$ (since, the sum of the transverse angles between two parallel lines is equal to ${180}^0$ )
It is given to us that,
 $\mathrm{\angle }PQR={110}^0$
By substituting this value in the above equation, we get
$\Rightarrow {110}^0+\mathrm{\angle }QRX={180}^0$
By rearranging it, we get
$\Rightarrow \mathrm{\angle }QRX={180}^0-{110}^0$
 $\therefore \mathrm{\angle }QRX={70}^{\circ }$
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
$\Rightarrow \mathrm{\angle }RST+\mathrm{\angle }SRY={180}^0$ (since, the sum of the transverse angles between two parallel lines is equal to ${180}^0$)
It is given to us that,
$\mathrm{\angle }RST={130}^0$
By substituting this value in the above equation, we get
$\Rightarrow {130}^0+\mathrm{\angle }SRY={180}^0$
By rearranging it, we get
$\mathrm{\angle }SRY={180}^0-{130}^0$
$\therefore \mathrm{\angle }SRY=50$
Now, we know that, a straight line always makes and angle of ${180}^0$
$\therefore \mathrm{\angle }XRY={180}^0$
But, $\mathrm{\angle }XRY=\mathrm{\angle }QRS+\mathrm{\angle }QRX+\mathrm{\angle }SRY$
Therefore, we can write
$\mathrm{\angle }QRS+\mathrm{\angle }QRX+\mathrm{\angle }SRY={180}^0$
By substituting the values of $\mathrm{\angle }QRX$ and $\mathrm{\angle }SRY$ , we get
$\mathrm{\angle }QRS+{70}^0+{50}^0={180}^0$
By re-arranging it, we get
$\mathrm{\angle }QRS={180}^0-{70}^0-{50}^0$
$\therefore \mathrm{\angle }QRS={60}^{\circ }$

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 $\mathrm{\angle }QRS$ . Drawing one line parallel to the given lines made the question very easy for us.