Question

In the adjacent figure PQ || ST, $\angle PQR={110}^o$ and $\angle RST={130}^o$, find $\angle QRS$.

• A. ${30}^o$
• B. ${130}^o$
• C. ${60}^o$
• D. ${160}^o$

Answer 1

The correct option is C ${60}^o$

Draw a line $XY$ parallel to $PQ\parallel ST$.

It is known that the sum of interior angles on the same side of the transversal is ${180}^{\circ }$.

So, $\angle PQR+\angle QRX={180}^{\circ }$

${110}^{\circ }+\angle QRX={180}^{\circ }$

$\angle QRX={70}^{\circ }$

Similarly,

$\angle RST+\angle SRY={180}^{\circ }$

${130}^{\circ }+\angle SRY={180}^{\circ }$

$\angle SRY={50}^{\circ }$

Now, by property of linear pair,

$\angle QRX+\angle QRS+\angle SRY={180}^{\circ }$

${70}^{\circ }+\angle QRS+{50}^{\circ }={180}^{\circ }$

$\angle QRS={60}^{\circ }$