In the given figure, if $P Q \| S T, \angle P Q R=110^{\circ}$ and $\angle R S T=130^{\circ}$, find $\angle Q R S$.


Let us draw a line $X Y$ parallel to $S T$ and passing through point $R$.

$\angle \mathrm{PQR}+\angle \mathrm{QRX}=180^{\circ}$ (Co-interior angles on the same side of transversal $\mathrm{QR}$ )

$\Rightarrow 110^{\circ}+\angle Q R X=180^{\circ}$

$\Rightarrow \angle Q R X=70^{\circ}$


$\angle \mathrm{RST}+\angle S R Y=180^{\circ}$ (Co-interior angles on the same side of transversal SR)

$130^{\circ}+\angle S R Y=180^{\circ}$

$\angle S R Y=50^{\circ}$

$X Y$ is a straight line. RQ and RS stand on it.

$\therefore \angle Q R X+\angle Q R S+\angle S R Y=180^{\circ}$

$70^{\circ}+\angle Q R S+50^{\circ}=180^{\circ}$

$\angle Q R S=180^{\circ}-120^{\circ}=60^{\circ}$

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