Mathematics

In Figure, if $PQ \parallel ST, \angle PQR=110^o$ and $\angle RST=130^o$, find $\angle QRS$.

$60^o$

SOLUTION
Given :$PQ || ST$, $\angle PQR = 110^o$ and $\angle RST = 130^o$

Construction: A line $XY$ parallel to $PQ$ and $ST$ is drawn.

$\angle PQR + \angle QRX = 180^o$ (Angles on the same side of transversal.)

$110^o + \angle QRX = 180^o$

$\angle QRX = 180^o - 110^o$

$\angle QRX = 70^o$

Also,$\angle RST + \angle SRY = 180^o$ (Angles on the same side of transversal.)

$130^o + \angle SRY = 180^o$

$\angle SRY = 50^o$

Now,$\angle QRX +\angle SRY + \angle QRS = 180^o$

$70^o + 50^o + \angle QRS = 180^o$

$\angle QRS = 60^o$

Hence, $\angle QRS = 60^o$

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Single Correct Medium Published on 09th 09, 2020
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