Mathematics

In the adjoining figure, AB$||$CD and PQ, QR intersects AB and CD both at E, F and G, H respectively. Given that $\angle$PEB$=80^o$, $\angle$QHD$=120^o$ and $\angle$PQR$=x^o$, find the value of x.

$20$

SOLUTION
$\angle PEB=\angle PGD \,\,(\because AB||CD)$
$\therefore \angle PGD=80^o$
$\angle QGD=180-\angle PGD=180-80=100^o$
$\angle QHC=180-\angle QHD=180-120=60^o$
In $\triangle QGH:$
$x+\angle QGD+\angle QHC=180^o\\ \Rightarrow x=180-100-60=20^o$

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Single Correct Medium Published on 09th 09, 2020
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