Question

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 $x$.

• A. $40$
• B. $20$
• C. $100$
• D. $30$

Answer 1

The correct option is B $20$

$\angle PEB=\angle PGD(\because AB||CD)$

$\therefore \angle PGD={80}^o$.

$\angle QGD=180-\angle PGD=180-80={100}^o$ (linear pairs).

$\angle QHC=180-\angle QHD=180-120={60}^o$ (linear pairs).

In $△QGH$, by angle sum property,

$x+\angle QGD+\angle QHC={180}^o\Rightarrow x=180-100-60={20}^o$.

Therefore, option $B$ is correct.