Question

In the adjoining figure $AB\parallel CD$ and PQ,QR intersects AB and CD both at EF, and G,H respectively. Given that $PER={80}^{\circ },QHD={120}^{\circ }$ and $PQR=x^{\circ }$, then the value of 'x' is

• A. ${40}^{\circ }$
• B. ${20}^{\circ }$
• C. ${100}^{\circ }$
• D. ${30}^{\circ }$

Answer 1

The correct option is B ${20}^{\circ }$

$\because AB||CD$

$\therefore \angle GAF=\angle QGH$ ( by corresponding angle)

$\angle QGH=180-{80}^{\circ }={100}^{\circ }$

In $\Delta QGH$

By exterior angle.

$x+{100}^{\circ }={120}^{\circ }$

$x={20}^{\circ }$