Question

In the given figure, if AB || CF and CD || FE, then the value of x is

• A. ${40}^{\circ }$
• B. ${75}^{\circ }$
• C. ${65}^{\circ }$
• D. ${105}^{\circ }$

Answer 1

The correct option is D ${105}^{\circ }$

Extend the line $EF$ to $H$ so that $HG||AC$

Now $HG||AC$ and $BC$ is transversal

$\angle BGH=\angle BCA$ (Corresponding angles)

$\angle BGH={65}^{\circ }$

Now exterior angle of triangle is equal to sum of interior opposite angles

Applying the same in $△BGH$

$\angle HBG+\angle BGH=\angle GHA$

${40}^{\circ }+{65}^{\circ }=yy={105}^{\circ }$

Now AB||FCF$and$HE$ is transversal

$\therefore x=y$ (corresponding angles)

$x={105}^{\circ }$