Question

In figure AB || CD. Find the value of x.

Answer 1

Draw EF parallel to both AB and CD.
Now, AB || EF and transversal AE cuts them at A and E respectively.
$\angle BAE+\angle FEA={180}^{\circ }$
$\Rightarrow {108}^{\circ }+\angle 1={180}^{\circ }\Rightarrow \angle 1={180}^{\circ }-{108}^{\circ }={72}^{\circ }$
Again, EF || CD and transversal CE cuts them at E and C respectively.
$\therefore \angle FEC+\angle ECD={180}^{\circ }$


$\Rightarrow \angle 2+{112}^{\circ }={180}^{\circ }$
$\Rightarrow \angle 2={180}^{\circ }-{112}^{\circ }$
$\Rightarrow \angle 2={68}^{\circ }$
Now, $x=\angle 1+\angle 2$
$\Rightarrow x={72}^{\circ }+{68}^{\circ }={140}^{\circ }$