Lines and Angles — MATHS STD 7 — Question

 
Construction: Draw a line $PQ$ parallel to $AB$ which is also parallel to $CD$
 Since, $PQ || AB$
$\therefore \angle \text{AME}+\angle \text{QEM}=180^\circ$ [Angles on the same side of a transversal line are supplementary]
$\Rightarrow 139^\circ+\angle \text{QEM}=180^\circ$
$\Rightarrow \angle \text{QEM}=41^\circ$
Now, $\angle \text{QEM}+\angle \text{DEQ}=\angle \text{MED}$
$\Rightarrow 41^\circ+\angle \text{DEQ}=67^\circ$
$\Rightarrow \angle \text{DEQ}=26^\circ$
Now, $\angle \text{PED}+\angle \text{DEQ}=180^\circ$ [Linear Pair angles]
$\Rightarrow \angle \text{PED}+26^\circ=180^\circ$
$\Rightarrow \angle \text{PED}=154^\circ$
Since, $ PQ || AB$
$\therefore \text{x}^\circ=\angle \text{PED}$ [Corresponding angles]
$\Rightarrow \text{x}^\circ-154^\circ$
$\Rightarrow \text{x}=154$
Hence, the correct answer is option $(a).$