Question

In the given figure $NQ\parallel PO$. Find the value of $a$.

• A. ${26}^{\circ }$
• B. ${14}^{\circ }$
• C. ${10}^{\circ }$
• D. ${13}^{\circ }$

Answer 1

The correct option is B ${14}^{\circ }$

As given in the question,
$QN=PN$
So,

$\angle PQN=\angle NPQ={78}^0$

In $△PQN,$
$\angle PQN+\angle NPQ+\angle QNP={180}^0$
${78}^0+{78}^0+\angle QNP={180}^0$
$\angle QNP={24}^0$

$\angle PQN+\angle MQN={180}^o$ (angles on a straight line add to ${180}^o$)

$\angle MQN={102}^0$

As given $NQ\parallel PO$,

So, $\angle MQN=\angle MPO$
Using angle sum property,
$\angle NMP+\angle MPO+a={180}^0$
${64}^0+{102}^0+a={180}^0$
$a={180}^0-{166}^0={14}^0$
Hence, option $B$ is correct.