Question

In Figure, if $ AB\parallel CD\parallel EF$, $ PQ\parallel RS$, $ \angle RQD=25°$ and $ \angle CQP=60°$, then $ \angle QRS$ is equal to

• A. $85°$
• B. $135°$
• C. $145°$
• D. $110°$

Answer 1

The correct option is C

$145°$

Step 1: According to the diagram

$AB\parallel CD\parallel EF$

$PQ\parallel RS$

$\angle RQD=25°$

$\angle CQP=60°$

$PQ\parallel RS.$

Step 2: Applying property of parallel lines.

From the property of parallel lines, we know that if a transversal intersects two parallel lines, then each pair of alternate exterior angles is equal.

Since $PQ\parallel RS$

$\Rightarrow \angle PQC=\angle BRS$

and $\angle PQC=60°$

$\angle BRS=60°$ ——-(i)

We also know that,

From the property of parallel lines, we know that if a transversal intersects two parallel lines, then each pair of alternate interior angles is equal.

Now again, since, $AB\parallel CD$

$\angle DQR=\angle QRA$

and $\angle DQR=25°$

$\angle QRA=25°$ ——-(ii)

Step 3: Applying linear pair axiom

From the linear pair axiom, we note that

$\angle ARS=\angle BRS=180°$

$\angle ARS=180°-\angle BRS$

$\Rightarrow \angle ARS=180°-60°$ (From (i), $\angle BRS=60°$)

$\Rightarrow \angle ARS=120°$ ———(iii)

Now, $\angle QRS=\angle QRA+\angle ARS$

From equations (ii) and (iii), we get,

$\angle QRA=25°and\angle ARS=120°$

Hence, the above equation can be expressed as:

$\angle QRS=25°+120°$

$\Rightarrow \angle QRS=145°$

Hence, the value of $\angle QRS$ is $145°$ and therefore the correct answer is option (C).