Question

In the given figure $PQ\parallel RS$, $\angle RSF={40}^o$, $\angle PQF={35}^o$ and $\angle QFP=x^o$. What is the value of $x$?

• A. ${75}^o$
• B. ${105}^o$
• C. ${135}^o$
• D. ${140}^o$

Answer 1

The correct option is B ${105}^o$

$\angle QPF=\angle RSP={40}^o$ ($RS\parallel PQ$, alt. $\angle s$)
In $△FQP$, $x+{35}^o+{40}^o={180}^o$
$\Rightarrow$ $x={180}^o-{75}^o={105}^o$