Question

In the given figure, $PQ\parallel RS$ and $EF\parallel QS.$ If $\angle PQS={60}^0,$ then what is the measure of $\angle RFE$ ?

Answer 1

$PQ\parallel RS$

$\Rightarrow$

$\angle PQS+\angle RSQ={180}^o$ ( Interior angles on same side are supplem)

$\Rightarrow$ $\angle RSQ={180}^o-\angle PQS$

$={180}^o-{60}^o$

$={120}^o$

Now $EF\parallel QS$

$\Rightarrow \angle EFR=\angle QSR$

$\Rightarrow \angle EFR$ or $\angle RFE={120}^o$ (corresponding angles)