Question

In figure, if $AB\parallel CD\parallel EF,PQ\parallel RS,\angle RQD={25}^o$ and $\angle CQP={60}^o$, then $\angle QRS$ is equal to

• A. ${85}^o$
• B. ${135}^o$
• C. ${145}^o$
• D. ${110}^o$

Answer 1

The correct option is C ${145}^o$

Given, RQD=${25}^o$ and CQP=${60}^o$
$RQD=QRA={25}^0$ (Alternate interior angles)
$CQP=BRS={60}^0$(Alternate Exterior angles)
$BRS+ARS={180}^0$ (Linear Pair)
$=>ARS={180}^0-BRS={180}^0-{60}^0={120}^0$
$QRS=ARS+QRA={120}^0+{25}^0={145}^0$