Question

In the adjoining figure , $AB\parallel CD\parallel EF,PQ\parallel RS,\angle RQD={25}^0$ and $\angle CQP={60}^0$, find the $\angle QRS$.

Answer 1

Given, $PQ\parallel RS$

$\angle PQC=\angle BRS={60}^{\circ }$ [alternate exterior angles and $\angle PQC={60}^{\circ }$ (given)]

and $\angle DQR=\angle QRA={25}^{\circ }$ [alternate interior angles]

[$\angle DQR={25}^{\circ }$, given]

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

$=\angle QRA+({180}^{\circ }–\angle BRS)$ [linear pair axiom]

$={25}^{\circ }+{180}^{\circ }–{60}^{\circ }$

$={205}^{\circ }–{60}^{\circ }={145}^{\circ }$