Question

Question 1
In figure, if $AB||CD||EF,PQ||RS,\angle RQD={25}^{\circ }and\angle CQP={60}^{\circ },then\angle QRS$ is equal to:


A) ${85}^{\circ }$
B) ${135}^{\circ }$
C) ${145}^{\circ }$
D) ${110}^{\circ }$

Answer 1

Given, PQ || RS
$\therefore \angle PQC=\angle BRS={60}^{\circ }$ [Alternate exterior angles and $\angle PQC={60}^{\circ }$ (given)]
$\angle DQR=\angle QRA={25}^{\circ }$ [Alternate interior angles]...(i)
$\angle ARS+\angle BRS={180}^{\circ }$ [linear pair axiom]
$\angle ARS={180}^{\circ }-\angle BRS={180}^{\circ }-{60}^{\circ }={120}^{\circ }...\left(ii\right)$
$\therefore \angle QRS=\angle QRA+\angle ARS={25}^{\circ }+{120}^{\circ }\left(Fromiandii\right)={145}^{\circ }$