Question

In the given figure, $PQR$ is a straight line and $QRS$ is an isosceles triangle. Find the value of $y$.

Answer 1

In $△QRS$,

$\angle QSR=\angle QRS=55°\left(\text{Angle opposite to the equal sides}\right)$

Now, in $△PRS$

$\angle PSR=(55+y)°$

Also,

$\angle PRS+\angle PSR+\angle RPS=180°\left(\text{Sum of interior angles}\right)$

$55°+(55+y)°+40°=180°$

$\Rightarrow y=180°-150°=30°$