Question

In the given figure, $PQR$ is an equilateral triangle and $QRST$ is a square. Then $\angle PSR=$

• A. ${60}^{\circ }$
• B. ${45}^{\circ }$
• C. ${30}^{\circ }$
• D. ${15}^{\circ }$

Answer 1

The correct option is D ${15}^{\circ }$

$QRST$ is a square
Let side $=a$.
Triangle $PQR$ is equilateral
$PQ=PR=QR=a$
$\angle PRQ={60}^{\circ }$ ($\Delta PQR$ is equilateral)
In $\Delta PRS$
$PR=RS$
$\angle PSR=\angle SPR=x$
(angle opp. to equal sides are also equal) $=x$
$\angle PRS={60}^{\circ }+{90}^{\circ }={150}^{\circ }$
$x+x+{150}^{\circ }={180}^{\circ }$ (sum of angle of $\Delta PRS$)
$2x={30}^{\circ }$
$x={15}^{\circ }$