Find the value of x and y in the following figure.

We can solve this using angle sum property. 

The angle sum property of a triangle states that the sum of the interior angles of any triangle is always ${\mathbf{180}}^{\mathbf{\circ }}$. 

Given: $\angle \mathrm{P}=40°$, $\angle \mathrm{S}=90°$, and $\angle \mathrm{R}=\mathrm{y}$

In $△$PSR

$\angle \mathrm{P}+\angle \mathrm{S}+\angle \mathrm{R}={180}^{\circ }$  [$\because$ Property sum of angle of triangle]

By putting the given value we get, 

$\angle \mathrm{P}+\angle \mathrm{S}+\angle \mathrm{R}={180}^{\circ }$

$\Rightarrow {40}^{\circ }+{90}^{\circ }+\mathrm{y}={180}^{\circ }$

$\Rightarrow {130}^{\circ }+\mathrm{y}={180}^{\circ }$

$\Rightarrow \mathrm{y}={180}^{\circ }-{130}^{\circ }$

$\Rightarrow \mathrm{y}={50}^{\circ }$

Similarly, In $△$PSQ 

$\angle \mathrm{P}+\angle \mathrm{S}+\angle \mathrm{Q}={180}^{\circ }$

By putting the values, ${\angle \mathrm{Q}=60}^{\circ }$, $\angle \mathrm{P}=\mathrm{x}$ and $\angle \mathrm{S}={90}^{\circ }$

$\therefore \angle \mathrm{P}+\angle \mathrm{S}+\angle \mathrm{Q}={180}^{\circ }$

$\Rightarrow {60}^{\circ }+{90}^{\circ }+\mathrm{x}$$={180}^{\circ }$

$\Rightarrow {150}^{\circ }+\mathrm{x}={180}^{\circ }$

$\Rightarrow \mathrm{x}={180}^{\circ }-{150}^{\circ }$

$\Rightarrow \mathrm{x }=\mathrm{ }{30}^{\circ }\mathrm{ }$

Hence, the value of x =${\mathrm{ }30}^{\circ }$ and y = ${50}^{\circ }$.