In the given problem, we have a quadrilateral $PQSR$ with the following information:

• $PQ \perp PS$ (Right angle at $P$)
• $PQ \parallel SR$
• $\angle SQR = 28^\circ$
• $\angle QRT = 65^\circ$

We are tasked with finding the values of $x$ and $y$, where:

• $x$ is the angle at $\angle PQR$
• $y$ is the angle at $\angle PSR$

Step-by-step solution:

1. Using the parallel lines property: Since $PQ \parallel SR$, $\angle PQR$ and $\angle QRT$ are alternate interior angles. Hence, $x = \angle PQR = \angle QRT = 65^\circ$

2. Using the triangle angle sum property: In $\triangle PSR$, the sum of the interior angles is $180^\circ$. We already know:

   • $\angle PSR = 90^\circ$ (since $PQ \perp PS$)
   • $\angle SQR = 28^\circ$

   Using the sum of angles in $\triangle PSR$: $y + 28^\circ + 90^\circ = 180^\circ$ Simplifying: $y = 180^\circ - 118^\circ = 62^\circ$

Final values:

• $x = 65^\circ$
• $y = 62^\circ$

Let me know if you need more details or have further questions.

Here are 5 related questions:

1. What is the sum of interior angles in any triangle?
2. What