In the given diagram, line segment $QR$ is parallel to line segment $ST$. Also, $|PQ| = |PR|$, which means that $\triangle PQR$ is an isosceles triangle. The angle $\angle PST = 75^\circ$. We need to find the value of the angle $y = \angle PRQ$.

1. Problem Description

In the given diagram, line segment $QR$ is parallel to line segment $ST$. Also, $|PQ| = |PR|$, which means that $\triangle PQR$ is an isosceles triangle. The angle $\angle PST = 75^\circ$. We need to find the value of the angle $y = \angle PRQ$.

2. Solution Steps

Since $QR \parallel ST$, we know that the corresponding angles are equal.

Therefore, $\angle PQR = \angle PST = 75^\circ$.

Since $\triangle PQR$ is an isosceles triangle with $|PQ| = |PR|$, we know that the angles opposite these sides are equal, i.e., $\angle PRQ = \angle PQR$.

Thus, $y = \angle PRQ = \angle PQR = 75^\circ$.

In $\triangle PQR$, the sum of the angles is $180^\circ$.

So, $\angle PQR + \angle PRQ + \angle QPR = 180^\circ$.

$75^\circ + 75^\circ + \angle QPR = 180^\circ$

$150^\circ + \angle QPR = 180^\circ$

$\angle QPR = 180^\circ - 150^\circ = 30^\circ$.

The angle $y$ is an exterior angle to the triangle $\triangle PRT$. Since $QR \parallel ST$, $\angle SRT$ is a straight line, so $\angle SRT = 180^\circ$.

Thus $\angle PRT = 180^\circ$. We know that $\angle PRQ = y$, so

$y + \angle QRT = 180^\circ$.

Since $QR \parallel ST$, $\angle PST$ and $\angle SQR$ are corresponding angles, which means that

$\angle SQR = \angle PST = 75^\circ$.

Also, $\angle SQR + \angle PQR = 180^\circ$, where $\angle PQS$ is a straight line and we have an error here.

Since $QR$ is parallel to $ST$, the alternate interior angles are equal.

$\angle S = \angle SQR = 75^\circ$. Since $SQR$ is exterior angle to $\triangle PQR$. We also know that $\angle SQR = \angle QPR + \angle PRQ$. Which means that

$75^\circ = \angle QPR + y$.

Since $\triangle PQR$ is isosceles with $|PQ|=|PR|$, then $\angle PQR = \angle PRQ = y$.

The sum of the angles in the triangle is $180^\circ$, so

$\angle PQR + \angle PRQ + \angle QPR = 180^\circ$.

$y + y + \angle QPR = 180^\circ$.

$2y + \angle QPR = 180^\circ$.

$\angle QPR = 180^\circ - 2y$.

We know that $\angle SQR = 75^\circ$ and $\angle SQR = \angle QPR + \angle PRQ$,

so $75^\circ = (180^\circ - 2y) + y$.

$75^\circ = 180^\circ - y$

$y = 180^\circ - 75^\circ$

$y = 105^\circ$.

3. Final Answer

105 degrees