In circle $PQRST$ with center $O$, $POR$ and $QOT$ are straight lines. $QT$ is parallel to $RS$. Angle $RTQ = 33^{\circ}$ and $|RS| = |ST|$. We need to find angle $RST$.

1. Problem Description

In circle $PQRST$ with center $O$, $POR$ and $QOT$ are straight lines. $QT$ is parallel to $RS$. Angle $RTQ = 33^{\circ}$ and $|RS| = |ST|$. We need to find angle $RST$.

2. Solution Steps

Since $QT \parallel RS$, $\angle TRS = \angle RTQ = 33^{\circ}$ (alternate angles).

Since $|RS| = |ST|$, triangle $RST$ is isosceles, so $\angle SRT = \angle STR$.

Let $\angle SRT = \angle STR = x$.

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

$\angle TRS + \angle SRT + \angle STR = 180^{\circ}$

$33^{\circ} + x + x = 180^{\circ}$

$2x = 180^{\circ} - 33^{\circ}$

$2x = 147^{\circ}$

$x = \frac{147^{\circ}}{2} = 73.5^{\circ}$

Therefore, $\angle SRT = \angle STR = 73.5^{\circ}$.

Since $QT \parallel RS$, $\angle QTR = \angle TRS = 33^{\circ}$ (alternate angles).

Since $QOT$ and $POR$ are straight lines, $QT$ and $PR$ are diameters.

Also $\angle PRQ = \angle PTQ = 90^{\circ}$ (angles in a semicircle).

$\angle QRS = \angle QTS$.

Since $QT \parallel RS$, we have $\angle TQR = \angle SRT$.

Angles in the same segment subtended by $QR$ are equal.

$\angle QTS = \angle QRS$

Also $\angle RST = x = 73.5^{\circ}$.

We need to find the value of $\angle RST$.

Since $RS = ST$, $\angle SRT = \angle STR$.

$\angle RTQ = 33^{\circ}$ and since $QT \parallel RS$, $\angle TRS = 33^{\circ}$ (alternate angles).

In triangle $RST$, $\angle TRS + \angle SRT + \angle STR = 180^{\circ}$.

$33^{\circ} + \angle SRT + \angle SRT = 180^{\circ}$.

$2 \angle SRT = 180^{\circ} - 33^{\circ} = 147^{\circ}$.

$\angle SRT = \frac{147}{2} = 73.5^{\circ}$.

Then $\angle RST = \angle SRT + \angle TRS = 73.5^{\circ}$.

In triangle $RST$, we found that $\angle SRT = \angle STR = \frac{147^{\circ}}{2} = 73.5^{\circ}$.

Therefore, $\angle RST = \angle SRT = 73.5^{\circ}$.

3. Final Answer

73.5 degrees