In a circle with center O, angle QOR = 146 degrees and angle QRT = 52 degrees, where RT is a tangent. Find angles QPR, QTR, and OQR.

The diagram shows a circle with center $O$.
$\angle QOR = 146^\circ$ (angle at the center).
$\angle QRT = 52^\circ$ (angle between tangent $RT$ and chord $QR$).
The angles to be found are $x = \angle QPR$, $y = \angle QTR$, and $z = \angle OQR$.

There is an inconsistency in the given information.
According to the angle at the center theorem, $\angle QPR = \frac{1}{2} \angle QOR$. If $\angle QOR = 146^\circ$, then $\angle QPR = 73^\circ$.
According to the alternate segment theorem, $\angle QPR = \angle QRT$. If $\angle QRT = 52^\circ$, then $\angle QPR = 52^\circ$.
Since $73^\circ \neq 52^\circ$, the given values for $\angle QOR$ and $\angle QRT$ are contradictory.
We will proceed by using $\angle QOR = 146^\circ$ as the primary information to determine the angles, as it defines the central arc.

The angle at the center $\angle QOR$ is twice the angle at the circumference $\angle QPR$ subtended by the same arc $QR$.

$$x = \angle QPR = \frac{1}{2} \angle QOR$$ $$x = \frac{1}{2} \times 146^\circ$$ $$x = 73^\circ$$

Reason: Angle at the center is twice the angle at the circumference.

$OQ$ and $OR$ are radii of the circle, so $OQ = OR$.
Therefore, $\triangle OQR$ is an isosceles triangle.
The base angles of an isosceles triangle are equal: $\angle OQR = \angle ORQ = z$.
The sum of angles in a triangle is $180^\circ$.

$$\angle OQR + \angle ORQ + \angle QOR = 180^\circ$$ $$z + z + 146^\circ = 180^\circ$$ $$2z = 180^\circ - 146^\circ$$ $$2z = 34^\circ$$ $$z = \frac{34^\circ}{2}$$ $$z = 17^\circ$$

Reason: Base angles of an isosceles triangle are equal; sum of angles in a triangle is $180^\circ$.

$OQ$ is a radius and $QT$ is a tangent at point $Q$. A radius is perpendicular to the tangent at the point of tangency.

$$\angle OQT = 90^\circ$$

Similarly, $OR$ is a radius and $RT$ is a tangent at point $R$.

$$\angle ORT = 90^\circ$$

Consider the quadrilateral $OQTR$. The sum of interior angles in a quadrilateral is $360^\circ$.

$$\angle QTR + \angle OQT + \angle ORT + \angle QOR = 360^\circ$$ $$y + 90^\circ + 90^\circ + 146^\circ = 360^\circ$$ $$y + 326^\circ = 360^\circ$$ $$y = 360^\circ - 326^\circ$$ $$y = 34^\circ$$

Reason: Radius is perpendicular to tangent at the point of contact; sum of angles in a quadrilateral is $360^\circ$.

The angles marked with letters are:
$x = \boxed{73^\circ}$
$y = \boxed{34^\circ}$
$z = \boxed{17^\circ}$