The diagram solve it

The angle subtended by an arc at the center of a circle is twice the angle subtended by the same arc at any point on the remaining part of the circle.
Arc $QR$ subtends $\angle QOR$ at the center and $\angle QPR$ at the circumference.

$$\angle QOR = 2 \times \angle QPR$$

Given $\angle QOR = 100^\circ$ and $\angle QPR = x$:

$$100^\circ = 2x$$ $$x = \frac{100^\circ}{2}$$ $$x = 50^\circ$$

The quadrilateral $PQRS$ is a cyclic quadrilateral because all its vertices lie on the circle. In a cyclic quadrilateral, opposite angles are supplementary (add up to $180^\circ$).
Angles $\angle QPS$ and $\angle QRS$ are opposite angles.

$$\angle QPS + \angle QRS = 180^\circ$$

We know $\angle QPS = x = 50^\circ$ and $\angle QRS = y$:

$$50^\circ + y = 180^\circ$$ $$y = 180^\circ - 50^\circ$$ $$y = 130^\circ$$

Angles $\angle PQR$ and $\angle PSR$ are opposite angles in the cyclic quadrilateral $PQRS$.

$$\angle PQR + \angle PSR = 180^\circ$$

We know $\angle PSR = z$, so we need to find $\angle PQR$.
From the diagram, $\angle QRS = \angle QRP + \angle PRS$.
We know $\angle QRS = y = 130^\circ$ and $\angle PRS = 30^\circ$:

$$130^\circ = \angle QRP + 30^\circ$$ $$\angle QRP = 130^\circ - 30^\circ$$ $$\angle QRP = 100^\circ$$

Now, consider $\triangle PQR$. The sum of angles in a triangle is $180^\circ$.

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

We know $\angle QPR = x = 50^\circ$ and $\angle PRQ = \angle QRP = 100^\circ$:

$$50^\circ + \angle PQR + 100^\circ = 180^\circ$$ $$150^\circ + \angle PQR = 180^\circ$$ $$\angle PQR = 180^\circ - 150^\circ$$ $$\angle PQR = 30^\circ$$

Finally, substitute $\angle PQR$ back into the cyclic quadrilateral property:

$$\angle PQR + z = 180^\circ$$ $$30^\circ + z = 180^\circ$$ $$z = 180^\circ - 30^\circ$$ $$z = 150^\circ$$

The values are:

$$x = 50^\circ, y = 130^\circ, z = 150^\circ$$