The image you uploaded involves a geometry question with a triangle and a straight line. The goal is to find the values of $x$, $y$, and $z$ based on the given angles and the configuration.

Given:

1. Triangle $ACD$.
2. A straight line $BCDE$.
3. Angle at $E$ is $142^\circ$.
4. Angle at $A$ is $271^\circ$.
5. Angle at $C$ is $53^\circ$.

Let’s start solving the values one by one.

Step 1: Find $z$

• The angle at $A$ is $271^\circ$, and it is inside the triangle. To find angle $z$, we know the angle at $A$ and the angle at $C$ should sum to $180^\circ$ since they are on a straight line.

Thus: $z = 180^\circ - 53^\circ = 127^\circ$

Step 2: Find $x$

• To solve for $x$, we use the fact that the interior angles of a triangle sum to $180^\circ$.
• In triangle $ACD$, the angles at $A$, $C$, and $D$ must sum to $180^\circ$. From the figure, angle $A = 271^\circ$ and angle $C = 53^\circ$, so:

$x = 180^\circ - (271^\circ + 53^\circ) = 180^\circ - 324^\circ = 180^\circ - 144^\circ = 36^\circ$

Step 3: Find $y$

• Finally, angle $y$ is located at the intersection of the line at $B$. It’s supplementary to $142^\circ$ as it is on the same straight line. Therefore:

$y = 180^\circ - 142^\circ = 38^\circ$

Final Answer:

• $x = 36^\circ$
• $y = 38^\circ$
• $z = 127^\circ$

Let me know if you need further clarification or more details on any part of the solution!