To solve for $x$ and $y$ in the given triangle, let's analyze the problem:

Given Information:

1. $\triangle ABC$:

   • $\angle B = 65^\circ$
   • $\angle A = 55^\circ$
   • $\angle C$ can be calculated since the sum of angles in a triangle is $180^\circ$.

2. $\triangle DQC$:

   • $QC = CD$ (isosceles triangle).
   • $\angle DQC = x$.
   • The interior angle adjacent to $x$ in $\triangle DQC$ corresponds to $\angle C$ from $\triangle ABC$.

Step-by-Step Solution:

Step 1: Calculate $\angle C$ in $\triangle ABC$

$\angle C = 180^\circ - \angle A - \angle B = 180^\circ - 55^\circ - 65^\circ = 60^\circ.$

Step 2: Analyze $\triangle DQC$

Since $QC = CD$, $\triangle DQC$ is an isosceles triangle. Therefore: $\angle DCQ = \angle QDC = \frac{180^\circ - x}{2}.$

Step 3: Relate $y$ to $\angle DCQ$

The angle at $P$ in $\triangle APQ$ corresponds to the supplementary angle of $\angle DCQ$ in $\triangle DQC$, because: $y = \angle DCQ = \frac{180^\circ - x}{2}.$

Step 4: Calculate $x$ using triangle rules

In $\triangle DQC$: $x + 2\left(\frac{180^\circ - x}{2}\right) = 180^\circ.$ Simplify: $x + (180^\circ - x) = 180^\circ.$ $x = 60^\circ.$

Step 5: Calculate $y$

Substitute $x = 60^\circ$ into $y = \frac{180^\circ - x}{2}$: $y = \frac{180^\circ - 60^\circ}{2} = \frac{120^\circ}{2} = 60^\circ.$

Final Answer:

$x = 60^\circ, \quad y = 60^\circ.$

Would you like a more detailed explanation or assistance with related problems?

Relative Questions:

1. How do you solve isosceles triangle problems when the base angles are unknown?
2. Can you derive the sum of angles in a triangle using geometric principles?
3. What properties define an isosceles triangle?
4. How do supplementary angles relate to this triangle's interior calculations?
5. What are some other methods to verify the correctness of $x$ and $y$?

Tip:

Always confirm relationships between angles by cross-referencing their geometric properties, such as symmetry or supplementary rules.