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This is a right triangle problem where you need to find the lengths of the sides \(x\) and \(y\). The given triangle has one angle of \(60^\circ\), one side length of \(13\sqrt{3}\), and a right angle.

### Step-by-Step Solution:

1. **Identify the Triangle Type**:
   - This triangle is a right triangle with one angle \(60^\circ\) and the side opposite the right angle \(13\sqrt{3}\). The other non-right angle will be \(30^\circ\) since the angles in a triangle sum to \(180^\circ\).

2. **Using Trigonometric Ratios**:
   - For a right triangle:
     \[
     \sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}}, \quad \cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}}, \quad \text{and} \quad \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}
     \]
   - Here, \(\theta = 60^\circ\), and the side \(13\sqrt{3}\) is opposite to the \(60^\circ\) angle.

3. **Find \(x\) (the hypotenuse)**:
   - Using the sine function:
     \[
     \sin(60^\circ) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{13\sqrt{3}}{x}
     \]
     Since \(\sin(60^\circ) = \frac{\sqrt{3}}{2}\),
     \[
     \frac{\sqrt{3}}{2} = \frac{13\sqrt{3}}{x}
     \]
     Solving for \(x\):
     \[
     x = \frac{13\sqrt{3} \times 2}{\sqrt{3}} = 26
     \]

4. **Find \(y\) (the adjacent side to \(60^\circ\))**:
   - Using the cosine function:
     \[
     \cos(60^\circ) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{y}{26}
     \]
     Since \(\cos(60^\circ) = \frac{1}{2}\),
     \[
     \frac{1}{2} = \frac{y}{26}
     \]
     Solving for \(y\):
     \[
     y = 13
     \]

### Final Answers:
- \(x = 26\)
- \(y = 13\)

Would you like more details or have any questions about this solution? Here are some additional related questions:

1. How do you find the length of the other side using the Pythagorean theorem in this context?
2. How would the problem change if the angle given was \(30^\circ\) instead of \(60^\circ\)?
3. What is the area of the triangle given the sides \(x\), \(y\), and the right angle?
4. Can you explain the significance of special right triangles in solving this problem?
5. How would you use the law of sines to solve for sides in non-right triangles?

**Tip:** Remember that in a 30-60-90 triangle, the sides are in the ratio \(1 : \sqrt{3} : 2\). This can help quickly determine side lengths.

Solve for the lengths of sides x and y in the right triangle with a 60° angle and a side length of 13√3 opposite to the 60° angle.

Learn how to find the lengths of sides x and y in a right triangle with a 60° angle and a side length of 13√3. This problem covers key trigonometric concepts, including the sine and cosine functions, as well as the 30-60-90 special right triangle ratio. Suitable for students in Grades 9-11.

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