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In the triangle \( \triangle ABC \), we are given that:

- \( AC = 2 \) (one leg),
- \( BC = 12 \) (the other leg),
- \( AB = 15 \) (the hypotenuse),
- It is a right triangle at \( C \).

To find \( x \), we assume that \( x \) refers to one of the angles in the triangle (either \( \angle A \) or \( \angle B \)).

Using trigonometry, let's find \( \angle A \) or \( \angle B \).

### Step 1: Use the Pythagorean theorem (for checking)
The Pythagorean theorem states:
\[
AB^2 = AC^2 + BC^2
\]
Substituting the known values:
\[
15^2 = 2^2 + 12^2
\]
\[
225 = 4 + 144
\]
\[
225 = 148 \quad (\text{this is incorrect, meaning the sides do not form a right triangle directly})
\]

There seems to be a discrepancy in the triangle's measurements. Could you clarify if this triangle might involve some adjustments, or should we proceed with trigonometric calculations (for example, calculating angles using sine, cosine, or tangent)?

Let me know how you'd like to proceed, and I can offer details or continue solving!

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**Follow-up questions:**

1. What does \( x \) refer to—an angle or another value?
2. Do you want to use trigonometry (sin, cos, tan) to find an angle?
3. Should we recheck the given side lengths for correctness?
4. Would you like a refresher on the Pythagorean theorem?
5. How can the tangent function be applied in right triangles?

**Tip:** In right triangles, the sine of an angle is the ratio of the opposite side to the hypotenuse.

Find x in the right triangle with sides AC = 2, BC = 12, and AB = 15.

Learn how to solve for x in a right triangle with sides AC = 2, BC = 12, and AB = 15 using the Pythagorean theorem and trigonometric ratios. This problem covers essential trigonometry and geometry concepts for high school students in Grades 9-11.

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