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To solve the problem of finding the angles of a triangle given its vertices, we first need to calculate the side lengths using the distance formula and then use those lengths to find the angles using trigonometric functions.

Given vertices:
- \( A(1, 0, -1) \)
- \( B(2, -5, 0) \)
- \( C(1, 4, 5) \)

### Step 1: Calculate the side lengths

1. **Length of \( AB \):**
   \[
   AB = \sqrt{(2 - 1)^2 + (-5 - 0)^2 + (0 - (-1))^2} = \sqrt{1 + 25 + 1} = \sqrt{27} \approx 5.2
   \]

2. **Length of \( BC \):**
   \[
   BC = \sqrt{(1 - 2)^2 + (4 - (-5))^2 + (5 - 0)^2} = \sqrt{1 + 81 + 25} = \sqrt{107} \approx 10.3
   \]

3. **Length of \( CA \):**
   \[
   CA = \sqrt{(1 - 1)^2 + (4 - 0)^2 + (5 - (-1))^2} = \sqrt{0 + 16 + 36} = \sqrt{52} \approx 7.2
   \]

### Step 2: Calculate the angles using the Law of Cosines

The Law of Cosines states that for any triangle with sides \(a\), \(b\), and \(c\), and corresponding opposite angles \(A\), \(B\), and \(C\):

\[
\cos A = \frac{b^2 + c^2 - a^2}{2bc}
\]
\[
\cos B = \frac{a^2 + c^2 - b^2}{2ac}
\]
\[
\cos C = \frac{a^2 + b^2 - c^2}{2ab}
\]

Let:
- \( AB = c \)
- \( BC = a \)
- \( CA = b \)

1. **Angle \( \angle BAC \) (at A):**
   \[
   \cos(\angle BAC) = \frac{AB^2 + AC^2 - BC^2}{2 \cdot AB \cdot AC}
   \]
   Plugging in the values:
   \[
   \cos(\angle BAC) = \frac{27 + 52 - 107}{2 \times 5.2 \times 7.2} = \frac{-28}{74.88} \approx -0.374
   \]
   \[
   \angle BAC = \cos^{-1}(-0.374) \approx 112^\circ
   \]

2. **Angle \( \angle ABC \) (at B):**
   \[
   \cos(\angle ABC) = \frac{AB^2 + BC^2 - CA^2}{2 \cdot AB \cdot BC}
   \]
   Plugging in the values:
   \[
   \cos(\angle ABC) = \frac{27 + 107 - 52}{2 \times 5.2 \times 10.3} = \frac{82}{107.12} \approx 0.765
   \]
   \[
   \angle ABC = \cos^{-1}(0.765) \approx 40^\circ
   \]

3. **Angle \( \angle BCA \) (at C):**
   \[
   \cos(\angle BCA) = \frac{BC^2 + CA^2 - AB^2}{2 \cdot BC \cdot CA}
   \]
   Plugging in the values:
   \[
   \cos(\angle BCA) = \frac{107 + 52 - 27}{2 \times 10.3 \times 7.2} = \frac{132}{148.32} \approx 0.89
   \]
   \[
   \angle BCA = \cos^{-1}(0.89) \approx 27^\circ
   \]

### Correct Angles:

- \( \angle CAB = 112^\circ \)
- \( \angle ABC = 40^\circ \)
- \( \angle BCA = 27^\circ \)

These calculations match the correct angles for the triangle.

Would you like further details on these calculations or have any questions? Here are some related questions you might find useful:

1. How do you find the side lengths of a triangle given its vertices in 3D space?
2. What is the Law of Cosines and when is it used?
3. How do you compute the cosine of an angle using dot products in vector form?
4. How can trigonometric functions help in calculating angles in 3D geometry?
5. What are some common mistakes to avoid when using the Law of Cosines?

**Tip:** When working with 3D coordinates, ensure that each component is correctly used in the distance and angle calculations to avoid mistakes.

Learn how to calculate the angles of a triangle in 3D space given its vertices using the Law of Cosines. This problem involves using the distance formula to find side lengths and applying trigonometric functions to compute angles accurately.

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