Find the measures of the three angles, in radians, of the triangle with the given vertices: D(1, 1, 1), E(1, −1, 2), and F(−5, 2, 6).

To find the measures of the angles in the triangle defined by the vertices \( D(1, 1, 1) \), \( E(1, -1, 2) \), and \( F(-5, 2, 6) \), we first need to calculate the lengths of the sides of the triangle using the distance formula in three-dimensional space:

\[
d(P_1, P_2) = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}
\]

1. **Calculate the lengths of the sides:**

   - Length \( DE \):
     \[
     DE = \sqrt{(1 - 1)^2 + (-1 - 1)^2 + (2 - 1)^2} = \sqrt{0 + 4 + 1} = \sqrt{5}
     \]

   - Length \( EF \):
     \[
     EF = \sqrt{(-5 - 1)^2 + (2 + 1)^2 + (6 - 2)^2} = \sqrt{(-6)^2 + 3^2 + 4^2} = \sqrt{36 + 9 + 16} = \sqrt{61}
     \]

   - Length \( FD \):
     \[
     FD = \sqrt{(1 + 5)^2 + (1 - 2)^2 + (1 - 6)^2} = \sqrt{6^2 + (-1)^2 + (-5)^2} = \sqrt{36 + 1 + 25} = \sqrt{62}
     \]

Now we have the lengths of the sides:
- \( a = EF = \sqrt{61} \)
- \( b = FD = \sqrt{62} \)
- \( c = DE = \sqrt{5} \)

2. **Use the cosine rule to find the angles:**

The cosine of an angle in a triangle can be calculated using the formula:
\[
\cos(A) = \frac{b^2 + c^2 - a^2}{2bc}
\]

### Calculate the angles:

- **Angle \( D \) (opposite side \( EF \))**:
\[
\cos(D) = \frac{(DE)^2 + (FD)^2 - (EF)^2}{2 \cdot DE \cdot FD} = \frac{(\sqrt{5})^2 + (\sqrt{62})^2 - (\sqrt{61})^2}{2 \cdot \sqrt{5} \cdot \sqrt{62}} = \frac{5 + 62 - 61}{2\sqrt{5}\sqrt{62}} = \frac{6}{2\sqrt{5}\sqrt{62}} = \frac{3}{\sqrt{5 \cdot 62}} = \frac{3}{\sqrt{310}}
\]
Thus,
\[
D = \cos^{-1}\left(\frac{3}{\sqrt{310}}\right)
\]

- **Angle \( E \) (opposite side \( FD \))**:
\[
\cos(E) = \frac{(DE)^2 + (EF)^2 - (FD)^2}{2 \cdot DE \cdot EF} = \frac{(\sqrt{5})^2 + (\sqrt{61})^2 - (\sqrt{62})^2}{2 \cdot \sqrt{5} \cdot \sqrt{61}} = \frac{5 + 61 - 62}{2\sqrt{5}\sqrt{61}} = \frac{4}{2\sqrt{5}\sqrt{61}} = \frac{2}{\sqrt{5 \cdot 61}} = \frac{2}{\sqrt{305}}
\]
Thus,
\[
E = \cos^{-1}\left(\frac{2}{\sqrt{305}}\right)
\]

- **Angle \( F \) (opposite side \( DE \))**:
\[
\cos(F) = \frac{(FD)^2 + (EF)^2 - (DE)^2}{2 \cdot FD \cdot EF} = \frac{(\sqrt{62})^2 + (\sqrt{61})^2 - (\sqrt{5})^2}{2 \cdot \sqrt{62} \cdot \sqrt{61}} = \frac{62 + 61 - 5}{2\sqrt{62}\sqrt{61}} = \frac{118}{2\sqrt{62}\sqrt{61}} = \frac{59}{\sqrt{62 \cdot 61}} = \frac{59}{\sqrt{3782}}
\]
Thus,
\[
F = \cos^{-1}\left(\frac{59}{\sqrt{3782}}\right)
\]

3. **Convert the angles to radians:**
The angles are already calculated in radians as the inverse cosine function returns angles in radians.

### Final Results:
- Angle \( D \) is \( \cos^{-1}\left(\frac{3}{\sqrt{310}}\right) \)
- Angle \( E \) is \( \cos^{-1}\left(\frac{2}{\sqrt{305}}\right) \)
- Angle \( F \) is \( \cos^{-1}\left(\frac{59}{\sqrt{3782}}\right) \)

If you would like, I can provide approximate decimal values for these angles or further details on how to evaluate them. Do you have any questions or need additional information? Here are a few related questions:

1. How do the angles relate to the lengths of the sides in a triangle?
2. What is the significance of the cosine rule in triangle calculations?
3. Can these angles be used to determine the area of the triangle?
4. What are the properties of triangles in three-dimensional space?
5. How would these calculations change if the triangle were in two-dimensional space? 

**Tip:** Always check if the triangle formed by three points is valid by ensuring the sum of the angles equals \( \pi \) radians.

Explore how to calculate the angles of a triangle in three-dimensional space defined by vertices D(1, 1, 1), E(1, −1, 2), and F(−5, 2, 6). This solution utilizes the distance formula and cosine rule to find angle measures in radians. Perfect for Grades 10-12.

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