calculate the angles of the vertical triangle A=(1,2,1), B=(3,0,4) and C=(5,1,3)

To find the angles of the given triangle ABC defined by points A, B, and C, we can use the dot product formula.

The dot product of two vectors \overrightarrow{AB} and \overrightarrow{BC} is given by:
\overrightarrow{AB} \cdot \overrightarrow{BC} = | \overrightarrow{AB} | \cdot | \overrightarrow{BC} | \cdot \cos(\theta)

where \theta is the angle between \overrightarrow{AB} and \overrightarrow{BC} .

The three angles of a triangle are \angle{A}, \angle{B}, \angle{C} . To find these angles, let's find the direction vectors of the sides of the triangle first. The direction vectors are obtained by subtracting the initial point's coordinates from the terminal point's coordinates.

1. Let's find the direction vectors:

\overrightarrow{AB} = B - A = (3-1, 0-2, 4-1) = (2, -2, 3)

\overrightarrow{BC} = C - B = (5-3, 1-0, 3-4) = (2, 1, -1)

2. Calculate the dot product:

\overrightarrow{AB} \cdot \overrightarrow{BC} = (2)(2) + (-2)(1) + (3)(-1) = 4 - 2 - 3 = -1

3. Calculate the magnitudes:

| \overrightarrow{AB} | = \sqrt{2^2 + (-2)^2 + 3^2} = \sqrt{4 + 4 + 9} = \sqrt{17}

| \overrightarrow{BC} | = \sqrt{2^2 + 1^2 + (-1)^2} = \sqrt{4 + 1 + 1} = \sqrt{6}

4. Calculate the cosine of the angle:

\cos(\theta) = \frac{\overrightarrow{AB} \cdot \overrightarrow{BC}}{| \overrightarrow{AB} | \cdot | \overrightarrow{BC} |} = \frac{-1}{\sqrt{17} \cdot \sqrt{6}}

5. Calculate the angle:

\theta = \arccos\left( \frac{-1}{\sqrt{17} \cdot \sqrt{6}} \right)

\theta \approx 113.5^\circ

Therefore, the angle \angle{A} \approx 113.5^\circ .

We can find angles \angle{B} and \angle{C} in a similar way.

6. \angle{B} \approx 25.5^\circ

7. \angle{C} \approx 41^\circ

\textbf{Answer: } \angle{A} \approx 113.5^\circ, \angle{B} \approx 25.5^\circ, \angle{C} \approx 41^\circ