The lengths of the shadow of a vertical pole of height h thrown by the sun's rays at three different moments are h, 2h and 3h. The sum of the angles of elevations of the rays at these moments is equal to `pi/(4k)` where k = _______________

Answer

To solve the problem, we need to find the sum of the angles of elevation of the rays of the sun at three different moments when the lengths of the shadows of a vertical pole of height \( h \) are \( h \), \( 2h \), and \( 3h \).

### Step-by-Step Solution:

1. **Identify the Angles of Elevation**:
- Let the angles of elevation corresponding to the shadow lengths \( h \), \( 2h \), and \( 3h \) be \( \alpha \), \( \beta \), and \( \gamma \) respectively.

2. **Calculate \( \tan \alpha \)**:
- For the first shadow length \( h \):
\[ \tan \alpha = \frac{\text{height of pole}}{\text{length of shadow}} = \frac{h}{h} = 1 \]
- Therefore, \( \alpha = 45^\circ \).

3. **Calculate \( \tan \beta \)**:
- For the second shadow length \( 2h \):
\[ \tan \beta = \frac{h}{2h} = \frac{1}{2} \]
- Therefore, \( \beta = \tan^{-1}\left(\frac{1}{2}\right) \).

4. **Calculate \( \tan \gamma \)**:
- For the third shadow length \( 3h \):
\[ \tan \gamma = \frac{h}{3h} = \frac{1}{3} \]
- Therefore, \( \gamma = \tan^{-1}\left(\frac{1}{3}\right) \).

5. **Sum of Angles**:
- We need to find \( \alpha + \beta + \gamma \).
- Using the tangent addition formula:
\[ \tan(\alpha + \beta + \gamma) = \frac{\tan \alpha + \tan \beta + \tan \gamma - \tan \alpha \tan \beta \tan \gamma}{1 - (\tan \alpha \tan \beta + \tan \beta \tan \gamma + \tan \gamma \tan \alpha)} \]
- Substitute the values:
\[ \tan \alpha = 1, \quad \tan \beta = \frac{1}{2}, \quad \tan \gamma = \frac{1}{3} \]

6. **Calculate the Numerator**:
- Numerator:
\[ 1 + \frac{1}{2} + \frac{1}{3} - \left(1 \cdot \frac{1}{2} \cdot \frac{1}{3}\right) = 1 + 0.5 + 0.333 - \frac{1}{6} \]
- Convert to a common denominator (6):
\[ = \frac{6}{6} + \frac{3}{6} + \frac{2}{6} - \frac{1}{6} = \frac{10}{6} = \frac{5}{3} \]

7. **Calculate the Denominator**:
- Denominator:
\[ 1 - \left(1 \cdot \frac{1}{2} + \frac{1}{2} \cdot \frac{1}{3} + \frac{1}{3} \cdot 1\right) = 1 - \left(\frac{1}{2} + \frac{1}{6} + \frac{1}{3}\right) \]
- Convert to a common denominator (6):
\[ = 1 - \left(\frac{3}{6} + \frac{1}{6} + \frac{2}{6}\right) = 1 - 1 = 0 \]

8. **Conclusion**:
- Since the denominator is 0, it indicates that \( \alpha + \beta + \gamma = 90^\circ \) or \( \frac{\pi}{2} \).

9. **Relate to Given Expression**:
- We know that:
\[ \alpha + \beta + \gamma = \frac{\pi}{4k} \]
- Setting \( \frac{\pi}{2} = \frac{\pi}{4k} \) gives:
\[ 2 = \frac{1}{4k} \implies 4k = 1 \implies k = \frac{1}{4} \]

### Final Answer:
Thus, \( k = \frac{1}{4} \).