A tower subtends angles, `alpha, 2 alpha, 3 alpha` respectively at points A, B and C all lying on a horizontal line through the foot of the tower. If `0 lt alpha lt (pi)/(6) and (AB)/(BC) = 8 cos^(2) alpha - 2k`, then k = ____________

Answer

To solve the problem, we need to find the value of \( k \) given the relationship between the segments \( AB \) and \( BC \) based on the angles subtended by the tower at points \( A \), \( B \), and \( C \).

### Step-by-Step Solution:

1. **Understanding the Geometry**:
- Let \( D \) be the foot of the tower.
- The angles subtended by the tower at points \( A \), \( B \), and \( C \) are \( \alpha \), \( 2\alpha \), and \( 3\alpha \) respectively.
- We denote the height of the tower as \( h \) and the distances from the foot of the tower to points \( A \), \( B \), and \( C \) as \( DA = x \), \( DB = y \), and \( DC = z \).

2. **Using Tangent Function**:
- From point \( A \):
\[ \tan(\alpha) = \frac{h}{x} \]
- From point \( B \):
\[ \tan(2\alpha) = \frac{h}{y} \]
- From point \( C \):
\[ \tan(3\alpha) = \frac{h}{z} \]

3. **Expressing Distances**:
- Rearranging the above equations gives:
\[ x = \frac{h}{\tan(\alpha)}, \quad y = \frac{h}{\tan(2\alpha)}, \quad z = \frac{h}{\tan(3\alpha)} \]

4. **Finding \( AB \) and \( BC \)**:
- The distance \( AB = y - x \) and \( BC = z - y \).
- Therefore:
\[ AB = \frac{h}{\tan(2\alpha)} - \frac{h}{\tan(\alpha)} = h \left( \frac{1}{\tan(2\alpha)} - \frac{1}{\tan(\alpha)} \right) \]
\[ BC = \frac{h}{\tan(3\alpha)} - \frac{h}{\tan(2\alpha)} = h \left( \frac{1}{\tan(3\alpha)} - \frac{1}{\tan(2\alpha)} \right) \]

5. **Finding the Ratio \( \frac{AB}{BC} \)**:
- Now, we can find the ratio:
\[ \frac{AB}{BC} = \frac{\frac{h}{\tan(2\alpha)} - \frac{h}{\tan(\alpha)}}{\frac{h}{\tan(3\alpha)} - \frac{h}{\tan(2\alpha)}} \]
- Simplifying this gives:
\[ \frac{AB}{BC} = \frac{\tan(3\alpha) \tan(2\alpha) - \tan(2\alpha) \tan(\alpha)}{\tan(3\alpha) \tan(2\alpha) - \tan(2\alpha) \tan(2\alpha)} \]

6. **Using the Sine Rule**:
- From the sine rule in triangles \( ABE \) and \( BCE \), we can derive:
\[ \frac{AB}{BC} = \frac{\sin(3\alpha)}{\sin(\alpha)} \]
- This leads to:
\[ \frac{AB}{BC} = 3 - 4 \cos^2(\alpha) \]

7. **Equating the Two Expressions**:
- We are given that:
\[ \frac{AB}{BC} = 8 \cos^2(\alpha) - 2k \]
- Setting the two expressions equal gives:
\[ 3 - 4 \cos^2(\alpha) = 8 \cos^2(\alpha) - 2k \]

8. **Solving for \( k \)**:
- Rearranging the equation:
\[ 2k = 8 \cos^2(\alpha) + 4 \cos^2(\alpha) - 3 \]
\[ 2k = 12 \cos^2(\alpha) - 3 \]
- Dividing by 2:
\[ k = 6 \cos^2(\alpha) - \frac{3}{2} \]

### Final Answer:
Thus, the value of \( k \) is:
\[ k = 6 \cos^2(\alpha) - \frac{3}{2} \]