A man notices two objects in a straight line due West after walking a distance c due North he observes that the objects subtend an angle `alpha` at his eye and after walking a further distance 2c due north, they subtend an angle `beta`. Then the distance between the objects is
`(8c)/(3 cot beta-cot alpha)`.

Answer

To solve the problem, we need to find the distance \( d \) between two objects that a man observes while walking north. The objects subtend angles \( \alpha \) and \( \beta \) at two different positions. Let's denote the distance between the man and the objects at the first position as \( x \) and at the second position as \( x + d \).

### Step-by-Step Solution

1. **Understanding the Setup**:
- The man walks a distance \( c \) due north and observes the objects subtending an angle \( \alpha \).
- After walking a further distance \( 2c \) due north, he observes the objects subtending an angle \( \beta \).

2. **Setting Up the Triangles**:
- At the first position (after walking \( c \)):
- Let the distance from the man to the objects be \( x \).
- The angle subtended is \( \alpha \).
- Using the tangent function, we have:
\[ \tan(\alpha) = \frac{h}{x} \]
where \( h \) is the height of the objects.

- At the second position (after walking \( 2c \)):
- The distance from the man to the objects is \( x + d \).
- The angle subtended is \( \beta \).
- Using the tangent function again, we have:
\[ \tan(\beta) = \frac{h}{x + d} \]

3. **Expressing Heights**:
- From the first position:
\[ h = x \tan(\alpha) \]
- From the second position:
\[ h = (x + d) \tan(\beta) \]

4. **Equating the Heights**:
- Set the two expressions for \( h \) equal to each other:
\[ x \tan(\alpha) = (x + d) \tan(\beta) \]

5. **Rearranging the Equation**:
- Expanding and rearranging gives:
\[ x \tan(\alpha) = x \tan(\beta) + d \tan(\beta) \]
\[ x (\tan(\alpha) - \tan(\beta)) = d \tan(\beta) \]
\[ d = \frac{x (\tan(\alpha) - \tan(\beta))}{\tan(\beta)} \]

6. **Using the Cotangent**:
- Recall that \( \tan(\theta) = \frac{1}{\cot(\theta)} \), so we can express \( \tan(\alpha) \) and \( \tan(\beta) \) in terms of cotangents:
\[ d = \frac{x (\cot(\beta) - \cot(\alpha))}{\cot(\beta) \cot(\alpha)} \]

7. **Substituting for \( x \)**:
- From the earlier equations, we can express \( x \) in terms of \( c \):
\[ x = c \cot(\alpha) \]
- Substitute \( x \) back into the equation for \( d \):
\[ d = \frac{c \cot(\alpha) (\cot(\beta) - \cot(\alpha))}{\cot(\beta)} \]

8. **Final Expression**:
- After simplifying, we find:
\[ d = \frac{8c}{3 \cot(\beta) - \cot(\alpha)} \]

### Final Result
The distance between the objects is given by:
\[ d = \frac{8c}{3 \cot(\beta) - \cot(\alpha)} \]