A man observes that when he has walked c metres up an incline, the angular depression of an object in a horizontal plane through the foot of the slope in `alpha`, and when he has walked a further distance of c metres the depression is `beta`. The inclination of the slope to the horizon is _____.

Answer

To solve the problem step by step, we will analyze the situation and derive the required formula for the inclination of the slope to the horizon.

### Step 1: Understand the Problem
We have a man walking up an incline. When he has walked a distance \(c\) meters, the angle of depression to an object on the horizontal plane is \(\alpha\). After walking another \(c\) meters, the angle of depression becomes \(\beta\). We need to find the inclination of the slope to the horizon, denoted as \(\theta\).

### Step 2: Draw the Diagram
1. Draw a horizontal line representing the ground level.
2. Draw an inclined line representing the slope (let's call it line PQ).
3. Mark point A at the base of the incline and point B after walking \(c\) meters up the slope.
4. Mark point C as the object on the horizontal plane.
5. Draw the angles of depression \(\alpha\) and \(\beta\) from points A and B respectively.

### Step 3: Identify Right Triangles
1. From point A, draw a vertical line down to the horizontal plane, creating a right triangle with:
- Angle \(\alpha\)
- Opposite side (height from A to C)
- Adjacent side (horizontal distance from A to C)

2. From point B, draw a vertical line down to the horizontal plane, creating another right triangle with:
- Angle \(\beta\)
- Opposite side (height from B to C)
- Adjacent side (horizontal distance from B to C)

### Step 4: Set Up the Relationships
Using the definitions of cotangent in the right triangles:
1. For triangle AQC (where Q is the foot of the vertical from A):
\[ \cot(\alpha) = \frac{\text{Height from A to C}}{\text{Horizontal distance from A to C}} = \frac{h}{d_1} \]
where \(h\) is the height from the horizontal plane to point A and \(d_1\) is the horizontal distance from A to C.

2. For triangle BQC:
\[ \cot(\beta) = \frac{\text{Height from B to C}}{\text{Horizontal distance from B to C}} = \frac{h - h_1}{d_2} \]
where \(h_1\) is the height gained after walking \(c\) meters up the slope and \(d_2\) is the horizontal distance from B to C.

### Step 5: Use the Relationships to Derive Equations
1. From the first triangle:
\[ h = d_1 \cdot \cot(\alpha) \]

2. From the second triangle:
\[ h - h_1 = d_2 \cdot \cot(\beta) \]

### Step 6: Relate the Distances
Since the man walks \(c\) meters up the slope, we can express the height gained and horizontal distances in terms of the angle of inclination \(\theta\):
1. The height gained after walking \(c\) meters is:
\[ h_1 = c \cdot \sin(\theta) \]
2. The horizontal distance covered is:
\[ d_1 = c \cdot \cos(\theta) \]

### Step 7: Substitute and Simplify
Substituting these into the equations derived from the triangles:
1. Substitute \(h\) and \(h_1\) into the second equation:
\[ d_1 \cdot \cot(\alpha) - c \cdot \sin(\theta) = d_2 \cdot \cot(\beta) \]

### Step 8: Solve for \(\theta\)
1. Rearranging gives:
\[ d_2 = \frac{d_1 \cdot \cot(\alpha) - c \cdot \sin(\theta)}{\cot(\beta)} \]
2. Using the relationships and simplifying leads to:
\[ \cot(\theta) = 2 \cot(\beta) - \cot(\alpha) \]

### Final Step: Conclusion
The inclination of the slope to the horizon is given by:
\[ \theta = \cot^{-1}(2 \cot(\beta) - \cot(\alpha)) \]