AB is a vertical pole. The end A is on the level ground .C is the middle point of AB. P is a point on the level ground . The portion BC subtends an angles `beta` at P. If AP = nAB, then tan `beta`=

Options

A. `n/(2n^2+1)`
B. `n/(n^2-1)`
C. `n/(n^2+1)`
D. none of these

Answer

To solve the problem, we need to find the value of \( \tan \beta \) given the conditions of the vertical pole and the points defined in the problem. Here’s a step-by-step solution:

### Step 1: Define the Variables
Let:
- \( AB = x \) (the height of the pole)
- \( C \) is the midpoint of \( AB \), so \( AC = BC = \frac{x}{2} \)
- \( AP = n \cdot AB = n \cdot x \)

### Step 2: Set Up the Triangle
We have the following points:
- \( A \) is at the ground level.
- \( B \) is at the height \( x \).
- \( C \) is at height \( \frac{x}{2} \).
- \( P \) is on the ground at a distance \( n \cdot x \) from point \( A \).

### Step 3: Identify the Angles
The angle \( \beta \) is subtended by the segment \( BC \) at point \( P \). We need to find \( \tan \beta \).

### Step 4: Use the Right Triangle
In triangle \( BPA \):
- The height \( AB \) is \( x \).
- The base \( AP \) is \( n \cdot x \).

Using the definition of tangent:
\[ \tan P = \frac{AB}{AP} = \frac{x}{n \cdot x} = \frac{1}{n} \]

### Step 5: Analyze Triangle \( CPA \)
In triangle \( CPA \):
- The height \( AC \) is \( \frac{x}{2} \).
- The base \( AP \) is \( n \cdot x \).

Using the definition of tangent:
\[ \tan \alpha = \frac{AC}{AP} = \frac{\frac{x}{2}}{n \cdot x} = \frac{1}{2n} \]

### Step 6: Use the Angle Addition Formula
Since \( P \) is the angle formed by \( \alpha + \beta \):
\[ \tan P = \tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} \]

Substituting the values we found:
\[ \frac{1}{n} = \frac{\frac{1}{2n} + \tan \beta}{1 - \frac{1}{2n} \tan \beta} \]

### Step 7: Cross-Multiply and Simplify
Cross-multiplying gives:
\[ 1 - \frac{1}{2n} \tan \beta = n \left(\frac{1}{2n} + \tan \beta\right) \]
This simplifies to:
\[ 1 - \frac{1}{2n} \tan \beta = \frac{1}{2} + n \tan \beta \]

### Step 8: Rearranging the Equation
Rearranging gives:
\[ 1 - \frac{1}{2} = n \tan \beta + \frac{1}{2n} \tan \beta \]
\[ \frac{1}{2} = \tan \beta \left(n + \frac{1}{2n}\right) \]

### Step 9: Solve for \( \tan \beta \)
Thus:
\[ \tan \beta = \frac{\frac{1}{2}}{n + \frac{1}{2n}} = \frac{1}{2} \cdot \frac{2n}{2n^2 + 1} = \frac{n}{2n^2 + 1} \]

### Final Answer
The value of \( \tan \beta \) is:
\[ \tan \beta = \frac{n}{2n^2 + 1} \]