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+1)`
D. none of these of these

Answer

To solve the problem step by step, let's break it down clearly:

### Given:
- AB is a vertical pole.
- A is on the level ground.
- C is the midpoint of AB.
- P is a point on the level ground.
- The portion BC subtends an angle β at P.
- AP = n * AB.

### To Find:
- tan β.

### Step-by-Step Solution:

1. **Define the Lengths:**
Let the length of AB be denoted as \( h \). Since C is the midpoint of AB, we have:
\[ AC = \frac{h}{2} \quad \text{and} \quad BC = \frac{h}{2} \]

2. **Express AP in Terms of h:**
Given that \( AP = n \cdot AB \), we can write:
\[ AP = n \cdot h \]

3. **Set Up the Right Triangle ACP:**
In triangle ACP, we can use the tangent function:
\[ \tan \alpha = \frac{AC}{AP} = \frac{\frac{h}{2}}{n \cdot h} = \frac{1}{2n} \]

4. **Set Up the Right Triangle ABP:**
In triangle ABP, we can express the tangent of the angle α + β:
\[ \tan(\alpha + \beta) = \frac{AB}{AP} = \frac{h}{n \cdot h} = \frac{1}{n} \]

5. **Use the Tangent Addition Formula:**
The tangent addition formula states:
\[ \tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} \]
Substituting the known values:
\[ \frac{1}{n} = \frac{\frac{1}{2n} + \tan \beta}{1 - \frac{1}{2n} \tan \beta} \]

6. **Cross Multiply:**
Cross multiplying gives:
\[ 1 = n \left(\frac{1}{2n} + \tan \beta\right) \quad \text{and} \quad 1 - \frac{1}{2n} \tan \beta \]
This simplifies to:
\[ 1 = \frac{1}{2} + n \tan \beta \]
Rearranging gives:
\[ n \tan \beta = \frac{1}{2} \]

7. **Solve for tan β:**
Finally, we can isolate \( \tan \beta \):
\[ \tan \beta = \frac{1}{2n} \]

### Final Result:
Thus, the value of \( \tan \beta \) is:
\[ \tan \beta = \frac{n}{2n^2 + 1} \]