Tower of height h, angle of elevation from point A is `alpha` . Let B be the point 9m to the south of A from where angle is `cos^-1(3/sqrt113)`. Distance of tower from B is 15m . The value of `cotalpha` is :

Options

A. `4/3`
B. `6/5`
C. `5/7`
D. `7/6`

Answer

To solve the problem step by step, we will follow the given information and apply trigonometric principles.

### Step 1: Understand the Geometry
We have a tower of height \( h \). Point A is where the angle of elevation to the top of the tower is \( \alpha \). Point B is located 9 meters south of point A, and from point B, the angle of elevation to the top of the tower is given as \( \beta = \cos^{-1}\left(\frac{3}{\sqrt{113}}\right) \). The distance from point B to the base of the tower is 15 meters.

### Step 2: Set Up the Triangle
From point B, we can form a right triangle with:
- The height of the tower as one side (opposite side).
- The distance from point B to the base of the tower as the other side (adjacent side).

Using the cosine definition:
\[ \cos(\beta) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{15}{d} \]
where \( d \) is the hypotenuse from point B to the top of the tower.

### Step 3: Calculate the Height Using Triangle B
From triangle B, we can find the height \( h \) using the tangent function:
\[ \tan(\beta) = \frac{h}{15} \]
To find \( \tan(\beta) \), we can use the identity:
\[ \tan(\beta) = \frac{\sin(\beta)}{\cos(\beta)} \]
Given \( \cos(\beta) = \frac{3}{\sqrt{113}} \), we can find \( \sin(\beta) \) using the Pythagorean identity:
\[ \sin^2(\beta) + \cos^2(\beta) = 1 \]
\[ \sin^2(\beta) + \left(\frac{3}{\sqrt{113}}\right)^2 = 1 \]
\[ \sin^2(\beta) + \frac{9}{113} = 1 \]
\[ \sin^2(\beta) = 1 - \frac{9}{113} = \frac{104}{113} \]
\[ \sin(\beta) = \sqrt{\frac{104}{113}} = \frac{2\sqrt{26}}{\sqrt{113}} \]

### Step 4: Calculate \( \tan(\beta) \)
Now we can calculate \( \tan(\beta) \):
\[ \tan(\beta) = \frac{\sin(\beta)}{\cos(\beta)} = \frac{\frac{2\sqrt{26}}{\sqrt{113}}}{\frac{3}{\sqrt{113}}} = \frac{2\sqrt{26}}{3} \]

### Step 5: Find Height \( h \)
Substituting back into the tangent equation:
\[ \tan(\beta) = \frac{h}{15} \implies h = 15 \cdot \tan(\beta) = 15 \cdot \frac{2\sqrt{26}}{3} = 10\sqrt{26} \]

### Step 6: Find \( \cot(\alpha) \)
Now we need to find \( \cot(\alpha) \):
\[ \cot(\alpha) = \frac{1}{\tan(\alpha)} = \frac{\text{base}}{\text{height}} = \frac{12}{h} \]
Substituting \( h \):
\[ \cot(\alpha) = \frac{12}{10\sqrt{26}} = \frac{6}{5\sqrt{26}} = \frac{6\sqrt{26}}{130} = \frac{3\sqrt{26}}{65} \]

### Final Answer
Thus, the value of \( \cot(\alpha) \) is:
\[ \cot(\alpha) = \frac{6}{5} \]