The foot of a tower of height h m is in a direct line between two observers A and B. If the angles of elevationof the top of the tower as seenfrom and B are `alpha` and beta` respectively and if AB = d m, then what is h/d equal to?

Options

A. `( tan (alpha + beta))/( ( cot alpha cot beta - 1)) `
B. `(cot ( alpha + beta))/(( cot alpha cot beta -1))`
C. `( tan (alpha + beta))/( (cot alpha cot beta + 1))`
D. `(cot (alpha + beta))/( (cot alpha cot beta+1))`

Answer

To solve the problem, we need to find the ratio \( \frac{h}{d} \) where \( h \) is the height of the tower and \( d \) is the distance between the two observers A and B.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- Let the height of the tower be \( h \).
- The distance between observers A and B is \( d \).
- The angles of elevation from A and B to the top of the tower are \( \alpha \) and \( \beta \) respectively.

2. **Labeling Points**:
- Let the foot of the tower be point C, the position of observer A be point A, and the position of observer B be point B.
- The distance from A to C is \( AC \) and from C to B is \( CB \).
- Therefore, \( AB = AC + CB = d \).

3. **Using Trigonometry**:
- From observer A, using the angle of elevation \( \alpha \):
\[ \tan(\alpha) = \frac{h}{AC} \]
Thus, we can express \( AC \) as:
\[ AC = \frac{h}{\tan(\alpha)} \]

- From observer B, using the angle of elevation \( \beta \):
\[ \tan(\beta) = \frac{h}{CB} \]
Thus, we can express \( CB \) as:
\[ CB = \frac{h}{\tan(\beta)} \]

4. **Relating Distances**:
- Since \( AB = AC + CB \), we can substitute the values we found:
\[ d = AC + CB = \frac{h}{\tan(\alpha)} + \frac{h}{\tan(\beta)} \]

5. **Factoring Out \( h \)**:
- Factoring \( h \) out of the equation gives:
\[ d = h \left( \frac{1}{\tan(\alpha)} + \frac{1}{\tan(\beta)} \right) \]

6. **Finding \( \frac{h}{d} \)**:
- Rearranging the equation to find \( \frac{h}{d} \):
\[ \frac{h}{d} = \frac{1}{\frac{1}{\tan(\alpha)} + \frac{1}{\tan(\beta)}} \]

7. **Using Cotangent**:
- We can rewrite \( \frac{1}{\tan(\theta)} \) as \( \cot(\theta) \):
\[ \frac{h}{d} = \frac{1}{\cot(\alpha) + \cot(\beta)} \]

8. **Final Expression**:
- The final expression for \( \frac{h}{d} \) can be simplified using the identity for cotangent:
\[ \frac{h}{d} = \frac{\cot(\alpha) \cdot \cot(\beta) - 1}{\cot(\alpha + \beta)} \]

### Final Result:
\[ \frac{h}{d} = \frac{\cot(\alpha + \beta)}{\cot(\alpha) \cdot \cot(\beta) - 1} \]