The top of a tower is observed from three point A,B,C on a straight line leading to the tower. If the angles of elevation are `theta,2theta,3theta` from them, then
`(AB)/(BC)=(cot theta-cot 2theta)/(cot 2theta-cot 3 theta)`

Answer

To solve the problem, we need to establish the relationship between the segments \( AB \) and \( BC \) based on the angles of elevation from points \( A \), \( B \), and \( C \) towards the top of the tower \( PQ \).

### Step-by-Step Solution:

1. **Understanding the Setup**:
- Let \( PQ \) be the height of the tower.
- Let \( A \), \( B \), and \( C \) be the points on the ground from which the angles of elevation to the top of the tower are \( \theta \), \( 2\theta \), and \( 3\theta \) respectively.

2. **Using Trigonometric Ratios**:
- From point \( A \):
\[ \cot \theta = \frac{PQ}{AQ} \quad \text{(where \( AQ = AB + BC + CQ \))} \]
- From point \( B \):
\[ \cot 2\theta = \frac{PQ}{BQ} \quad \text{(where \( BQ = BC + CQ \))} \]
- From point \( C \):
\[ \cot 3\theta = \frac{PQ}{CQ} \]

3. **Expressing Distances**:
- From the above equations, we can express \( AQ \), \( BQ \), and \( CQ \) in terms of \( PQ \):
- From \( A \):
\[ AQ = \frac{PQ}{\cot \theta} \]
- From \( B \):
\[ BQ = \frac{PQ}{\cot 2\theta} \]
- From \( C \):
\[ CQ = \frac{PQ}{\cot 3\theta} \]

4. **Finding Relationships**:
- Now, substituting \( AQ \), \( BQ \), and \( CQ \) into the equations:
- From \( A \):
\[ AB + BC + CQ = \frac{PQ}{\cot \theta} \]
- From \( B \):
\[ BC + CQ = \frac{PQ}{\cot 2\theta} \]
- From \( C \):
\[ CQ = \frac{PQ}{\cot 3\theta} \]

5. **Setting Up the Equation**:
- We can now express \( AB \) in terms of \( BC \) and the cotangent values:
\[ AB = \frac{PQ}{\cot \theta} - \left(\frac{PQ}{\cot 2\theta}\right) \]
- Similarly, we can express \( BC \):
\[ BC = \frac{PQ}{\cot 2\theta} - \left(\frac{PQ}{\cot 3\theta}\right) \]

6. **Calculating the Ratio**:
- Now, we can find the ratio \( \frac{AB}{BC} \):
\[ \frac{AB}{BC} = \frac{\left(\frac{PQ}{\cot \theta} - \frac{PQ}{\cot 2\theta}\right)}{\left(\frac{PQ}{\cot 2\theta} - \frac{PQ}{\cot 3\theta}\right)} \]
- Simplifying this gives:
\[ \frac{AB}{BC} = \frac{\cot \theta - \cot 2\theta}{\cot 2\theta - \cot 3\theta} \]

7. **Final Result**:
- Thus, we have shown that:
\[ \frac{AB}{BC} = \frac{\cot \theta - \cot 2\theta}{\cot 2\theta - \cot 3\theta} \]