From the top of a pole of height 150 m, the angles of depression of another pole’s upper and lower end are `alpha` and `beta` respectively. If `tan alpha = 4//3, tan beta = 5//2`, then the distance of the top of two poles is :

Options

A. 100m
B. 110 m
C. 120 m
D. 130 m

Answer

To solve the problem, we need to find the distance between the tops of two poles given the angles of depression and the height of the first pole. Let's break down the solution step by step.

### Given:
- Height of the first pole (CD) = 150 m
- \( \tan \alpha = \frac{4}{3} \)
- \( \tan \beta = \frac{5}{2} \)

### Step 1: Find the distance from the top of the first pole to the top of the second pole (AC)

1. **Using the angle of depression for the lower end (beta)**:
- In triangle BCE, we have:
\[ \tan \beta = \frac{BE}{CE} \]
- Given \( \tan \beta = \frac{5}{2} \):
\[ \frac{5}{2} = \frac{150}{CE} \]
- Rearranging gives:
\[ CE = \frac{150 \times 2}{5} = 60 \text{ m} \]

### Step 2: Find the distance from the top of the first pole to the top of the second pole (AC) using angle alpha

2. **Using the angle of depression for the upper end (alpha)**:
- In triangle ACE, we have:
\[ \cos \alpha = \frac{CE}{AC} \]
- First, we need to find \( \cos \alpha \) using \( \tan \alpha \):
\[ \tan \alpha = \frac{4}{3} \implies \sec^2 \alpha = 1 + \tan^2 \alpha = 1 + \left(\frac{4}{3}\right)^2 = 1 + \frac{16}{9} = \frac{25}{9} \]
- Therefore, \( \sec \alpha = \frac{5}{3} \) and \( \cos \alpha = \frac{1}{\sec \alpha} = \frac{3}{5} \).

3. **Substituting into the cosine equation**:
\[ \frac{3}{5} = \frac{60}{AC} \]
- Rearranging gives:
\[ AC = \frac{60 \times 5}{3} = 100 \text{ m} \]

### Conclusion:
The distance between the tops of the two poles is **100 m**.

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