A transmitting antenna at the top of the tower has a height 18 m and the height of the receiving antenna is 32m. The maximum distance between them for satisfactory communciation in line of sight mode is (Radius of earth`= 6.4xx10^(6)m`)

Options

A. 15.15km
B. 21.25km
C. 30.45km
D. 35.42km.

Explanation

To solve the problem of finding the maximum distance for satisfactory communication in line of sight mode between a transmitting antenna and a receiving antenna, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Height of the transmitting antenna, \( H_T = 18 \, \text{m} \)
- Height of the receiving antenna, \( H_R = 32 \, \text{m} \)
- Radius of the Earth, \( R_E = 6.4 \times 10^6 \, \text{m} \)

2. **Use the Formula for Maximum Distance**:
The formula for the maximum distance \( D_{max} \) in line of sight communication is given by:
\[ D_{max} = \sqrt{2 H_T R_E} + \sqrt{2 H_R R_E} \]

3. **Calculate Each Term**:
- Calculate \( \sqrt{2 H_T R_E} \):
\[ \sqrt{2 H_T R_E} = \sqrt{2 \times 18 \times 6.4 \times 10^6} \]
\[ = \sqrt{230.4 \times 10^6} = \sqrt{230.4} \times 10^3 \approx 15.2 \times 10^3 \, \text{m} \]

- Calculate \( \sqrt{2 H_R R_E} \):
\[ \sqrt{2 H_R R_E} = \sqrt{2 \times 32 \times 6.4 \times 10^6} \]
\[ = \sqrt{409.6 \times 10^6} = \sqrt{409.6} \times 10^3 \approx 20.2 \times 10^3 \, \text{m} \]

4. **Add the Two Distances**:
Now, add the two distances calculated:
\[ D_{max} = \sqrt{2 H_T R_E} + \sqrt{2 H_R R_E} \approx (15.2 \times 10^3 + 20.2 \times 10^3) \, \text{m} \]
\[ = 35.4 \times 10^3 \, \text{m} \]

5. **Convert to Kilometers**:
To convert meters to kilometers:
\[ D_{max} = 35.4 \, \text{km} \]

6. **Final Result**:
The maximum distance for satisfactory communication in line of sight mode is:
\[ D_{max} = 35.4 \, \text{km} \]