A transmitting antenna at top of a tower has a height of 50 m, and the height of receiving antenna is 80 m. What is the range of communication for the line of sight (LOS) mode?
[use radius of the earth = 6400 km]

The line-of-sight (LOS) range between two antennas is obtained from the well-known empirical relation

$$d \;=\;\sqrt{2\,R\,h_t}\;+\;\sqrt{2\,R\,h_r}$$

where

$$R = 6.4 \times 10^6 \text{ m}$$ is the radius of the Earth,

$$h_t = 50 \text{ m}$$ is the height of the transmitting antenna, and

$$h_r = 80 \text{ m}$$ is the height of the receiving antenna.

We first evaluate the contribution from the transmitting antenna:

$$ \sqrt{2\,R\,h_t} \;=\;\sqrt{2 \times (6.4 \times 10^6)\times 50}. $$

Inside the square root we have

$$ 2 \times 6.4 \times 50 = 12.8 \times 50 = 640, $$

so

$$ 2\,R\,h_t = 640 \times 10^6 = 6.4 \times 10^8. $$

Taking the square root:

$$ \sqrt{6.4 \times 10^8} = \sqrt{6.4}\;\times\;\sqrt{10^8} = 2.5298 \times 10^4 \text{ m} \approx 2.53 \times 10^4 \text{ m}. $$

In kilometres this is

$$ 2.53 \times 10^4 \text{ m} = 25.3 \text{ km}. $$

Next we evaluate the contribution from the receiving antenna:

$$ \sqrt{2\,R\,h_r} \;=\;\sqrt{2 \times (6.4 \times 10^6)\times 80}. $$

Now

$$ 2 \times 6.4 \times 80 = 12.8 \times 80 = 1024, $$

hence

$$ 2\,R\,h_r = 1024 \times 10^6 = 1.024 \times 10^9. $$

Taking the square root:

$$ \sqrt{1.024 \times 10^9} = \sqrt{1.024}\;\times\;\sqrt{10^9} = 1.012 \times 10^{4.5} = 1.012 \times 31622.8 \text{ m} \approx 3.20 \times 10^4 \text{ m}. $$

Thus, in kilometres,

$$ 3.20 \times 10^4 \text{ m} = 32.0 \text{ km}. $$

Adding the two distances gives the LOS range:

$$ d = 25.3 \text{ km} + 32.0 \text{ km} = 57.3 \text{ km}. $$

The option closest to this calculated value is 57.28 km.

Hence, the correct answer is Option C.