Question:
in a line of sight ratio communication, a distance of about $50 \mathrm{~km}$ is kept between the transmitting and receiving antennas. If the height of the receiving antenna is $70 \mathrm{~m}$, then the minimum height of the transmitting antenna should be :
(Radius of the Earth $=6.4 \times 10^{6} \mathrm{~m}$ ).
Correct Option: , 3
Solution:
(3) $\mathrm{LOS}=\sqrt{2 h_{T} R}+\sqrt{2 h_{R} R}$
or $50 \times 10^{3}=\sqrt{2 h_{T} \times 6.4 \times 10^{6}}+\sqrt{2 \times 70 \times 6.4 \times 40^{6}}$
On solving, $h_{T}=32 \mathrm{~m}$