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(C) The range of $LOS$ communication is given by $d = \sqrt{2Rh_T} + \sqrt{2Rh_R}$,where $h_T$ and $h_R$ are the heights of the transmitting and recei

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$64$

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(C) The range of $LOS$ communication is given by $d = \sqrt{2Rh_T} + \sqrt{2Rh_R}$,where $h_T$ and $h_R$ are the heights of the transmitting and receiving antennas respectively.Given $h_T + h_R = 160 \, m$. Let $h_T = x$,then $h_R = 160 - x$.$d = \sqrt{2R}(\sqrt{h_T} + \sqrt{h_R}) = \sqrt{2R}(\sqrt{x} + \sqrt{160 - x})$.To maximize $d$,we differentiate with respect to $x$ and set it to zero: $\frac{d}{dx}(\sqrt{x} + \sqrt{160 - x}) = 0$.$\frac{1}{2\sqrt{x}} - \frac{1}{2\sqrt{160 - x}} = 0 \implies \sqrt{x} = \sqrt{160 - x} \implies x = 80 \, m$.Thus,$h_T = 80 \, m$ and $h_R = 80 \, m$.Converting heights to km: $h_T = h_R = 0.08 \, km$.$d_{max} = \sqrt{2 	imes 6400 	imes 0.08} + \sqrt{2 	imes 6400 	imes 0.08} = 2 	imes \sqrt{12800 	imes 0.08} = 2 	imes \sqrt{1024} = 2 	imes 32 = 64 \, km$.

If the sum of the heights of transmitting and receiving antennas in the line of sight $(LOS)$ communication is fixed at $160 \, m$,then the maximum range of $LOS$ communication is $.... \, km$. (Take radius of Earth $R = 6400 \, km$)

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