The height of a transmitting antenna at the top of a tower is $$25$$ m and that of receiving antenna is, $$49$$ m. The maximum distance between them, for satisfactory communication in LOS (Line-Of-Sight) is $$K\sqrt{5} \times 10^2$$ m. The value of $$K$$ is ______.
(Assume radius of Earth is $$64 \times 10^5$$ m) [Calculate upto nearest integer value]

The height of the transmitting antenna is $$h_T = 25$$ m and the receiving antenna is $$h_R = 49$$ m, while the radius of Earth is $$R = 64 \times 10^5$$ m. Using the LOS communication distance formula, the maximum line-of-sight distance between the two antennas is: $$d = \sqrt{2Rh_T} + \sqrt{2Rh_R}$$

First, we calculate $$\sqrt{2Rh_T}$$. Substituting gives $$2Rh_T = 2 \times 64 \times 10^5 \times 25 = 3200 \times 10^5 = 32 \times 10^7$$, and then $$\sqrt{32 \times 10^7} = \sqrt{32} \times \sqrt{10^7} = 4\sqrt{2} \times 10^3 \times \sqrt{10} = 4\sqrt{20} \times 10^3 = 4 \times 2\sqrt{5} \times 10^3 = 8\sqrt{5} \times 10^3 \text{ m} = 80\sqrt{5} \times 10^2 \text{ m}$$.

Next, we calculate $$\sqrt{2Rh_R}$$. We have $$2Rh_R = 2 \times 64 \times 10^5 \times 49 = 6272 \times 10^5$$, so $$\sqrt{6272 \times 10^5} = \sqrt{6272} \times \sqrt{10^5} = \sqrt{6272} \times 10^2 \times \sqrt{10}$$. Simplifying $$\sqrt{6272}$$ by factoring yields $$6272 = 4 \times 1568 = 4 \times 4 \times 392 = 16 \times 392 = 16 \times 4 \times 98 = 64 \times 98$$ and hence $$\sqrt{6272} = 8\sqrt{98} = 8 \times 7\sqrt{2} = 56\sqrt{2}$$. Therefore, $$\sqrt{2Rh_R} = 56\sqrt{2} \times 10^2 \times \sqrt{10} = 56\sqrt{20} \times 10^2 = 56 \times 2\sqrt{5} \times 10^2 = 112\sqrt{5} \times 10^2 \text{ m}$$.

Finally, combining these results gives $$d = 80\sqrt{5} \times 10^2 + 112\sqrt{5} \times 10^2 = 192\sqrt{5} \times 10^2 \text{ m}$$. Comparing with the given form $$K\sqrt{5} \times 10^2$$ m, we identify $$K = 192$$.

The value of K is 192.