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In NUMBAS, to insert the equation $A = BC + \\frac{\\sqrt{D}}{E^2}$, the correct format is as follows: A = B*C + sqrt(D)/E^2. Please make a note of this.

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A cellular system that comprises of hexagonal cells with radius is shown in Figure 1.

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Part A:

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The adjacent side of the right-angled triangle has a length of $i_hd\\sqrt{3}+j_hd\\sqrt{3}\\cos{(60^\\circ)}$ and the opposite side has a length of $j_hd\\sqrt{3}\\sin{(60^\\circ)}$.

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Therefore:

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$U=\\sqrt{(i_hd\\sqrt{3}+j_hd\\sqrt{3}\\cos{(60^\\circ)})^2+(j_hd\\sqrt{3}\\sin{(60^\\circ)})^2}$

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$U=\\sqrt{3i_h^2d^2+3i_hj_hd^2+3j_h^2d^2(\\cos^2{(60^\\circ)}+\\sin^2{(60^\\circ)})}$

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$U=\\sqrt{3}\\sqrt{i_h^2+i_hj_h+j_h^2}$

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Part B:

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If $K=i_h^2+i_hj_h+j_h^2$ and $U=3\\sqrt{d}\\sqrt{i_h^2+i_hj_h+j_h^2}$

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$U=d\\sqrt{3K}$

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$U/d=\\sqrt{3K}$

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Starting from the original cell, the destination cell is reached by moving $i_h$ cells vertically, rotating $60^\\circ$, and then moving $j_h$ cells in this direction

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If the distance between neighbouring cell centres is $d\\sqrt{3}$, derive the distance $U$ between the source and destination cell in terms of $i_h$ and $j_h$.

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Answer here: [[0]]

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\"Figure1\"/

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Given that the number of cells in a cluster, $K$, is defined as $K=i_h^2+i_hj_h+j_h^2$, derive an expression for the co-channel reuse distance $U/d$ in terms of $K$. 

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Answer here: [[0]]

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