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(C) Let the heights of the two poles be $h_1 = 5 \, m$ and $h_2 = 10 \, m$. Let the distance between them be $x$. By drawing a horizontal line from th

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$5 \, (2 + \sqrt{3})$

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(C) Let the heights of the two poles be $h_1 = 5 \, m$ and $h_2 = 10 \, m$. Let the distance between them be $x$. By drawing a horizontal line from the top of the shorter pole to the taller pole,we form a right-angled triangle with a vertical side of length $h_2 - h_1 = 10 - 5 = 5 \, m$. The angle of elevation from the top of the shorter pole to the top of the taller pole is $15^o$. In the right-angled triangle,we have: $	an(15^o) = \frac{	ext{opposite}}{	ext{adjacent}} = \frac{5}{x}$ Since $	an(15^o) = 2 - \sqrt{3}$,we have: $2 - \sqrt{3} = \frac{5}{x}$ $x = \frac{5}{2 - \sqrt{3}}$ Rationalizing the denominator: $x = \frac{5(2 + \sqrt{3})}{(2 - \sqrt{3})(2 + \sqrt{3})} = \frac{5(2 + \sqrt{3})}{4 - 3} = 5(2 + \sqrt{3}) \, m$.

Two poles standing on a horizontal ground are of heights $5 \, m$ and $10 \, m$ respectively. The line joining their tops makes an angle of $15^o$ with the ground. Then the distance (in $m$) between the poles is:

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