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(B) Let the height of the pole be $H = 10\ell$. The pole is divided into two parts of lengths $3\ell$ and $7\ell$.Let the point on the ground be $P$,a

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$12 \sqrt{10}$

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(B) Let the height of the pole be $H = 10\ell$. The pole is divided into two parts of lengths $3\ell$ and $7\ell$.Let the point on the ground be $P$,at a distance of $18 \ m$ from the base of the pole.Let the lower part subtend an angle $\alpha$ at $P$. Then $	an \alpha = \frac{3\ell}{18} = \frac{\ell}{6}$.Let the total height subtend an angle $2\alpha$ at $P$. Then $	an 2\alpha = \frac{10\ell}{18} = \frac{5\ell}{9}$.Using the formula $	an 2\alpha = \frac{2 	an \alpha}{1 - 	an^2 \alpha}$,we have:$\frac{5\ell}{9} = \frac{2(\ell/6)}{1 - (\ell/6)^2} = \frac{\ell/3}{1 - \ell^2/36} = \frac{12\ell}{36 - \ell^2}$.Since $\ell 
eq 0$,we can divide by $\ell$:$\frac{5}{9} = \frac{12}{36 - \ell^2}$ $\Rightarrow 5(36 - \ell^2) = 108$ $\Rightarrow 180 - 5\ell^2 = 108$ $\Rightarrow 5\ell^2 = 72$ $\Rightarrow \ell^2 = \frac{72}{5}$.Thus,$\ell = \sqrt{\frac{72}{5}} = \frac{6\sqrt{2}}{\sqrt{5}} = \frac{6\sqrt{10}}{5}$.The total height of the pole is $10\ell = 10 	imes \frac{6\sqrt{10}}{5} = 12\sqrt{10} \ m$.

$A$ vertical pole fixed to the horizontal ground is divided in the ratio $3:7$ by a mark on it,with the lower part shorter than the upper part. If the two parts subtend equal angles at a point on the ground $18 \ m$ away from the base of the pole,then the height of the pole (in $meters$) is:

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