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Two poles are $10 \mathrm{~m}$ and $20 \mathrm{~m}$ high. The line joining their tips makes an angle of $15^{\circ}$ with the horizontal. What is the distance between the poles? $\quad$
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The correct answer is:
$5(4+2 \sqrt{3}) \mathrm{m}$
Let $\mathrm{AB}$ and $\mathrm{CD}$ be two poles of height $10 \mathrm{~m}$ and $20 \mathrm{~m}$ respectively. $\operatorname{In} \Delta \mathrm{AEC}$
$\frac{\mathrm{CE}}{\mathrm{AE}}=\tan 15^{\circ}$
$\frac{10}{\mathrm{AE}}=\tan \left(45^{\circ}-30^{\circ}\right)$
$=\frac{\tan 45^{\circ}-\tan 30^{\circ}}{1+\tan 45^{\circ} \tan 30^{\circ}}$
$=\frac{1-\frac{1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}}$
$\frac{10}{\mathrm{AE}}=\frac{\sqrt{3}-1}{\sqrt{3}+1}$
$\mathrm{AE}=\frac{10(\sqrt{3}+1)}{\sqrt{3}-1} \times \frac{\sqrt{3}+1}{\sqrt{3}+1}=5(4+2 \sqrt{3})$
$\frac{\mathrm{CE}}{\mathrm{AE}}=\tan 15^{\circ}$
$\frac{10}{\mathrm{AE}}=\tan \left(45^{\circ}-30^{\circ}\right)$
$=\frac{\tan 45^{\circ}-\tan 30^{\circ}}{1+\tan 45^{\circ} \tan 30^{\circ}}$
$=\frac{1-\frac{1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}}$
$\frac{10}{\mathrm{AE}}=\frac{\sqrt{3}-1}{\sqrt{3}+1}$
$\mathrm{AE}=\frac{10(\sqrt{3}+1)}{\sqrt{3}-1} \times \frac{\sqrt{3}+1}{\sqrt{3}+1}=5(4+2 \sqrt{3})$
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