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A vertical tower stands on a horizontal plane and is surmounted by a vertical flag staff of height $h$. At a point $P$ on the plane, the angle of elevation of the bottom of the flag staff is $\beta$ and that of the top is $\alpha$. What is the height of the tower?
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The correct answer is:
$\frac{h \tan \beta}{\tan \alpha-\tan \beta}$
Let $\mathrm{BC}$ be the vertical tower and $\mathrm{CD}$ be the flagstaff so that $\mathrm{CD}=h$
Let $\mathrm{P}$ be the point of observation on the plane. Then, $\angle \mathrm{BPC}=\beta$ and $\angle \mathrm{BPD}=\alpha$
Let $\mathrm{BC}=x$
Now, $\frac{\mathrm{PB}}{x}=\cot \beta \Rightarrow \mathrm{PB}=x \cot \beta$
and $\frac{\mathrm{PB}}{x+h}=\cot \alpha \Rightarrow \mathrm{PB}=(x+h) \cot \alpha$
From (1) and (2), we get $x \cot \beta=(\mathrm{x}+\mathrm{h}) \cot \alpha \Rightarrow x(\cot \beta-\cot \alpha)=h \cot \alpha$
$\therefore$ Height of the tower,
$\mathrm{x}=\frac{h \cot \alpha}{\cot \beta-\cot \alpha}=\frac{h \tan \beta}{\tan \alpha-\tan \beta}$
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