A tree standing on a horizontal plane is leaning towards east. At two points situated at distances $ a $ and $ b $ exactly due west on it, the angles of elevation of the top are respectively $ \alpha $ and $ \beta $. Prove that the height of the top from the ground is $ \frac{(b-a) \tan \alpha \tan \beta}{\tan \alpha-\tan \beta} $
Given:
A tree standing on a horizontal plane is leaning towards east. At two points situated at distances \( a \) and \( b \) exactly due west on it, the angles of elevation of the top are respectively \( \alpha \) and \( \beta \).
To do:
We have to prove that the height of the top from the ground is \( \frac{(b-a) \tan \alpha \tan \beta}{\tan \alpha-\tan \beta} \).
Solution:
Let $AB$ be the tree and $C$ and $D$ are two points situated at distances \( a \) and \( b \) exactly due west on it.
Draw a perpendicular to the ground from the top of the tree.
From the figure,
$\angle BCE=\alpha, \angle BDE=\beta$
Let the height of the perpendicular be $h$, the distance between $A$ and $E$ be $x$.
In right $\Delta \mathrm{BCE}$,
$\tan \alpha=\frac{BE}{CE}$
$=\frac{h}{x+a}$
$x+a=\frac{h}{\tan \alpha}$
$x=\frac{h}{\tan \alpha}-a$..........(i)
Similarly,
In right $\Delta \mathrm{BDE}$,
$\tan \beta=\frac{BE}{DE}$
$=\frac{h}{x+b}$
$h=(x+b)\tan \beta$
$x=\frac{h}{\tan \beta}-b$............(ii)
From (i) and (ii), we get,
$\frac{h}{\tan \alpha}-a=\frac{h}{\tan \beta}-b$
$\Rightarrow h(\frac{1}{\tan \alpha}-\frac{1}{\tan \beta})=a-b$
$h(\frac{\tan \beta-\tan \alpha}{\tan \alpha \tan\ \beta})=a-b$
$h=\frac{(b-a) \tan \alpha \tan \beta}{\tan \alpha-\tan \beta}$
Hence proved.
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