Question

A tree standing on a horizontal plane is leaning towards east. At two points situated at distance $a$ 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 $\frac{(b-a)\tan \alpha \tan \beta }{\tan \alpha -\tan \beta }$.

Answer 1

Let $AB$ be the tree, $P$ and $Q$ be the points from where it is observed.

Draw $AC$ perpendicular to the ground.

Here, $BP=a,BQ=b$

Let $AC=h$ and $BC=x$

$\Rightarrow$ In $△ACP,\tan \alpha =\frac{AC}{PC}$

$\Rightarrow$ $\tan \alpha =\frac{h}{x+a}$

$\therefore$ $x+a=\frac{h}{\tan \alpha }$

$\therefore$ $x=\frac{h}{\tan \alpha }-a$ -------- ( 1 )

Similarly, in $△ACQ,\tan \beta =\frac{AC}{QC}$

$\Rightarrow$ $\tan \beta =\frac{h}{x+b}$

$\therefore$ $x+b=\frac{h}{\tan \beta }$

$\therefore$ $x=\frac{h}{\tan \beta }-b$ --------- ( 2 )

$\Rightarrow$ $\frac{h}{\tan \alpha }-a=\frac{h}{\tan \beta }-b$ [ Comparing ( 1 ) and ( 2 ) ]

$\therefore$ $\frac{h}{\tan \alpha }-\frac{h}{\tan \beta }=-b+a$

$\therefore h\left(\frac{\tan \beta -\tan \alpha }{\tan \alpha \times \tan \beta }\right)=-(b-a)$

$\therefore$ $h=-(b-a)\times \frac{\tan \alpha \times \tan \beta }{\tan \beta -\tan \alpha }$

$\therefore$ $h=\frac{(b-a)\tan \alpha \times \tan \beta }{\tan \alpha -\tan \beta }$ [ Hence proved ]