Question

PQ is a post of given height , and AB is a tower at some distance. If $\alpha$ and $\beta$ are the angles of elevation from the top of the tower, at P and Q respectively. Find the height of the tower and its distance from the post.

Answer 1

$PQ$ is the post of height, $a$ and $AB$ is the tower,

Let the height of the tower be $h$.

$PQ=AR=a$
$\therefore$ $BR=AB-AR=h-a$

In Right angled $△BAP$,

$\tan \alpha =\frac{AB}{AP}$

$\therefore$ $\tan \alpha =\frac{h}{AP}$

$\Rightarrow$ $AP=\frac{h}{\tan \alpha }$ ------ ( 1 )

In right angled $△BRQ,$

$\tan \beta =\frac{BR}{QR}$

$\therefore$ $\tan \beta =\frac{h-a}{AP}$ [ Since, $AP=RQ$ ]

$\Rightarrow$ $AP=\frac{h-a}{\tan \beta }$ ------- ( 2 )

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

$\Rightarrow$ $h\tan \alpha -a\tan \alpha =h\tan \beta$

$\Rightarrow$ $h(\tan \alpha -\tan \beta )=a\tan \alpha$

$\Rightarrow$ $h=\frac{a\tan \alpha }{\tan \alpha -\tan \beta }$

$\therefore$ Height of the tower $=\frac{a\tan \alpha }{\tan \alpha -\tan \beta }$

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

$\therefore$ $AP=\frac{a}{\tan \alpha -\tan \beta }$

$\therefore$ Distance of the tower from post is $\frac{a}{\tan \alpha -\tan \beta }$