#### A vertical tower stands on a horizontal plane and is surmounted by a vertical flagstaff of height h. At a point on the plane, the angles of elevation of the bottom and the top of the flagstaff are α and β, respectively. Prove that the height of the tower is$\left ( \frac{h\tan \alpha }{\tan \beta -\tan \alpha } \right )$

According to question

Here h is the height of flagstaff AD.
Let $\l$ is the height of the tower
$\alpha$ and $\beta$ be the angle of elevation of the bottom and the top of the flagstaff.
In $\bigtriangleup$BDC
$\tan \alpha = \frac{\l }{BC}$  $\left [ \because \tan \theta = \frac{Perpendicular}{Base} \right ]$
$BC= \frac{\l }{\tan \alpha }\cdots \left ( 1 \right )$
In $\bigtriangleup$ABC
$\tan \beta = \frac{AB}{BC}$
$\tan \beta = \frac{h+\l }{BC}$
$BC= \frac{h+\l }{\tan \beta }\cdots \left ( 2 \right )$
equate equation (1) and (2) we get
$\frac{\l }{\tan \alpha }=\frac{h+\l }{\tan \beta }$
$\l \tan \beta = h\tan \alpha +\l \tan \alpha$ [by cross multiplication]
$\l \tan \beta -\l \tan \alpha = h\tan \alpha$
$\l \left ( \l \tan \beta - \tan \alpha \right ) = h\tan \alpha$
$\l = \frac{h\tan \alpha }{ \tan \beta - \tan \alpha}$
Hence Proved