The top of a tower, the angle of an object are on the same side of the tower end to be $x$ and $b(x>b)$. If the distance objects is 'h' meters, show that the height of 'h' is given by: 

Let $AB$ be the tower of height $h$ meters and C and D are two objects which are$y$ meter apart

Then,

$AB=hmeter$

$BC=xmeter$

$CD=ymeter$

Then,

$In\Delta ABC,$

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

$\tan \alpha =\frac{h}{x}$

$x=\frac{h}{\tan \alpha }.......\left(1\right)$

Now, $In\Delta ABD$$\tan \beta =\frac{h}{x+y}$

$h=\left(x+y\right)\tan \beta$

$h=x\tan \beta +y\tan \beta .......\left(2\right)$

By equation (1) and we get,

$h=y\tan \beta +\frac{h\tan \beta }{\tan \alpha }$

$h-\frac{h\tan \beta }{\tan \alpha }=y\tan \beta$

$h\left(1-\frac{\tan \beta }{\tan \alpha }\right)=y\tan \beta$

$h\left(\frac{\tan \alpha -\tan \beta }{\tan \alpha }\right)=y\tan \beta$

$h=\frac{y\tan \alpha \tan \beta }{\tan \alpha -\tan \beta }$

$h=\frac{CD\tan \alpha \tan \beta }{\tan \alpha -\tan \beta }$