A bridge across a valley is h metres lomg. There is a temple in the valley diresctly below the bridge. The angles of depression of the temple from the two ends of the bridge  have measures $\alpha and\beta .$ Prove that the height of the bridge above the top of the temple is $\frac{h(\tan \alpha .\tan \beta )}{\tan \alpha +\tan \beta }m.$

We have

Let $AP=\text{h}\text{mtr}.$

Makes an angle $\alpha$ and $\beta$ from point A and P

And meet the down point D.

Then makes triangle $\Delta APD$

According to the question,

$\angle PAD=\alpha$

$\angle APD=\beta$

Draw a line

$AB=PQ=\text{y}\text{mtr}\text{.}$

Then,

$In\Delta DAB$

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

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

$In\Delta DPQ$

$\tan \beta =\frac{PQ}{DQ}$

$\tan \beta =\frac{y}{h-x}.......\left(2\right)$

Dividing equation (2) by (1) and we get,

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

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

$h\tan \beta -xtan\beta =xtan\alpha$

$h\tan \beta =xtan\alpha +xtan\beta$

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

$x=\frac{h\tan \beta }{\left(\tan \alpha +\tan \beta \right)}$

Put the value of x in equation (1) and we get,

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