Question

From the top light house, the angles of the depression of two ship on the opposite side of it are observed to be $\alpha$ and $\beta$. If the height of the light house be h metres and the line joining the ship passes through the foot of the light house, show that the distance between the ship is $\frac{h(\tan \alpha +\tan \beta )}{\tan \alpha \tan \beta }$ meters.

Answer 1

Let $AB$ be the light house and $P$ and $Q$ be the position of two ships.

$\Rightarrow$ In $△APB,\tan \alpha =\frac{AB}{PB}$

$\Rightarrow$ $\tan \alpha =\frac{h}{PB}$

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

$\Rightarrow$ In $△ABQ,\tan \beta =\frac{AB}{BQ}$

$\Rightarrow$ $\tan \beta =\frac{h}{BQ}$

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

Now the distance between two ships $PQ=PB+BQ$

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

$\Rightarrow$ $PQ=\frac{h\tan \beta +h\tan \alpha }{\tan \alpha \tan \beta }=\frac{h(\tan \beta +\tan \alpha )}{\tan \alpha \tan \beta }m$