Two boats approach a light house in mid-sea from opposite directions. The angles of elevation of the top of the light house from two boats are 30° and 45° respectively. If the distance between two boats is 100 m, find the height of the light house.
Let
be the height of light house
. Angle of elevation of the top of light house from two boats are 30° and 45°. Let
,
and it is given that
m. So
. And
,![]()
Here we have to find height of light house.
The corresponding figure is as follows

So we use trigonometric ratios.
In
,
$\Rightarrow \quad \tan 45^{\circ}=\frac{B D}{B C}$
$\Rightarrow \quad 1=\frac{h}{x}$
$\Rightarrow \quad x=h$
Again in ![]()
$\Rightarrow \quad \tan 30^{\circ}=\frac{D B}{A B}$
$\Rightarrow \quad \frac{1}{\sqrt{3}}=\frac{h}{100-x}$
$\Rightarrow \quad \sqrt{3} h=100-x$
$\Rightarrow \quad \sqrt{3 h}=100-h$
$\Rightarrow(\sqrt{3}+1) h=100$
$\Rightarrow \quad h=\frac{100}{\sqrt{3}+1} \times \frac{\sqrt{3}-1}{\sqrt{3}-1}$
$\Rightarrow \quad h=50(\sqrt{3}-1)$
Hence the height of light house is $50(\sqrt{3}-1) \mathrm{m}$.