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 thatm. 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 $50(\sqrt{3}-1) \mathrm{m}$.