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^{\circ} $ and $ 45^{\circ} $ respectively. If the distance between two boats is $ 100 \mathrm{~m} $, find the height of the light house.

Given:

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 \( 30^{\circ} \) and \( 45^{\circ} \) respectively.

The distance between two boats is \( 100 \mathrm{~m} \).

To do:

We have to find the height of the light house.

Solution:

 Let $C$ and $D$ be two boats and $AB$ be the light house whose height is $h\ m$.

Let the distance between the boat $C$ and the lighthouse be $x$ meters, then the distance of the other boat from the light house is $(100-x)\ m$. 

In right-angled $\vartriangle BAD$, we have 

$tan\ 45^{o}=\frac{AB}{AD}$

$\Rightarrow 1=\frac{h}{100-x}$

$\Rightarrow 100-x=h\ m$

$\Rightarrow x=100-h\ m$......(i)

In right-angled $\vartriangle BAC$,

$tan\ 30^{o}=\frac{AB}{AC}$

$\Rightarrow \frac{1}{\sqrt{3}} =\frac{h}{100-h}$                   [From (i)]

$\Rightarrow 100-h=h\sqrt3$

$h(\sqrt3+1)=100$

$\Rightarrow h=\frac{100}{\sqrt3+1}$

$\Rightarrow h=\frac{100(\sqrt{3}-1)}{(\sqrt{3}+1)(\sqrt3-1)}$

$\Rightarrow h=\frac{100(\sqrt3-1)}{3-1)}$

$\Rightarrow h=50(\sqrt3-1)$

Therefore, the height of the light house is $50(\sqrt3-1)\ m$ approximately. 

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Updated on: 10-Oct-2022

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