Question

Two boats approach a light house in mid-sea from opposite direction. The angle of elevation of the top of the lighthouse from two boats is ${30}^{\circ }$ and ${45}^{\circ }$respectively. If the distance between two boats is 100m, find the height of the light house.

Answer 1

Let $AP=x$ then $BP=100-x$

In $△ADC,\tan {30}^o=\frac{1}{\sqrt{3}}=\frac{DC}{AD}$

$\frac{1}{\sqrt{3}}=\frac{DC}{x}$

$\Rightarrow \frac{x}{\sqrt{3}}=DC---\left(1\right)$

In $△BDC,\tan {45}^o=1\frac{DC}{BD}$

$1=\frac{DC}{100-x}$

$\Rightarrow 100-xDC----\left(2\right)$

Equation $\left(1\right)$ & $\left(2\right)$

$100-x=\frac{x}{\sqrt{3}}$

$100\sqrt{3}=x\sqrt{3}=x$

$100\sqrt{3}=x(\sqrt{3}+1)$

$\frac{100\sqrt{3}}{\sqrt{3}+1}=x$

Then Height of house $=DC=100-\frac{100\sqrt{3}}{\sqrt{3}+1}=36.60m$