Question

Two boats approached a lighthouse in the midsea from opposite directions the angle of elevation of the top of the lighthouse are 30 and 45 degree respectively if the distance between two boats is 100m find the highest of the lighthouse

Answer 1

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

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

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

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

In

$\Delta BDC,\tan {45}^{\circ }=1=\frac{DC}{BD}$

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

$\Rightarrow DC=100-x----\left(ii\right)$

From equation

$\left(i\right)$

and

$\left(ii\right)$

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

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

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

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

And, we know that,

$\sqrt{3}=1.732$ ,

$\Rightarrow x=\frac{100\times 1.732}{1.732+1}$

$\Rightarrow x=36.60m$

Hence, the height of the house is $36.60m$.