Question

A ship of height $24m$ is sighted from a lighthouse. From the top of the lighthouse, the angles of depression to the top of the mast and base of the ship are ${30}^o$ and ${45}^o$ respectively. How far is the ship from the lighthouse? $(\sqrt{3}=1.73)$

Answer 1

Let $AB$ be the lighthouse and $CD$ be the ship. Let the height of the lighthouse be '$h$' and let the distance from the ship of the lighthouse is '$x$' $m$. The height of the ship is $24m$. The angle of depression to the top of the mast is ${30}^o$ and the angle of depression to the base of the ship is ${45}^o$.
$CE\perp AB$.
In $\Delta ABD$,
$\tan {45}^o=\frac{h}{x}$
$1=\frac{h}{x}$
$\therefore h=x$ ....(i)
In $\Delta AEC$,
$\tan {30}^o=\frac{AE}{EC}=\frac{h-24}{x}$
$\frac{1}{\sqrt{3}}=\frac{h-24}{x}$
$h-24=\frac{x}{\sqrt{3}}$ ...(ii)
From equations (i) and (ii), we get
$x-24=\frac{x}{\sqrt{3}}$
$x-\frac{x}{\sqrt{3}}=24$
$(\sqrt{3}-1)x=24\sqrt{3}$
$x=\frac{24\sqrt{3}}{\sqrt{3}-1}$
$=\frac{24\sqrt{3}(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}$
$=\frac{24(3+\sqrt{3})}{3-1}$
$=12(3+\sqrt{3})$
$=12(3+1.73)$
$=12\times 4.73$
$=56.76$
$\therefore$ The ship is at a distance of $56.76m$ from the lighthouse.