Question

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

Answer 1

Let $AB=24m$ be the height of the ship and $OC$ be the height of the lighthouse

Let the height of the lighthouse, from above the ship be $y$ and the distance between the bases of the lighthouse and the ship be $x$

In $△OAD,$

$\Rightarrow \tan {30}^{\circ }=\frac{y}{x}$

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

$\Rightarrow y=\frac{x}{\sqrt{3}}$

In $△OBC$

$\Rightarrow \tan {45}^o=\frac{y+24}{x}$

$\Rightarrow x=y+24$

$\Rightarrow x=\frac{x}{\sqrt{3}}+24$

$\Rightarrow x(1-\frac{1}{\sqrt{3}})=24$

$\Rightarrow x=\frac{24\sqrt{3}}{\sqrt{3}-1}$

$\Rightarrow x=\frac{24\times 1.73}{0.73}$

$\Rightarrow x=56.88m$