Question

From the top of lighthouse, an observer looks at a ship and finds the angle of depression to be ${60}^{\circ }$. If the height of the lighthouse is $90$ metres, then find how far is that ship from the lighthouse? $(\sqrt{3}=\mathrm{1.7.3})$

Answer 1

Let $AB=90m$ represents the height of lighthouse. Point $C$ represents the position of the ship.
Draw ray $AM\parallel$ seg $BC$
$\angle ACB$ is the angle of depression,
$\angle MAC={60}^o$
Also, $\angle ACB\simeq \angle MAC$ (Alernate angles)
$\angle ACB={60}^o$
Let the distance of ship from the light-house be $x$ metres.
In right angled $\Delta ABC$,
$\tan {60}^o=\frac{90}{x}$
$\Rightarrow x=\frac{90}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}$
$=\frac{90\sqrt{3}}{3}$
$=30\sqrt{3}$
$=30\times 1.73=51.9$
$\therefore$ Distance of the ship from the lighthouse is $51.9m.$