Question

As observed form the top of a lighthouse, 100 m above sea level, the angle of depression of a ship, sailing directly towards it, changes from 30° to 60°. Determine the distance travelled by the ship during the period of observation.

Answer 1

Let $OA$ be the lighthouse and B and C be the two positions of the ship.
Thus, we have:
$OA\hspace{0.17em}=100$ m, ∠$OBA\hspace{0.17em}={30}^o$ and ∠$OCA={60}^o$



Let:
$OC=x$ m and $BC\hspace{0.17em}=y$ m
In the right ∆$OAC$, we have:
$\frac{OA}{OC}=\tan {60}^o=\sqrt{3}$

⇒$\frac{100}{x}=\sqrt{3}$
⇒ $x=\frac{100}{\sqrt{3}}$m
Now, in the right ∆$OBA$, we have:
$\frac{OA}{OB}=\tan {30}^o=\frac{1}{\sqrt{3}}$

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

On putting $x=\frac{100}{\sqrt{3}}$ in the above equation, we get:
$y=100\sqrt{3}-\frac{100}{\sqrt{3}}=\frac{300-100}{\sqrt{3}}=\frac{200}{\sqrt{3}}=115.47$ m

∴ Distance travelled by the ship during the period of observation = $BC=y=115.47$ m