Question

As observed form the top of a 75-m-tall lighthouse, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.

Answer 1

Let $AB$ be the lighthouse such that $AB=75$m and $C$and $D$ be the positions of the two ships.
Thus, we have:
∠$ACB={45}^o$ and ∠$ADB={30}^o$
Let $BD=x$ m and $BC=y$ m such that $CD=(x-y)$ m.
In the right ∆​$ABC$, we have:
$\frac{AB}{BC}=\tan {45}^o=1$

⇒ $\frac{75}{y}=1$
⇒ $y=75$ m

In the right ∆$ABD$, we have:
$\frac{AB}{BD}=\tan {30}^o=\frac{1}{\sqrt{3}}$

⇒ $\frac{75}{x}=\frac{1}{\sqrt{3}}$
= $x=75\sqrt{3}$ m

∴ Distance between the two ships = $CD=(x-y)=(75\sqrt{3}-75)=75(\sqrt{3}-1)=54.90$ m