Question

As observed from the top of a 75 m tall lighthouse, the angle of depression of two ships is 30° and 45°. If one ship is exactly behind the other on the same side of the lighthouse, then the distance between the ships is
(a) $75(\sqrt{3}+1)\mathrm{m}$
(b) $75(\sqrt{3}-1)\mathrm{m}$
(c) $25(\sqrt{3}+1)\mathrm{m}$
(d) $25(\sqrt{3}-1)\mathrm{m}$

Answer 1

(b) $75(\sqrt{3}-1)\mathrm{m}$
Let $OB$ be the lighthouse. From O, the angle of depression is observed.
We have:
∠$OAB={45}^o$, ∠$OCB={30}^o$ and $OB=75$ m
Let $BC=y$ m and $AB=x$ m such that $AC=(BC-AB)=(y-x)$ m.

In ∆$OBA$, we have:

$\frac{OB}{AB}=\tan {45}^o=1$

⇒$\frac{75}{x}=1\Rightarrow x=75$ m

In ∆$OBC$, we have:

$\frac{OB}{BC}=\tan {30}^o=\frac{1}{\sqrt{3}}$

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