Question

As observed from 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 and the two ships be at point $C$ and $D$ respectively.

In $△ABC,$

$\Rightarrow$ $\tan {45}^o=\frac{AB}{BC}$

$\Rightarrow$ $1=\frac{75}{BC}$

$\therefore$ $BC=75m$

In $△ABD$,

$\Rightarrow$ $\tan {30}^o=\frac{AB}{BD}$

$\Rightarrow$ $\frac{1}{\sqrt{3}}=\frac{75}{BC+CD}$

$\Rightarrow$ $\frac{1}{\sqrt{3}}=\frac{75}{75+CD}$

$\Rightarrow$ $75\sqrt{3}=75+CD$

$\therefore$ $CD=75(\sqrt{3}-1)m$

$\therefore$ $CD=54.75m$ $(Take\sqrt{3}=1.73)$

$\therefore$ The distance between the two ships is $54.75m$.