Question

As observed from the top of a 100 m high light house from the sea-level, the angles of depression of two ships are ${30}^o$ and ${45}^o$. If one ship is exactly behind the other on the same side of the light house, find the distance between the two ships. [Use $\sqrt{3}=1.732$]

Answer 1

Let AB be the light house and two ships be at C and D.

In $\Delta ABC,$
$\frac{BC}{AB}=cot{45}^o$
$\Rightarrow \frac{x}{100}=1$
$\Rightarrow x=100$ ........ (i)
Similarly, in $\Delta ABD$,
$\frac{BD}{AB}=cot{30}^o$
$\Rightarrow \frac{y}{100}=\sqrt{3}$
$\Rightarrow y=100\sqrt{3}$ ....... (ii)
Distance between two ships = y - x
= 100 $\sqrt{3}-100$ [from (i) and (11)]
$=100(\sqrt{3}-1)$
$=100(1.732-1)$
$=100\left(0.732\right)$
$=73.2$ m