Question

As observed from the top of a $100m$ 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, height of light house from sea level $\left(AB\right)=100m$

Let, two ships be at the positions be $C$ and $D$

In $△ABC$,

$tan{45}^{\circ }=\frac{AB}{BC}$

$⟹1=\frac{100}{BC}$

$⟹BC=100m$

In $△ABD$,

$tan{30}^{\circ }=\frac{AB}{BD}$

$⟹\frac{1}{\sqrt{3}}=\frac{AB}{BD}$

$BD=AB\times \sqrt{3}$

$=100\times \sqrt{3}$

$=173.2[\sqrt{3}=1.732]$

$\therefore CD=BD-BC=173.2=100=73.2$m

$\therefore$ the distance between two ships $73.2$ m