Question

As observed from the top of a $75m$ high lighthouse 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 lighthouse, find the distance between the two ships.

Answer 1

Lines PA and BD are parallel and AB is the transversal

$\therefore \angle ABD=\angle PAB$ [Alternate angles]

So,$\angle ABD={30}^o$

Similarly,

Lines PA and BD are parallel and AC is the transversal

$\therefore \angle ACD=\angle PAC$ [Alternate angles]

So,$\angle ABD={45}^o$

Since lighthouse is perpendicular to ground

$\angle ADB={90}^o$

In right angled triangle ACD

$\tan 45=\frac{AD}{CD}$

$1=\frac{75}{CD}$

$CD=75m$

Similarly,

In a right angled triangle ABD

$\tan 30=\frac{AD}{BD}$

$\frac{1}{\sqrt{3}}=\frac{75}{BD}$

$BD=75\sqrt{3}m$

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

$BC+75=75\sqrt{3}$

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

Hence the distance between two ships id $75(\sqrt{3}-1)m$