Question

As observed from the top of a 75 m high lighthouse from the sea-level, 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,

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

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

$BC = 75\,m$

In ∆ABD,

$\frac{AB}{ BD}= tan 60^{\degree}$

$\frac{75}{ BC +CD} = \frac{1}{\sqrt3}$

$\frac{75}{ 75 + CD} = \frac1{ \sqrt3}$

$75 \sqrt3 = 75 + CD$
$75 (\sqrt3 -1)m = CD$

Therefore, the distance between the two ships is $75(\sqrt3 -1) \,m$.