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 the lighthouse be represented by AB, with a height of 75 m. Let the two ships be located at points C and D, respectively.

Considering the triangle ABC, where the angle of depression is 45°:

\(\frac{AB}{ BC} = tan 45^{\degree}\)

\(\frac{75}{ BC} = 1\)

\(BC = 75\,m\)

Considering the triangle ABD, where the angle of depression is 30°:

\(\frac{AB}{ BD}= tan 30^{\degree}\)

\(\frac{75}{ BC +CD} = \frac{1}{\sqrt3}\)

\(\frac{75}{ 75 + CD} = \frac1{ \sqrt3}\)

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

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