Question

As observed from the top of a 100 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. [Use `sqrt3` = 1.732]

Answer 1

Let AB be the lighthouse and two ships be at point C and D.

In ΔABC,

tan 45° = `(AB)/(BC)`

⇒ 1 = `100/(BC)` 

⇒ BC = 100 m

In ΔABD,

⇒ tan 30° = `(AB)/(BD)`

⇒ `1/sqrt(3) = 100/(100 + CD)`

⇒ 100 + CD = `100sqrt(3)`

⇒ CD = `100(sqrt(3) - 1)`

⇒ CD = 100(1.732 – 1)

⇒ CD = 100 × 0.732

⇒ CD = 73.2 m

Therefore, the distance between the two ships is 73.2 m.