Question

From a light house the angles of depression of two ships on opposite sides of the light house are observed to be 30° and 45°. If the height of the light house is h metres, the distance between the ships is

A. \((\sqrt3+1)\) h metres

B.   \((\sqrt3-1)\) h metres

C.   \(\sqrt3\) h metres

D.1 + \((1+\frac{1}{\sqrt3})\) h metres

Answer 1

Here AB is the light house of height h meters 

The angles of depression from the light house are 30° and 45° 

The distance between the two ships is CD 

In Δ ABC 

tan 45° = \(\frac{AB}{BC}\)

⇒ BC = AB (∵ tan 45° = 1) 

⇒ BC = h 

In Δ ABD 

tan 30° = \(\frac{AB}{BD}\)

\(\frac{1}{\sqrt3}\) = \(\frac{AB}{BD}\)

⇒ (h + CD) = √3 h (AB = BC = h) 

⇒ CD = (√3 -1) h