Question

From the top of a 45 m high light house, the angles of depression of two ships, on the opposite side of it, are observed to be 30° and 60°. If the line joining the ships passes through the foot of the light house, find the distance between the ships. (Use `sqrt3` =1.73) 

Answer 1

Given: Height of a light house AB = 45 m

To find: CD = ?

Solution: Distance between the ships = CD

∠ACB = 30° and ∠ADB = 60°       ....(alternate angles are equal) 

Now, In right angle ΔABC

tan 30° = `(AB)/(BC)`

`1/sqrt3 = (45)/(BC)`

BC = `45sqrt3`        ...(i)

In right angle ΔABD,

tan 60° = `(AB)/(BD)`

`sqrt3 = (45)/(BD)`

BD = `(45 xx sqrt3)/(sqrt3 xx sqrt3)`

BD = `(45sqrt3)/3`

BD = `15sqrt3`    ...(ii)

From (i) and (ii) we get, 

CD = BC + BD

= `45sqrt3 + 15sqrt3 = 60sqrt3`

= 60 × 1.73

= 103.8 m