Question

As observed from the top of a lighthouse, 100 m above sea level, the angle of depression of a ship, sailing directly towards it, changes from 30° to 60°. Determine the distance travelled by the ship during the period of observation. [Use `sqrt(3) = 1.732`.]

Answer 1

Let OA be the lighthouse and B and C be the positions of the ship.
Thus, we have:
OA = 100m, ∠OBA = 30° and  ∠OCA = 60°

Let OC =  xm and BC = ym
In the right ΔOAC,we have

`(OA)/(OC) = tan 60° = sqrt(3) `

`⇒100/x = sqrt(3)`

`⇒ x = 100/sqrt(3) m`

Now, in the right ΔOBA,we have:

`(OA)/(OB) =tan 30° = 1/ sqrt(3)`

`⇒  100/(x+y) = 1/ sqrt(3)`

`⇒  x+ y = 100 sqrt(3) `

On putting `x = 100/ sqrt(3)` in the above equation, we get:

`y = 100 sqrt(3) - 100/sqrt(3) = (300-100)/ sqrt(3) = 200/sqrt(3) = 115.47 m` 

∴ Distance travelled by the ship during the period of observation = B = y = 115.47m