As observed form the top of a lighthouse, 100m above sea level, the angle of depression of a ship, sailing directly towards it, changes from 30° and 60° . Determine the distance travelled by the ship during the period of observation.
Solution
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 = xmand 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
