Question

A man standing on a cliff observes a ship at an angle of depression of the ship is 30°, approaching the shore just beneath him. Three minutes later, the angle of depression of the ship is 60°. How soon will it reach the shore? 

Answer 1

Let AB be the tower.

Initial position of ship is C , which changes to D after 3 minutes.

In ΔADB

`"AB"/"DB" = tan60^circ`

`"AB"/"DB" = sqrt(3)`

`"DB" = "AB"/sqrt(3)`

In ΔABC

`"AB"/"BC" = tan30^circ`

`"AB"/"BD + DC" = 1/sqrt(3)`

`"AB"sqrt(3) = "BD + DC"`

`"AB"sqrt(3) = "AB"/sqrt(3) + "DC"`

 `"DC" = "AB"sqrt(3) - "AB"/sqrt(3) = "AB"(sqrt(3) - 1/sqrt(3))`

= `"2AB"/sqrt(3)`

Time taken by car to travel DC distance `("i.e". "2AB"/sqrt(3))` = 3 minutes

Time taken by car to travel DB distance `("i.e". "AB"/sqrt(3))`

= `3/((2AB)/sqrt(3)) xx "AB"/sqrt(3) = 3/2` = 1 min 30 secs

Thus , the total time taken is 3 minutes + 1 minute 30 seconds = 4 minutes 30 seconds.