Question

A man in a boat rowing away from a lighthouse 180 m high takes 2 minutes to change the angle of elevation of the top of the lighthouse from 60° and 30°. Find the speed of the boat. 

Answer 1

Let AB be the lighthouse. 
Initial position of boat is C, which changes to D after 2 minutes. 
In ΔADB

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

`180/"x" = sqrt(3)`

`"x" = 180/sqrt(3)`

In ΔABC
`"AB"/"BC" = tan30^circ`

`180/("x + y") = 1/sqrt(3)`

`180sqrt(3) = "x + y"`

`180sqrt(3) = 180/sqrt(3) + "y"`

`"y" = 180(sqrt(3) - 1/sqrt(3)) = 180(2/sqrt(3)) = 360/sqrt(3)`

Time taken by car to travel DC distance `("i.e". 360/sqrt(3))` = 2 minutes = 120 seconds

Speed of the boat = `"Distance"/"Time" = (360/sqrt(3))/120 = 3/sqrt(3) = 3/sqrt(3) xx sqrt(3)/sqrt(3) = (3sqrt(3))/3 = sqrt(3) = 1.732`

Thus , the speed of the boat is 1.732 m/sec.