Question

A man in a boat rowing away from a lighthouse 100 m high takes 2 minutes to change the angle of elevation of the top of the lighthouse from 60° to 30°. Find the speed of the boat in metres per minute. [Use `sqrt3` = 1.732]

Answer 1

AB is a lighthouse of height 100m. Let the speed of boat be x metres per minute. And CD is the distance which man travelled to change the angle of elevation.

Therefore,
CD = 2 x     ...[Distance = speed × time]

tan(60°) = `(AB)/(BC)`

`sqrt3 = 100/(BC)`

`=> BC = 100/sqrt3`

tan(30°) = `(AB)/(BD)`

`=> 1/sqrt3 = 100/(BD)`

BD = 100`sqrt3`

CD = BD − BC

`2x = 100 sqrt3 - 100/sqrt3`

`2x = (300 - 100)/sqrt3`

`=> x = 200/(2sqrt3)`

`=> x = 100/sqrt3`

Using,

`sqrt3 = 1.73`

`x = 100/1.73 = 57.80`

Hence, the speed of the boat is 57.80 metres per minute.