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 `sqrt(3` = 1.732]Use 3=1.732">

#### Solution

AB is a lighthouse of height 100m.

Let the speed of the boat be x metres per minute.

And CD is the distance which man travelled to change the angle of elevation.

So, CD = 2x ...[∵ Distance = Speed x Time]

tan (60°) = `("AB")/("BC")`

`sqrt(3) = (100)/("BC")`

⇒ `"BC" = (100)/(sqrt(3)`

tan (30°) = `("AB")/("BD")`

`(1)/sqrt(3) = (100)/("BD")`

`"BD" = 100 sqrt(3)`

`"CD"= "BD" - "BC"`

`2"x" = 100sqrt(3) - (100)/sqrt(3)`

`2"x" = (300-100)/sqrt(3)`

⇒ `"x" = (200)/(2sqrt(3)`

⇒ `"x" = (100)/(sqrt3)`

Using `sqrt(3)` = 1.73

`"x" = (100)/(1.73) ≈ 57.80`

Hence, the speed of the boat is 57.80">57.8057.80 metres per minute.