From the top of a lighthouse, an observer looking at a boat makes an angle of depression of `60^(@)`. If the height of the lighthouse is `90m`, then find how far is the boat from the lighthouse.
`(sqrt(3)=1.73)`

Explanation

Let seg AB represent the lighthouse.
C is the position of the boat
AD is the horizonta `/_DAC` is the angle of depression.

`AB=90m` and `/_DAC=60^(@)`……(Given)
`/_ACB` and `/_DAC` are alternate angles.
`:./_ACB=60^(@)`
In right angled `DeltaABC`
`tan 60^(@)=(AB)/(BC)`..........(By definition)
`:.sqrt(3)=90/(BC)` ....`[tan 60^(@)=sqrt(3)]`
`:.BC=90/(sqrt(3))`
`:.BC=90/(sqrt(3))xx(sqrt(3))/(sqrt(3))`
`:.BC=(90sqrt(3))/3`
`:.BC=30sqrt(3)`
`:.BC=30xx1.73`
`:.BC+51.9m`
Distance of boat from the lighthouse is `51.9m`.