Question

A moving boat is observed from the top of a 150 m high cliff moving away from the cliff. The angle of depression of the boat changes from 60° to 45° in 2 minutes. Find the speed of the boat in m/min.

Answer 1

Let AO be the cliff of height 150 m.
Let the speed of the boat be x meters per minute.
And BC is the distance which man travelled.

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

tan(60°) = `"AO"/"OB"`

`sqrt3` = `150/"OB"`

⇒ OB = `(150sqrt3)/3` = `50sqrt3`

tan(45°) = `"AO"/"OC"`

⇒1 = `150/"OC"`

⇒ OC = 150
Now OC = OB + BC

⇒ 150 = `50sqrt3` + 2x

⇒ x = `(150 − 50sqrt3)/2`

⇒ x = 75 − `25sqrt3`

Using `sqrt3 = 1.73`
x = 75 − 25 x 1.732 ≈ 32 m/min 
Hence, the speed of the boat is 32 metres per minute.