#### 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/h.

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#### Solution

Let the distance BC be *x* m and CD be *y* m.

In ΔABC

`tan 60^@ = (AB)/(BC) = 150/x`

`=> sqrt3 = 150/x`

`=> x = 150/sqrt3 m` ....(1)

In ΔABD

`tan 45^@ = (AB)/(BD) = 150/(x + y)`

`=> 1 = 150/(x + y)`

=> x + y = 150

`=> y = 150 - x`

Using 1 we get

`=> y = 150 - 150/sqrt3 = (150(sqrt3 - 1))/sqrt3 m`

Time taken to move from the point C to point D is 2 min = `2/60 h = 1/30 h`

Now.

Speed = `"Distance"/"Time" = y/(1/30)`

`=(150((sqrt3-1))/sqrt3)/(1/30) = 1500 sqrt3 (sqrt3 - 1) "m/h"`

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#### Reference Material

Solution for 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/H concept: null - Heights and Distances. For the course CBSE