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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.
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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