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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/min.
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Solution
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.
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