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Question
A man in a boat rowing away from a lighthouse 180 m high takes 2 minutes to change the angle of elevation of the top of the lighthouse from 60° and 30°. Find the speed of the boat.
Solution
Let AB be the lighthouse.
Initial position of boat is C, which changes to D after 2 minutes.
In ΔADB
`"AB"/"DB" = tan60^circ`
`180/"x" = sqrt(3)`
`"x" = 180/sqrt(3)`
In ΔABC
`"AB"/"BC" = tan30^circ`
`180/("x + y") = 1/sqrt(3)`
`180sqrt(3) = "x + y"`
`180sqrt(3) = 180/sqrt(3) + "y"`
`"y" = 180(sqrt(3) - 1/sqrt(3)) = 180(2/sqrt(3)) = 360/sqrt(3)`
Time taken by car to travel DC distance `("i.e". 360/sqrt(3))` = 2 minutes = 120 seconds
Speed of the boat = `"Distance"/"Time" = (360/sqrt(3))/120 = 3/sqrt(3) = 3/sqrt(3) xx sqrt(3)/sqrt(3) = (3sqrt(3))/3 = sqrt(3) = 1.732`
Thus , the speed of the boat is 1.732 m/sec.
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