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प्रश्न
A man in a boat rowing away from a lighthouse 100 m high takes 2 minutes to change the angle of elevation of the top of the lighthouse from 60° to 30°. Find the speed of the boat in metres per minute [Use `sqrt3` = 1.732]
उत्तर
AB is a lighthouse of height 100m. Let the speed of boat be x metres per minute. And CD is the distance which man travelled to change the angle of elevation.
Therefore,
CD = 2x [Distance = speed x time]
tan(60°) = `("AB")/("BC")`
`sqrt3 = 100/"BC"`
`=> "BC" = 100/sqrt3`
tan(30°) = `"AB"/"BD"`
`=> 1/sqrt3 = 100/"BD"`
BD = 100`sqrt3`
CD = BD - BC
`2"x" = 100 sqrt3 - 100/sqrt3`
`2"x" = (300 - 100)/sqrt3`
`=> "x" = 200/(2sqrt3)`
`=> x = 100/sqrt3`
Using,
`sqrt3 = 1.73`
`"x" = 100/1.73 = 57.80`
Hence, the speed of the boat is 57.80 metres per minute.
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