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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 `sqrt(3` = 1.732]Use 3=1.732">
उत्तर
AB is a lighthouse of height 100m.
Let the speed of the boat be x metres per minute.
And CD is the distance which man travelled to change the angle of elevation.
So, CD = 2x ...[∵ Distance = Speed x Time]
tan (60°) = `("AB")/("BC")`
`sqrt(3) = (100)/("BC")`
⇒ `"BC" = (100)/(sqrt(3)`
tan (30°) = `("AB")/("BD")`
`(1)/sqrt(3) = (100)/("BD")`
`"BD" = 100 sqrt(3)`
`"CD"= "BD" - "BC"`
`2"x" = 100sqrt(3) - (100)/sqrt(3)`
`2"x" = (300-100)/sqrt(3)`
⇒ `"x" = (200)/(2sqrt(3)`
⇒ `"x" = (100)/(sqrt3)`
Using `sqrt(3)` = 1.73
`"x" = (100)/(1.73) ≈ 57.80`
Hence, the speed of the boat is 57.80">57.8057.80 metres per minute.
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