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Question
A man in a boat rowing away from a lighthouse 150 m high, takes 2 minutes to change the angle of elevation of the top of the lighthouse from 60° to 45°. Find the speed of the boat.
Solution
Let AB be the lighthouse and C and D be the two positions of the boat such that AB = 150 m, ∠ADB = 45° and ∠ACB = 60°.
Let speed of the boat be x metre per minute.
Therefore, CD = 2x m;
In ΔADB,
`(AB)/(DB) = tan 45^circ`
`=>` BD = 150 m
In ΔABC,
`(AB)/(BC) = tan 60^circ = sqrt(3)`
`=> 150/(BC)= sqrt(3)`
`=> BC = 150/ sqrt(3)`
= `150/1.732`
= 86.605 m
∴ CD = BD – BC
= 150 – 86.605
= 63.395 m
`=>` 2x = 63.395
`=> x = 63.395/2`
= 31.6975 m/min
= `31.6975/60` m/sec
= 0.53 m/sec
Hence, the speed of the boat is 0.53 m/sec
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