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Question
The angle of elevation of the top of an unfinished tower at a distance of 75m from its base is 30° .How much higher must the tower be raised so that the angle of elevation of its top at the same point may be 60 .
Solution
Let AB be the unfinished tower, AC be the raised tower and O be the point of observation
We have:
OA = 75m,∠AOB = 30° and ∠AOC = 60°
Let AC = H m such that BC = (H -h)m.
In ΔAOB,we have:
`(AB)/(OA) = tan 30° = 1/ sqrt(3)`
`⇒ h/ 75 = 1/ sqrt(3)`
`⇒ = 75/ sqrt(3) m = (75 xx sqrt(3)) /(sqrt(3) xx sqrt(3)) = 25 sqrt(3) m`
In ΔAOC,we have:
`(AC)/(OA) = tan 60° = sqrt(3)`
`⇒ H/75 = sqrt(3)`
`⇒ H = 75 sqrt(3 )m`
`∴"Required height" =(H - h) = (75 sqrt(3) - 25 sqrt(3)) = 50 sqrt(3)m = 86.6m`
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