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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 .

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#### 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`

Concept: Heights and Distances

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