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Question
The angle of elevation of the top of a vertical tower PQ from a point X on the ground is 60°. At a point Y, 40 m vertically above X, the angle of elevation of the top is 45°. Calculate the height of the tower.
Solution
Let PQ be the tower of height H m and an angle of elevation of the top of tower PQ from point X is 60°. An angle of elevation at 40 m vertical from point X is 45°.
Let PQ = H m and SX = 40m. OX = x, ∠PST = 45°, ∠PXQ = 60°
Here we have to find height of tower.
The corresponding figure is as follows
We use trigonometric ratios.
In ΔPST
`=> tan 45^@ = h/x`
`=> 1 = h/x`
`=> x = h`
Again in ΔPST
`=> tan 456@ = h/x`
`=> 1 = h/x`
`=> x = h`
Again in ΔPXQ
`=> tan 60^@ = (h + 40)/x`
`=> sqrt3 = (h + 40)/x`
`=> h + 40 = sqrt3h`
`=> h(sqrt3 - ) = 40`
`=> h = 40/(sqrt3 - 1)`
`=> h = 54.64`
Therefore `H = 54.64 + 40`
`=> H = 94.64`
Hence the height of tower uis 94.64 m
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