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Question
The angle of elevation of the top Q of a vertical tower PQ from a point X on the ground is 60°. At a point Y, 40m vertically above X, the angle of elevation is 45°. Find the height of the tower PQ and the distance XQ.
Solution
In the figure, PQ is the tower.
In ΔPQX,
∴ `"h"/"x" = tan60^circ = sqrt(3)`
⇒ h = `sqrt(3)`x ...(1)
In ΔQRY,
`("h" - 40)/"x" = tan 45^circ = 1`
⇒ h = 40 + x ...(2)
From (1) and (2),
`sqrt(3)`x = 40 + x
⇒ `(sqrt(3) - 1)"x" = 40`
⇒ `"x" = 40/(sqrt(3) - 1) = (40(sqrt(3) + 1))/((sqrt(3) - 1)(sqrt(3) + 1)) = 40/2(sqrt(3) + 1) = 20(sqrt(3) + 1)`
∴ `"h" = 40 + 20(sqrt(3) + 1) = 20sqrt(3) + 60 = 20(sqrt(3) + 3) = 20 xx 4.732 = 94.64`
Thus , the height of the tower PQ is 94.64 m.
Again, in ΔPQX,
∴ `"h"/"XQ" = sin60^circ = 1/sqrt(2)`
⇒ `"XQ" = sqrt(2)"h" = 1.414 xx 94.64 = 109.3`m
Thus , the distance XQ is 109.3m.
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