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Question
The angle of elevation of a tower from a point in line with its base is `45^circ` . On moving 20m towards the tower , the angle of elevation changes to `60^circ` . Find the height of the tower.
Solution
Let the height of the tower (AB) be h .
In ΔABC,
`tan60^circ = "AB"/"BC"`
⇒ `sqrt(3) = "h"/"BC"`
⇒ `"BC" = "h"/sqrt(3)`
In ΔABD,
`tan 45^circ = "AB"/"BD"`
⇒ `1 = "h"/"BC + 20"`
⇒ BC + 20 = h
⇒ `"h"/sqrt(3) + 20 = "h"`
⇒ `"h"(1-1/sqrt(3)) = 20`
⇒ `"h" = 20 xx sqrt(3)/(sqrt(3) - 1)`
⇒ h = `20 xx sqrt(3)/(sqrt(3) - 1) xx (sqrt(3) + 1)/(sqrt(3) + 1) `
= `20 xx (sqrt(3)(sqrt(3) + 1))/(3 - 1) = 10(3 + sqrt(3)) = 10 xx 4.732 = 47.32`
Thus, the height of the tower is 47.32 m.
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