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.

Answer 1

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.