Question

On a horizontal plane there is a vertical tower with a flagpole on the top of the tower. At a point, 9 metres away from the foot of the tower, the angle of elevation of the top and bottom of the flagpole are 60° and 30° respectively. Find the height of the tower and the flagpole mounted on it. [Take `sqrt(3) = 1.73`.]

Answer 1

Let OX be the horizontal plane, AD be the tower and CD be the vertical flagpole
We have:

AB = 9m, ∠DBA= 30° and ∠CBA = 60°
Let:
AD = hm and CD = xm

In the right ΔABD,we have:

`(AD)/(AB) = tan 30° = 1/ sqrt(3)`

`⇒ h/9 = 1/sqrt(3)`

`⇒ h = 9/sqrt(3) = 5.19 `m

Now, in the right  ΔABC,we have

`(AC)/(BA) = tan 60° = sqrt(3)`

`⇒  (h+x ) /9 = sqrt(3)`

`⇒  h + x = 9 sqrt(3)`

By putting h= `9/sqrt(3)` in the above equation, we get:

`9/sqrt(3) + x = 9 sqrt(3) `

`⇒   x = 9 sqrt(3) - 9/ sqrt(3) `

`⇒  x = (27-9)/ sqrt(3) = 18/ sqrt(3) = 18/ 1.73 = 10.4`

Thus, we have:
Height of the flagpole = 10. 4 m
Height of the tower = 5. 19 m