Question

A vertical tower stands on a horizontal plane and is surmounted by a vertical flagstaff of height 6 m. At a point on the plane, the angle of elevation of the bottom of the flagstaff is 30° and that of the top of the flagstaff is 60°. Find the height of the tower. [Use `sqrt(3) = 1.732`.]

Answer 1

Let AB be the tower and BC be the flagstaff,
We have,
BC = 6m, ∠AOB = 30° and ∠AOC - 60°
Let AB = h
In  ΔAOB

 `tan 30° = (AB)/(OA)`

`⇒ 1/ sqrt(3) = h/( OA)`

` ⇒  OA = h sqrt(3)`                ..............(i) 

Now in Δ AOC,

` tan 60° = (AC)/(OA)`

`⇒   sqrt(3) = (AB +BC)/ (h sqrt(3))`               [ Using (i)]

⇒ 3h = h + 6

⇒ 3h  - h = 6

⇒ 2h = 6

`⇒  h= 6/2`

⇒  h = 3m

So, the height of the tower is 3 m.