Question

On the level ground, the angle of elevation of a tower is 30°. On moving 20 meters nearer, the angle of elevation is 60°. The height of the tower is ______.

Answer 1

On the level ground, the angle of elevation of a tower is 30°. On moving 20 meters nearer, the angle of elevation is 60°. The height of the tower is `underlinebb(10sqrt(3)  m)`.

Explanation:

Let the height be h and the initial horizontal distance from the tower be x.

`tan 30^circ = h/x`

= `1/sqrt(3)` 

⇒ `h = x/sqrt(3)`

After moving 20 m nearer the distance is x – 20 and `tan 60^circ = h/(x - 20) = sqrt(3)`

⇒ `h = sqrt(3)(x - 20)`

Equate: `x/sqrt(3) = sqrt(3)(x - 20)` 

⇒ x = 3(x – 20)

⇒ x = 30 m

So `h = x/sqrt(3)`

= `30/sqrt(3)` 

= `10sqrt(3)` m ≈ 17.32 m