Question

The angle of elevation of a tower from a distance 200 m from its foot is 30 degree. height of the tower will be???????

Answer 1

Let the height of the tower = BC = h m

Distance from foot to the point

= AB = 200m

Angle of elevation =<BAC = 30°

Now,

From ∆ABC,

<B = 90°

tan<BAC = BC/AB

=> tan 30° = h/200

=> 1/√3 = h/200

=> h = 200/√3

=> h = (200×√3)/(√3×3)

= (200√3/3) m

••••

Answer 2

this is your answer is 115.4m