Question

The angle of elevations of the top of the tower from two points distance a and b from the base of tower (a > b) and 30° and 60° then height of tower is :

(A) √(a + b)

(B) √(a - b)

(C) √(ab)

(D) √(a/b)

Answer 1

Answer is (C) √(ab)

Let CD is a tower with its height h. 

Let b is the distance of point B from the base of tower C and a is a distance of point A.

So, ∠CAD = 30°, ∠CBD = 60°

From right angled ∆ACD,

tan 30° = h/ac

⇒ 1/√3 = h/a

⇒ h = a/√3 …..(i)

From right angled ∆BCD,

tan 60° = h/b

⇒ √3 = h/b …(ii)

⇒ h = √3b

From equation (i) and (ii),

h² = ab

or h = √(ab)