On the ground level the angle of elevation of the top of a tower is 30°, On moving 20 m nearer the tower, the angle of elevation found to be 60°. The height of the tower is?

Let us draw a figure below from the given question.
Let AB = h meter be the height of the towers. B and C are two points such that BC = 20 m; ∠ADB = 30° and ∠ACB = 60° BC = x meter (let us assume)
Now, from right triangle ABC,
x = h cot 60°
⇒ x = h/√3 meter
Again, from right triangle ABD,
h = (20 + x) tan 30°
put the value of x in above equation.
∵ x = h/√3
⇒ h = (20 + h/√3) x 1/√3 ( ∵ tan 30° = 1/√3 )
⇒ h - h/3 = 20/√3 ⇒ 2h/3 = 20/√3
∴ h = 20 x 3/2√3 = 10√3 m