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