For a man , the angle of elevation of the highest point of a tower situated west to him is `60^@` . On walking 240 meters to north , the angle of elevation reduces to `30^@` . The height of the tower is

Options

A. `50sqrt3m`
B. `30sqrt6m`
C. `60sqrt6m`
D. 60m

Explanation

Let OA the tower of height h. Also , let B amnd C be the initial and final position , respectivily , of the man .

In triangle AOB ,OB =`hcot60^@=h/(sqrt3)`
in triangle AOC ,OC =`hcot30^@=sqrt3h`
Now in triangle OBC ,
`OB^2+BC^2=OC^2`
`therefore h^2/(3)+(240)^2=3h^2`
`rArr (8h^2)/3=(240)^2`
`rArr h= 60sqrt6m`