# A person observed the angle of elevation of the top of a tower as 30°. He walked 50 m towards the foot of the tower along level ground and found the angle of elevation of the top of the tower as 60°. Find the height of the tower.

Let the height of the tower = h (m)

In ∆ABD,

tan 60° =

√3 =

h = ---------- (1)

In ∆ABC,

tan 30° =

tan 30° =

------(2)

on substituting value of h from equn. (1) In equn. (2)

On cross multiplication

√3×√3 =

3 =

3

2

25 m.

Now substituting value of in eqn. (1)

h = 25√3 43.3m.

Therefore height of tower is 43.3 m.

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