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.