Given,
The angle of elevation of the tower before he started walking = 30o
Distance walked by the person towards the tower = 50m
The angle of elevation of the tower after he walked = 60o
Let height of the tower (AB) = h m
Let the distance BC = x m
From the fig. in ΔABC,
tan 60o = AB/ BC
√3 = h/x
x = h/√3 ….(i)
Now, in ΔABD
tan 30o = AB/ BD
1/√3 = h/ (50 + x)
√3h = 50 + x
√3h = 50 + (h/√3) [using (i)]
3h = 50√3 + h
2h = 50√3
h = 25√3 = 25(1.73) = 43.25 m
Therefore, the height of the tower = 43.25 m
