The angle of elevation of the top of a tower as observed from a point in a horizontal plane through the foot of the tower is 32°. When the observer moves towards the tower a distance of 100 m, he finds the angle of elevation of the top to be 63°. Find the height of the tower and the distance of the first position from the tower. [Take tan 32° = 0.6248 and tan 63° = 1.9626]
Let the height of the tower = h (m)
Let the distance of point from the foot of the tower = 
(m)
In ∆ABC,
tan 32° = 
tan 32° = 
0.6248 = 
h = 0.6248(100+
) -----------(1)
In ∆ABD,
tan 63° = 
1.9626 = 
h = 1.9626
-----------(2)
Substituting value of h from eqn. (2) in eqn. (1)
1.9626 
= 0.6248(100+
1.9626 
= 62.48+0.6248 
1.9626 
-0.6248
= 62.48
1.3378
= 62.48
= 46.70 m.
on substituting value of x in eqn.(2)
1.9626× 46.7
91.66 m.
Distance of the first position from tower=
= CD+DB
100+46.7
height of tower is 91.66 m.