# 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.

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