# 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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