Q. 153.9( 20 Votes )

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]

Answer :

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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