Q13 of 35 Page 219

Watching from a window 40 m high of a multi – storeyed building, the angle of elevation of the top of a tower is found to have measure 45. The angle of elevation of the top of the same tower from the bottom of the building is found to have measure 60. Find the height of the tower.

Let CD be the window, AB be the tower and D the point of observation.



CD = 40 m


Let the height of the tower = AB = h


Horizontal line DE intersects AB in E.


BE = CD = 40 m


AE = AB – BE = (h – 40)


∠AED = ∠ABC = 90⁰


Now, the angle of elevation of A from D is 45⁰ and the angle of elevation of A from C is 60⁰.


∠ADE = 45⁰, ∠ACB = 60⁰


In ∆AED and in ∆AEC,




DE = h – 40 = BC ………..(1)




H = (h – 40)√3


H = √3h – 40√3


H(√3 – 1) = 40√3







= 60 + 20(1.73)


= 60 + 34.6


= 94.6 m


The height of tower is 94.6 m.


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