Q. 284.4( 31 Votes )

From the top a 7 m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the height of the tower.

Answer :

Let the distance between the foots of building and cable tower is (m).


The height of cable tower = AB = AE+EB (h+7)m.


In ∆AED,


tan 60° =


√3 =


h = √3 ---------(1)


The height of cable tower = AB = AE+EB (h+7)m.


In ∆DEB,


tan 45° =


1 =


----------(2)


On substituting value of in eqn. (1)


h = 7√3



Height of cable tower is (h+7)m.


7√3+7


7(√3+1)m.


Therefore height of cable tower is 7(√3+1)m.


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