The angle of elevation of the top of the building from the foot of the tower is 30° and the angle of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.
Let AB be the building of height 50 m. and tower of height h (m.)
In ∆ABC
tan 60° = 
√3 = 
------(1)
Now in ∆DCB
tan 30° = 
-----(2)
On substituting value of 
in eqn. (1)
√3h = 
h = 
h = 
Therefore height of tower is 
m.
30