From the top of a 10 m high building, the angle of elevation of the top of a tower is 60° and the angle of depression of the foot of the tower is 45°. What is the height of the tower ? (Take $\sqrt{3}$ = 1.73)

27.3 m

As, BD = CE = 10 m

tan 45° = 1

\(\frac{BC}{CE}\) = 1

BC = CE = 1

BC = 10

In triangle ABC ,

tan 60° = √3

\(\frac{AC}{BC}\)= √3 

AC = 10√3

Height of tower = ( 10 + 10√3 )

= 10 + 10(1.73)

= 27.3 m