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)
Answer & explanation
Correct answer: option 1
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