An observer 1.6 m tall is 20√3 m away from a tower. The angle of elevation from his eye to the top of the tower is 30°. The height of the tower is:
Answer: A
Let H be the total height of the tower. The height of the tower above the observer's eye level is h'.
The distance is d = 20√3 m.
\(\tan 30° = \frac{h'}{d} = \frac{h'}{20\sqrt{3}}\)
\(\frac{1}{\sqrt{3}} = \frac{h'}{20\sqrt{3}} \Rightarrow h' = 20\) m.
The total height of the tower H = h' + height of observer = 20 + 1.6 = 21.6 m.