An observer 1.6 m tall is \(20\sqrt{3}\) 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 & explanation
Correct answer: option 1
The correct answer is Option (1) → 21.6 m
The angle of elevation is measured from the observer’s eye, so the right triangle gives only the part of the tower above eye level; the observer’s height (1.6 m) is then added to get the whole tower.
Height above eye level: \(\displaystyle \tan30^\circ=\frac{h}{20\sqrt3}\), so \(\displaystyle h=20\sqrt3\times\frac{1}{\sqrt3}=20\) m
Height of the tower \(=20+1.6=21.6\) m