An observer 2 m tall is \(24\sqrt{3}\) m away from a building. The angle of elevation of the top of the building as seen by him is 30°. What is the height of the building?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 26 m
The angle of elevation is measured from the observer’s eye, 2 m above the ground. So the right triangle is formed above eye level, and \(\tan30^\circ=\frac{\text{height above eye level}}{\text{distance}}\); the 2 m is added at the end.
Height above eye level \(=24\sqrt{3}\times\tan30^\circ\)
\(\displaystyle =24\sqrt{3}\times\frac{1}{\sqrt{3}}=24\) m
Height of the building \(=24+2=26\) m
Tip: When the observer’s height is given, add it back at the end. Option (1), 24 m, is only the part of the building above his eye level.