A tower is 50 m high. The angle of elevation of the top of the tower from a point on the ground is 30°. Find the distance of the point from the base of the tower.
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 50\(\sqrt{3}\) m
The tower, the ground and the line of sight form a right triangle. The angle of elevation at the point P links the height AB (the side opposite it) and the distance PB (the side next to it) through the tangent of 30°, which is \(\frac{1}{\sqrt{3}}\).
In the right triangle PBA, the tangent of the angle at P is opposite side over adjacent side: \(\displaystyle \tan30^{\circ}=\frac{AB}{PB}=\frac{50}{PB}\)
Putting \(\tan30^{\circ}=\frac{1}{\sqrt{3}}\): \(\displaystyle \frac{1}{\sqrt{3}}=\frac{50}{PB}\), so the distance \(PB=50\sqrt{3}\) m
