The angle of elevation from a point situated at a distance of 70 meter from the base of a tower is 60°. The height of the tower is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(70\sqrt{3}\) meter
The tower, the ground and the line of sight form a right triangle. The height is opposite the angle of elevation and the 70 m distance is next to it, so \(\tan(\text{angle})=\frac{\text{height}}{\text{distance}}\).
In the triangle APQ, right-angled at the foot P of the tower: \(\displaystyle \tan60^\circ=\frac{h}{70}\)
Since \(\tan60^\circ=\sqrt{3}\), the height is \(h=70\sqrt{3}\) m