Question

The angles of elevation of the top of a tower at a distance of $\frac{50\sqrt{3}}{3}$ meters from the foot is ${60}^{\circ }$ . Find the height of the tower.

• A. $50\sqrt{3}$ meters
• B. $\frac{20}{\sqrt{3}}$ meters
• C. -50 meters
• D. 50 meters

Answer 1

The correct option is D 50 meters

Let the height of the tower be h

In $\Delta ABC\tan \theta =\frac{AB}{BC}\Rightarrow \tan {60}^o=\frac{AB}{BC}\tan {60}^o=\frac{h}{\frac{50\sqrt{3}}{3}}\sqrt{3}=\frac{3\times h}{50\sqrt{3}}\Rightarrow 3h=\sqrt{3}\times 50\sqrt{3}3h=50\times 3\therefore h=50m$

Option D is the correct answer.