An observer 1.6m tall is 20√3m 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:

The observer’s eye height is 1.6 m, distance to tower is (20\sqrt{3}) m, and elevation angle is 30°. Using trigonometry, height above eye level = (20\sqrt{3} \cdot \tan(30°) = 20\sqrt{3} \cdot 1/\sqrt{3} = 20). Total tower height = (20 + 1.6 = 21.6) m. Studying this highlights trigonometric applications, providing lessons on using angles for real-world measurements in engineering and surveying.