An observer 2.8 m tall is 14√3 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-

Question: An observer 2.8 m tall is 14√3 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- Solution: Here, Height = AB Now, tan∠c = AE/CE ⇒ tan30° = AE/14√3 ⇒ 1/√3 = AE/14√3 ∴ AE = 14 ∴ AB = AE + BE = 14 + 2.8 = 16.8m