An observer 2.7 m tall is 18√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 2.7 m tall is 18√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:
Let AB be the observer and CD be the tower.

Draw BE ⊥ CD.
Then, CE = AB = 2.7 m,
And BE = AC = 18√3 m.

∴ DE/BE = tan 30° = 1/√3
⇒ DE = 18√3/√3 = 18 m
∴ CD = CE + DE = (2.7 + 18) m = 20.7 m.