An observer 1.6 m tall is 20√3 away from a tower. The angle of elevation from his eye to the top of the tower is 30º. The height of the tower is:

Question: An observer 1.6 m tall is 20√3 away from a tower. The angle of elevation from his eye to the top of the tower is 30º. The height of the tower is:

Solution:

Let AB be the observer and CD tower
Draw BE perpendicular to CD

Then CE = AB = 1.6 m
And BE = AC = 20√3 m

Then right angle triangle DEB
∴ tan30° = DE/BE
⇒ 1/√3 = DE/20√3
⇒ DE = 20√3m

Then CD = CE + DE = 1.6 + 20 = 21.6 m