An observer 1.6 m tall is 203away 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

Let AB be the observer and CD be the tower.Draw BECD.Then, CE = AB = 1.6 m,BE = AC = 203m.DE= tan 30° =1BE3DE =203m = 20 m.3CD = CE + DE = (1.6 + 20) m = 21.6 m.

Accepted Answer

Answer

21.6 m

Answer

23.2 m

Answer

24.72 m

Answer

None of these

Question 162: An observer 1.6 m tall is 203away from a tower. The angle of elevation from his eye to the top of the tower is 30°. The height of the tower is: