An observer 1.6 m tall is 203 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:
Answer: Option A
Let AB be the observer and CD be the tower.
Draw BE CD.
Then, CE = AB = 1.6 m,
BE = AC = 203 m.
CD = CE + DE = (1.6 + 20) m = 21.6 m.