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 heights of the tower is:

Answer: Option A

Let AB be the observer and CD be the tower.


Draw BE ⊥ CD
Then CE = AB = 1.6m
BE = AC = 20√3m
$$\frac{DE}{BE}=\tan 30^{\circ }=\frac{1}{\sqrt{3}}$$
$$\Rightarrow DE=\frac{BE}{\sqrt{3}}=\frac{20\sqrt{3}}{\sqrt{3}}=20$$
∴ CD = CE + DE = (1.6 + 20) m = 21.6 m