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:
a) 21.6 m
b) 23.2 m
c) 24.72 m
d) None of these

Answer: a
Explanation: 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