# 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: Option A
$\begin{array}{l}\text{Let AB be the observer and CD tower.}\\ \text{Draw BE perpendicular to CD}\\ \text{Then CE}=\text{AB}=1.6\ \text{m}\\ \text{And}\ \text{BE}=\text{AC}=20\sqrt{3}\text{m}\\ \text{Then right angle triangle DEB}\\ \therefore\tan30^{\circ}=\frac{\mathrm{DE}}{\mathrm{BE}}\\ \Rightarrow\frac{1}{\sqrt{3}}=\frac{\mathrm{DE}}{20\sqrt{3}}\\ \Rightarrow\mathrm{DE}=20\sqrt{3}\mathrm{~m}\\ \text{Then, CD}=\text{CE+DE}\\ =1.6+20\\ =21.6\ \text{m}\end{array}$