Question

Two men standing on opposite sides of a flagstaff measure the angles of the top of the flagstaff as 30° and 60°. If the height of the flagstaff is 18 m, the distance between the men is
(a) 24 m
(b) $24\sqrt{3}\mathrm{m}$
(c) $\frac{24}{\sqrt{3}}\mathrm{m}$
(d) 31.2 m

Answer 1

(b) $24\sqrt{3}\mathrm{m}$
Let $AB$ be the flagstaff and $C\mathrm{and}D$ be the positions of the two men. Thus, we have:
$AB=18$ m,∠$ACB={30}^o$ and ∠$ADB={60}^o$

In ∆$ACB$, we have:

$\frac{AC}{AB}=cot{30}^o=\sqrt{3}$

⇒ $\frac{AC}{18}=\sqrt{3}$
⇒ $AC=18\sqrt{3}$ m

In ∆$ADB$, we have:

$\frac{AD}{AB}=cot{60}^o=\frac{1}{\sqrt{3}}$

⇒ $\frac{AD}{18}=\frac{1}{\sqrt{3}}$

⇒ $AD=\frac{18}{\sqrt{3}}$ m
∴ $CD=(AC+AD)=(18\sqrt{3}+\frac{18}{\sqrt{3}})=\frac{72}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}=\frac{72\sqrt{3}}{3}=24\sqrt{3}$ m

Hence, the distance between the two men is $24\sqrt{3}$ m.