Question

Two men on either side of the cliff $90m$ height observes the angles of elevation of the top of the cliff to be ${30}^o$ and ${60}^o$ respectively. Find the distance between the two men.

Answer 1

In $\Delta ABD$,

$\tan {60}^{\circ }=\frac{BD}{AB}$

$\sqrt{3}=\frac{80}{AB}$

$AB=\frac{80}{\sqrt{3}}$

In $\Delta CBD$,

$\tan {30}^{\circ }=\frac{BD}{BC}$

$\frac{1}{\sqrt{3}}=\frac{80}{BC}$

$BC=80\sqrt{3}$

The distance AC will be,

$AC=AB+BC$

$=\frac{80}{\sqrt{3}}+80\sqrt{3}$

$=\frac{80+80\left(3\right)}{\sqrt{3}}$

$=\frac{320}{\sqrt{3}}$

$=\frac{320}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}$

$=\frac{320\sqrt{3}}{3}$

$=184.53m$

Therefore, the distance between the two men is $184.53m$.