Question

Two men on either side of a 75 m high building and in line with base of the building, observe the angles of elevation of the top of the building as ${30}^0$ and ${60}^0$ respectively. Find the distance between the two men. $(use\sqrt{3}=1.73)$

Answer 1

In $△ABD$,

$\tan {30}^o=\frac{BD}{AD}$

$\frac{1}{\sqrt{3}}=\frac{75}{AD}$

$AD=75\sqrt{3}m$

In $△BCD$

$\tan {60}^o=\frac{BD}{CD}$

$\sqrt{3}=\frac{75}{CD}$

$CD=\frac{75}{\sqrt{3}}m$

$AC=AD+CD=(75+\frac{75}{\sqrt{3}})m$

$=\frac{75\left(4\right)}{\sqrt{3}}=\frac{300}{1.73}=173.41m$