Question

A vertical tower subtends an angle of $60^o$ at a point on the same level as the foot of the tower. On moving $100m$ further from the first point in line with the tower, it subtends an angle of $30^o$ at the point. If the height of the tower is $Hm$ , then the value of $\frac{H}{25\sqrt{3}}$ (in meters) is

Answer 1

Answer: 2

$tan60^o=\frac{H}{x},tan30^o=\frac{H}{x+100}$
$\Rightarrow$ $\frac{H}{tan60^{{}^{\circ }}}=\frac{H}{tan30^{{}^{\circ }}}-100$
$\Rightarrow$ $\frac{H}{\sqrt{3}}=\frac{H}{1/\sqrt{3}}-100$
$\Rightarrow$ $\sqrt{3}H-\frac{H}{\sqrt{3}}=100\Rightarrow 3H-H=100\sqrt{3}$
$\Rightarrow H=50\sqrt{3}$
$\frac{H}{25\sqrt{3}}=2$