A vertical tower subtends an angle of 60 ^ o at a point on the same...

$tan {60}^{o} = \dfrac{H}{x}, tan {30}^{o} = \dfrac{H}{x + 100}$
$⇒$ $\dfrac{H}{\tan 6{0}^{∘}} = \dfrac{H}{\tan 3{0}^{∘}} - 100$
$⇒$ $\dfrac{H}{\sqrt{3}} = \dfrac{H}{\dfrac{1}{\sqrt{3}}} - 100$
$⇒$ $\sqrt{3}H - \dfrac{H}{\sqrt{3}} = 100 ⇒ 3H - H = 100\sqrt{3}$
$⇒ H = 50\sqrt{3}$
$\dfrac{H}{25\sqrt{3}} = 2$