A vertical tower subtends an angle of $60^{0}$ at a point on the same level as the foot of the tower. On moving 100 mts further from the first point in line with the tower, it subtends an angle of $30^{0}$ at the point. If the height of the tower is H mts, then the value of $\frac{H}{25 \sqrt{3}}$ (in meters) is

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Detailed Solution

$\tan 60^{0} = \frac{H}{x}, \tan 30^{0} = \frac{H}{x + 100}$

$\Rightarrow \frac{H}{\tan 60^{0}} = \frac{H}{\tan 30^{0}} - 100$

$\Rightarrow \frac{H}{\sqrt{3}} = \frac{H}{1 / \sqrt{3}} - 100 \Rightarrow \sqrt{3} H - \frac{H}{\sqrt{3}} = 100 \Rightarrow 3 H - H = 100 \sqrt{3}$ $\Rightarrow H = 50 \sqrt{3}$ $\frac{H}{25 \sqrt{3}} = 2$

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