The angle of elevation of a tower from a point is 30⁰. At a point on the horizontal line passing through the foot of 'the tower and 50 metros nearer it, the angle of elevation is 60⁰. The distance of the first point from the tower is ________.

Let AC = x meter and AB = h  meter 

From $∆\mathrm{BCA}$, $\tan 30°=\frac{\mathrm{AB}}{\mathrm{AC}}$ 

=>  $\frac{1}{\sqrt{3}}=\frac{\mathrm{h}}{\mathrm{x}}$ 

=>  $\mathrm{h}=\frac{\mathrm{x}}{\sqrt{3}}$

From $∆\mathrm{BDA}, \tan 60°=\frac{\mathrm{AB}}{\mathrm{AD}}$

=>  $\sqrt{3}=\frac{\mathrm{h}}{\mathrm{x}-50}$ 

=>  $\mathrm{h}=(\mathrm{x}-50)\sqrt{3} \mathrm{meter}$

Equating both the value of h, we get:

$\frac{\mathrm{x}}{\sqrt{3}}=(\mathrm{x}-50)\sqrt{3}$

=> 2x = 150

=> x = 75 m

Thus, The distance of the first point from the tower is 75 m.