A vertical pole fixed to the horizontal ground is divided in the...

Given that

A vertical pole fixed to the horizontal ground is divided in the ratio $3: 7$ by a mark on it with lower part shorter than the upper part. If the two parts subtend equal angles at a point on the ground $18m$ away from the base of the pole.

$tan\theta = \dfrac{x}{6}$
$\tan 2\theta = \dfrac{10x}{18} = \dfrac{5x}{9}$

We know that $\tan 2\theta = \dfrac{2tan\theta}{1 - \tan^{2}\theta}$

$\dfrac{5x}{9} = \dfrac{\dfrac{x}{3}}{1 - \dfrac{{x}^{2}}{36}}$

$⇒ 5 = \dfrac{3}{1 - \dfrac{{x}^{2}}{36}}$

$1 - \dfrac{{x}^{2}}{36} = \dfrac{3}{5}$

$\dfrac{{x}^{2}}{36} = \dfrac{2}{5}$

$⇒ {x}^{2} = \dfrac{72}{5} ⇒ x = \dfrac{6\sqrt{2}}{\sqrt{5}}$

From the figure ;

Length of pole $= 10x = \dfrac{60\sqrt{2}}{\sqrt{5}} \times \dfrac{\sqrt{5}}{\sqrt{5}}$

$= \dfrac{60\sqrt{10}}{5} = 12\sqrt{10}m$