Triangle DCB, \( \Large \tan \theta = \frac{h}{x} \)



and in \( \Large \triangle ECA \tan \theta = \frac{h+6}{2\sqrt{3}+x} \)

=> \( \Large \frac{h}{x} = \frac{h+6}{2\sqrt{3}+x} \) [from (i)]

=> \( \Large 2\sqrt{3}h + hx = hx + 6x \)

=> \( \Large 2\sqrt{3}h = 6x => h = \frac{6x}{2\sqrt{3}} \)

From equation (i)

\( \Large \tan \theta = \frac{6x}{2\sqrt{3x}} = \sqrt{3} => \theta = 60 ^{\circ} \)