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