B C ‾ = k , B D ‾ = x \displaystyle \overline{BC} = k , \overline{BD} = x B C = k , B D = x A B ‾ = 2 k \displaystyle \overline{AB} = 2 k A B = 2 k A C ‾ = 3 k , \displaystyle \overline{AC} = \sqrt{3} k , A C = 3 k , C D ‾ = k − x \displaystyle \overline{CD} = k - x C D = k − x △ A C D \triangle ACD △ A C D k − x 3 k = tan ( π 6 − θ ) \displaystyle \frac{k - x}{\sqrt{3} k} = \tan \left( \frac{\pi}{6} - \theta \right) 3 k k − x = tan ( 6 π − θ ) 1 3 − x 3 k = tan π 6 − tan θ 1 + tan π 6 tan θ \displaystyle \frac{1}{\sqrt{3}} - \frac{x}{\sqrt{3} k} = \frac{\tan \frac{\pi}{6} - \tan \theta}{1 + \tan \frac{\pi}{6} \tan \theta} 3 1 − 3 k x = 1 + tan 6 π tan θ tan 6 π − tan θ = 1 3 − tan θ 1 + 1 3 tan θ \displaystyle = \frac{\frac{1}{\sqrt{3}} - \tan \theta}{1 + \frac{1}{\sqrt{3}} \tan \theta} = 1 + 3 1 tan θ 3 1 − tan θ = 1 − 3 tan θ 3 + tan θ \displaystyle = \frac{1 - \sqrt{3} \tan \theta}{\sqrt{3} + \tan \theta} = 3 + tan θ 1 − 3 tan θ ∴ 1 − x k = 3 − 3 tan θ 3 + tan θ \displaystyle \therefore 1 - \frac{x}{k} = \frac{\sqrt{3} - 3 \tan \theta}{\sqrt{3} + \tan \theta} ∴ 1 − k x = 3 + tan θ 3 − 3 tan θ x k = 1 − 3 − 3 tan θ 3 + tan θ \displaystyle \frac{x}{k} = 1 - \frac{\sqrt{3} - 3 \tan \theta}{\sqrt{3} + \tan \theta} k x = 1 − 3 + tan θ 3 − 3 tan θ = 4 tan θ 3 + tan θ \displaystyle = \frac{4 \tan \theta}{\sqrt{3} + \tan \theta} = 3 + tan θ 4 tan θ = 4 sin θ cos θ 3 + sin θ cos θ \displaystyle = \frac{4 \frac{\sin \theta}{\cos \theta}}{\sqrt{3} + \frac{\sin \theta}{\cos \theta}} = 3 + c o s θ s i n θ 4 c o s θ s i n θ = 4 sin θ 3 cos θ + sin θ \displaystyle = \frac{4 \sin \theta}{\sqrt{3} \cos \theta + \sin \theta} = 3 cos θ + sin θ 4 sin θ ∴ B D ‾ A B ‾ = x 2 k \displaystyle \therefore \frac{\overline{BD}}{\overline{AB}} = \frac{x}{2 k} ∴ A B B D = 2 k x = 2 sin θ 3 cos θ + sin θ \displaystyle = \frac{2 \sin \theta}{\sqrt{3} \cos \theta + \sin \theta} = 3 cos θ + sin θ 2 sin θ
