삼각형 넓이 극한

점$C$에서 선분$AD$에 내린 수선의 발을 점$H$라 하면 $\displaystyle \angle ACH = \frac{\pi}{2} - 2 \theta$, $\displaystyle \angle DCH = 2 \theta + \frac{\pi}{6}$, $\displaystyle \angle DCB = \frac{5}{6} \pi$ $\displaystyle {\overline{BC}} = \sin \theta$, $\displaystyle {\overline{AC}} = \cos \theta$, $\displaystyle {\overline{CH}} = \cos \theta \sin 2 \theta$ $\displaystyle \therefore {\overline{CD}} = \frac{\cos \theta \sin 2 \theta}{\cos \left( 2 \theta + \frac{\pi}{6} \right)}$ $\displaystyle \therefore S ( \theta ) = \frac{1}{2} {\overline{BC}} {\overline{CD}} \sin \frac{5}{6} \pi = \frac{\sin \theta \cos \theta \sin 2 \theta}{4 \cos \left( 2 \theta + \frac{\pi}{6} \right)}$ $\displaystyle \lim\limits _{\theta \rightarrow 0} {\frac{S ( \theta )}{\theta ^{2}}}$ $\displaystyle = \lim\limits _{\theta \rightarrow 0} {\frac{\sin \theta}{\theta} \times \frac{\sin 2 \theta}{\theta} \times \frac{\cos \theta}{4 \cos \left( 2 \theta + \frac{\pi}{6} \right)}}$ $\displaystyle = \frac{1}{\sqrt{3}} = p$ ∴ $300 p ^{2} = 100$