삼각형 각과 탄젠트

[출제의도] 삼각함수의 덧셈정리를 이용하여 추론하기 삼각형 $\mathrm{BCD}$는 이등변삼각형이므로 $\angle \mathrm{CBD}$$= \angle \mathrm{DCB}$$= \alpha$이고 $\angle \mathrm{CDA}$$= 2 \alpha$ 삼각형 $\mathrm{ADC}$에서 $\beta$$\displaystyle = \frac{\pi}{3} - 2 \alpha$ $\cos 2 \alpha$$= \cos \alpha \cos \alpha - \sin \alpha \sin \alpha$ $= \cos ^{2} \alpha - \sin ^{2} \alpha$ $= 2 \cos ^{2} \alpha - 1$$\displaystyle = \frac{\sqrt{21}}{7}$ $\sin ^{2} 2 \alpha$$= 1 - \cos ^{2} 2 \alpha$$\displaystyle = \frac{28}{49}$이고 $0 < 2 \alpha < \pi$이므로 $\sin 2 \alpha$$\displaystyle = \frac{2 \sqrt{7}}{7}$ $\tan 2 \alpha$$\displaystyle = \frac{\sin 2 \alpha}{\cos 2 \alpha}$$\displaystyle = \frac{2 \sqrt{3}}{3}$ $\tan \beta$$\displaystyle = \tan \left( \frac{\pi}{3} - 2 \alpha \right)$ $\displaystyle = \frac{\tan \frac{\pi}{3} - \tan 2 \alpha}{1 + \tan \frac{\pi}{3} \times \tan 2 \alpha}$ $\displaystyle = \frac{\sqrt{3} - \frac{2 \sqrt{3}}{3}}{1 + \sqrt{3} \times \frac{2 \sqrt{3}}{3}}$$\displaystyle = \frac{\sqrt{3}}{9}$ 따라서 $\displaystyle 54 \sqrt{3} \times \tan \beta$$= 18$