사인법칙과 코사인법칙으로 삼각형 둘레 구하기

삼각형 $\mathrm{ABC}$에서 사인법칙에 의하여 $\displaystyle \frac{\mathrm{\overline{BC}}}{\sin \frac{\pi}{3}} = 2 \times 4$ $\therefore$ $\displaystyle \mathrm{\overline{BC}} \mathit{=} 4 \sqrt{3}$ 삼각형 $\mathrm{ABC}$의 넓이가 $\displaystyle 4 \sqrt{3}$이므로 $\displaystyle \frac{1}{2} \times \mathrm{\overline{CA}} \times \overline{AB} \times \sin \frac{\pi}{3} = 4 \sqrt{3}$ $\therefore$ $\displaystyle \mathrm{\overline{CA}} \times \overline{AB} = 16$ 삼각형 $\mathrm{ABC}$에서 코사인법칙에 의하여 $48$$\displaystyle = \mathrm{\overline{CA}} ^{2} + \mathrm{\overline{AB}} ^{2} - 2 \times \overline{CA} \times \overline{AB} \times \cos \frac{\pi}{3}$ $\displaystyle = \left( \mathrm{\overline{CA} + \overline{AB} ) ^{2} - 3 \overline{CA} \times \overline{AB}} \right.$ $\displaystyle = \left( \mathrm{\overline{CA}} + \overline{AB} \right) ^{2} - 3 \times 16$ $\therefore$ $\displaystyle \mathrm{\overline{CA}} + \overline{AB} = 4 \sqrt{6}$ $\therefore$ $\displaystyle \mathrm{\overline{AB}} + \mathrm{\overline{BC}} + \overline{CA} = 4 \sqrt{3} + 4 \sqrt{6} = 4 \left( \sqrt{3} + \sqrt{6} \right)$