정육각형과 코사인법칙

$\displaystyle {\mathrm{\overline{AG}}} = \sqrt{5}$, $\displaystyle {\mathrm{\overline{AB}}} = 2$이므로 직각삼각형 $\mathrm{GBA}$에서 $\displaystyle {\mathrm{\overline{BG}}} = 3$이고 삼각형 $\mathrm{BGH}$는 정삼각형이므로 $\displaystyle {\mathrm{\overline{BH}}} = 3$ $\angle \mathrm{GBA} \mathit{=} \theta$라 하면 $\displaystyle \cos \theta = \frac{2}{3}$, $\displaystyle \angle \mathrm{CBH} = 2 \pi - \left( \frac{\pi}{3} \times 3 + \theta \right) = \pi - \theta$ 삼각형 $\mathrm{CBH}$에서 코사인법칙에 의하여 $\displaystyle {\mathrm{\overline{CH}}} ^{2}$$\displaystyle = {\mathrm{\overline{BC}}} ^{2} + {\mathrm{\overline{BH}}} ^{2} - 2 \times {\mathrm{\overline{BC}}} \times {\mathrm{\overline{BH}}} \times \cos \left( \angle \mathrm{CBH} \right)$ $\displaystyle = \mathrm{\overline{BC}} ^{2} + \mathrm{\overline{BH}} ^{2} - 2 \times \mathrm{\overline{BC}} \times \mathrm{\overline{BH}} \times \cos \left( \pi - \theta \right)$ $\displaystyle = 2 ^{2} + 3 ^{2} + 2 \times 2 \times 3 \times \frac{2}{3}$ $= 21$ $\therefore$$\displaystyle \mathrm{\overline{CH}} = \sqrt{21}$