두 정사각형 코사인법칙

$\angle \mathrm{DCG} = \theta$ ($0 < \theta < \pi$), $\angle \mathrm{BCE} = \pi - \theta$ $\displaystyle \sin \theta = \frac{\sqrt{11}}{6}$이므로 $\displaystyle \cos ^{2} \theta = 1 - \sin ^{2} \theta = \frac{25}{36}$ 코사인법칙에 의하여 $\displaystyle {\overline{\mathrm{DG}}} ^{2} = 3 ^{2} + 4 ^{2} - 2 \times 3 \times 4 \times \cos \theta$$= 25 - 24 \cos \theta$ $\displaystyle {\overline{\mathrm{BE}}} ^{2} = 3 ^{2} + 4 ^{2} - 2 \times 3 \times 4 \times \cos ( \pi - \theta )$ $= 25 - 24 \cos ( \pi - \theta )$$= 25 + 24 \cos \theta$ 따라서 $\displaystyle \overline{\mathrm{DG}} \times \overline{\mathrm{BE}} = \sqrt{25 ^{2} - 24 ^{2} \times \cos ^{2} \theta}$ $\displaystyle = \sqrt{25 ^{2} - 24 ^{2} \times \frac{25}{36}}$ $\displaystyle = 5 \sqrt{25 - 16} = 15$