직각삼각형과 탄젠트

$\angle \mathrm{EAB} \mathit{=} \alpha$, $\angle \mathrm{CDB} \mathit{=} \beta$, $\displaystyle {\overline{\mathrm{BE}}} = x \left( 0 < x < \frac{1}{2} \right)$이라 하면 $\displaystyle {\overline{\mathrm{AD}}} = 2 x$, $\displaystyle {\overline{\mathrm{DB}}} = 1 - 2 x$ $\tan \alpha = x$, $\displaystyle \tan \beta = \frac{1}{1 - 2 x}$ $\tan \left( \angle \mathrm{CFE} \right)$$= \tan \left( \beta - \alpha \right)$ $\displaystyle = \frac{\tan \beta - \tan \alpha}{1 + \tan \beta \times \tan \alpha}$ $\displaystyle = \frac{\frac{1}{1 - 2 x} - x}{1 + \frac{1}{1 - 2 x} \times x}$ $\displaystyle = \frac{1 - x \left( 1 - 2 x \right)}{\left( 1 - 2 x \right) + x}$ $\displaystyle = \frac{2 x ^{2} - x + 1}{1 - x} = \frac{16}{15}$ $15 \left( 2 x ^{2} - x + 1 \right)$$= 16 \left( 1 - x \right)$ $30 x ^{2} + x - 1 = 0$ $\left( 5 x + 1 \right) \left( 6 x - 1 \right) = 0$ $\displaystyle 0 < x < \frac{1}{2}$이므로 $\displaystyle x = \frac{1}{6}$ 따라서 $\displaystyle \tan \left( \angle \mathrm{CDB} \right) = \frac{1}{1 - 2 x} = \frac{1}{1 - \frac{1}{3}} = \frac{3}{2}$