\(\theta=\)?
問い(\(\tan{\left(\theta\right)=}\))
\(\overline{AC}\)に垂直な補助線\(\overline{BD}\)を引く
$$\begin{eqnarray}
\angle C &=& \pi - \left(\angle A + \angle B\right)
\\ &=& \pi-\left\{\theta+\left(\frac{\pi}{2}+\theta\right)\right\}
\\ &=& \frac{\pi}{2}-2\theta
\end{eqnarray}$$
$$\begin{eqnarray}
\overline{BD}&=&2\sin{\left(\frac{\pi}{2}-2\theta\right)}
\\\overline{CD}&=&2\cos{\left(\frac{\pi}{2}-2\theta\right)}
\\\overline{AD}&=&\overline{AC}-\overline{CD}
\\&=&5-2\cos{\left(\frac{\pi}{2}-2\theta\right)}
\end{eqnarray}$$
$$\begin{eqnarray}
\tan{\left(\theta\right)}&=&\frac{\overline{BD}}{\overline{AD}}
\\&=&\frac{2\sin{\left(\frac{\pi}{2}-2\theta\right)}}{5-2\cos{\left(\frac{\pi}{2}-2\theta\right)}}
\\&=&\frac{2\cos{\left(2\theta\right)}}{5-2\sin{\left(2\theta\right)}}
\;\cdots\;\cos{\left(\frac{\pi}{2}-x\right)}=\sin{\left(x\right)},\;\sin{\left(\frac{\pi}{2}-x\right)}=\cos{\left(x\right)}
\\&=&\frac{2}{5}\;\cdots\;\tan{\left(\theta\right)}=\frac{B\cos{\left(2\theta\right)}}{A-B\sin{\left(2\theta\right)}}ならば\tan{\left(\theta\right)}=\frac{B}{A}(下記)
\end{eqnarray}$$
$$\begin{eqnarray}
\tan{\left(\theta\right)}&=&\frac{B\cos{\left(2\theta\right)}}{A-B\sin{\left(2\theta\right)}}
\\\frac{1}{\tan{\left(\theta\right)}}&=&\frac{A-B\sin{\left(2\theta\right)}}{B\cos{\left(2\theta\right)}}
\\&=&\frac{1}{\cos{\left(2\theta\right)}}\left\{\frac{A}{B}-\frac{\cancel{B}}{\cancel{B}}\sin{\left(2\theta\right)}\right\}
\\\frac{\cos{\left(2\theta\right)}}{\tan{\left(\theta\right)}}&=&\frac{A}{B}-\sin{\left(2\theta\right)}
\\\frac{\cos{\left(2\theta\right)}}{\tan{\left(\theta\right)}}+\sin{\left(2\theta\right)}&=&\frac{A}{B}
\\\frac{\cos{\left(2\theta\right)}+\tan{\left(\theta\right)}\sin{\left(2\theta\right)}}{\tan{\left(\theta\right)}}&=&\frac{A}{B}
\\\frac{\cos^2{\left(\theta\right)}-\sin^2{\left(\theta\right)}+2\tan{\left(\theta\right)}\sin{\left(\theta\right)}\cos{\left(\theta\right)}}{\tan{\left(\theta\right)}}&=&\frac{A}{B}
\;\cdots\;\cos{\left(2\theta\right)}=\cos^2{\left(\theta\right)}-\sin^2{\left(\theta\right)},\;\sin{\left(2\theta\right)}=2\sin{\left(\theta\right)}\cos{\left(\theta\right)}
\\\frac{\cos^2{\left(\theta\right)}-\sin^2{\left(\theta\right)}+2\frac{\sin{\left(\theta\right)}}{\cancel{\cos{\left(\theta\right)}}}\sin{\left(\theta\right)}\cancel{\cos{\left(\theta\right)}}}{\tan{\left(\theta\right)}}&=&\frac{A}{B}
\;\cdots\;\tan{\left(\theta\right)}=\frac{\sin{\left(\theta\right)}}{\cos{\left(\theta\right)}}
\\\frac{\cos^2{\left(\theta\right)}-\sin^2{\left(\theta\right)}+2\sin^2{\left(\theta\right)}}{\tan{\left(\theta\right)}}&=&\frac{A}{B}
\\\frac{\cos^2{\left(\theta\right)}+\sin^2{\left(\theta\right)}}{\tan{\left(\theta\right)}}&=&\frac{A}{B}
\\\frac{1}{\tan{\left(\theta\right)}}&=&\frac{A}{B}\;\cdots\;\cos^2{\left(\theta\right)}+\sin^2{\left(\theta\right)}=1
\\\tan{\left(\theta\right)}&=&\frac{B}{A}
\end{eqnarray}$$
