回復 1# thankyou 的帖子
第1題
\(\begin{align}
& \sin \angle POE=\sin \left( 60{}^\circ -\angle POF \right) \\
& \frac{a}{r}=\frac{\sqrt{3}}{2}\times \frac{\sqrt{{{r}^{2}}-{{b}^{2}}}}{r}-\frac{1}{2}\times \frac{b}{r} \\
& ...... \\
\end{align}\)
第2題
令\(\angle ABP=\angle ACP=\theta \)
\(\begin{align}
& \angle PAB=60{}^\circ -\theta ,\angle PAC=120{}^\circ -\theta \\
& \\
& \frac{\overline{PB}}{\sin \left( 60{}^\circ -\theta \right)}=\frac{6}{\sin 120{}^\circ } \\
& \frac{\overline{PC}}{\sin \left( 120{}^\circ -\theta \right)}=\frac{6}{\sin 60{}^\circ } \\
& \\
& \overline{PB}=4\sqrt{3}\sin \left( 60{}^\circ -\theta \right) \\
& \overline{PC}=4\sqrt{3}\sin \left( 120{}^\circ -\theta \right) \\
& \\
& \Delta APB+\Delta APC \\
& =\frac{1}{2}\times 6\times 4\sqrt{3}\sin \left( 60{}^\circ -\theta \right)\times \sin \theta +\frac{1}{2}\times 6\times 4\sqrt{3}\sin \left( 120{}^\circ -\theta \right)\times \sin \theta \\
& =...... \\
\end{align}\)
[ 本帖最後由 thepiano 於 2016-10-9 09:09 PM 編輯 ]
