回覆 30# lovejade 的帖子
第 6 題
\(\begin{align}
& {{a}^{2}}={{b}^{2}}+{{c}^{2}}-2bc\cos A \\
& \\
& \overline{AD}=d \\
& {{b}^{2}}+{{c}^{2}}=2{{\left( \frac{1}{2}a \right)}^{2}}+2{{d}^{2}} \\
& {{d}^{2}}=\frac{{{b}^{2}}+{{c}^{2}}-\frac{1}{2}{{a}^{2}}}{2} \\
& =\frac{{{b}^{2}}+{{c}^{2}}-\frac{1}{2}\left( {{b}^{2}}+{{c}^{2}}-2bc\cos A \right)}{2} \\
& =\frac{{{b}^{2}}+{{c}^{2}}+2bc\cos A}{4} \\
& \\
& {{\left( \Delta ABC \right)}^{2}}=4{{\left( \Delta ADC \right)}^{2}} \\
& \frac{1}{4}{{b}^{2}}{{c}^{2}}{{\sin }^{2}}A=4{{\left( \frac{1}{4}ad\sin \angle ADC \right)}^{2}}=\frac{1}{4}{{a}^{2}}{{d}^{2}}{{\sin }^{2}}\angle ADC \\
& {{\sin }^{2}}\angle ADC=\frac{{{b}^{2}}{{c}^{2}}{{\sin }^{2}}A}{{{a}^{2}}{{d}^{2}}} \\
& =\frac{{{b}^{2}}{{c}^{2}}{{\sin }^{2}}A}{\left( {{b}^{2}}+{{c}^{2}}-2bc\cos A \right)\left( \frac{{{b}^{2}}+{{c}^{2}}+2bc\cos A}{4} \right)} \\
& =\frac{4{{b}^{2}}{{c}^{2}}{{\sin }^{2}}A}{{{\left( {{b}^{2}}+{{c}^{2}} \right)}^{2}}-4{{b}^{2}}{{c}^{2}}{{\cos }^{2}}A} \\
& =\frac{4{{b}^{2}}{{c}^{2}}-4{{b}^{2}}{{c}^{2}}{{\cos }^{2}}A}{{{\left( {{b}^{2}}+{{c}^{2}} \right)}^{2}}-4{{b}^{2}}{{c}^{2}}{{\cos }^{2}}A}\quad \left( 0\le {{\cos }^{2}}A<1\ ,\ {{\left( {{b}^{2}}+{{c}^{2}} \right)}^{2}}\ge 4{{b}^{2}}{{c}^{2}} \right) \\
& \le \frac{4{{b}^{2}}{{c}^{2}}}{{{\left( {{b}^{2}}+{{c}^{2}} \right)}^{2}}} \\
& \\
& \sin \angle ADC\le \frac{2bc}{{{b}^{2}}+{{c}^{2}}} \\
& \\
\end{align}\)
[ 本帖最後由 thepiano 於 2023-4-11 18:13 編輯 ]
