PART 15：餘弦定律-例題

於 \(\vartriangle ABC\) ，若 \(\overline {BC}  = 4\sqrt 2 \) ， \(\overline {CA}  = 5\) ，

則 \(\cos A = ?\)  \(\sin A = ?\)  \(\vartriangle ABC\) 之外接圓半徑 \(R\) 為?

SOL:

\(\cos A = \frac{{{b^2} + {c^2} - {a^2}}}{{2bc}}\)  \( = \frac{{{5^2} + {7^2} - {{\left( {4\sqrt 2 } \right)}^2}}}{{2 \cdot 5 \cdot 7}}\)  \( = \frac{{25 + 49 - 32}}{{70}} = \frac{3}{5}\) ，

三角恆等式 \({\sin ^2}A + {\cos ^2}A = 1\) ， \({\sin ^2}A = \frac{{16}}{{25}}\) \(  \Rightarrow \sin A =  \pm \frac{4}{5}\) (取正)

\(\frac{a}{{\sin A}} = 2R\) ， \( \Rightarrow \frac{{4\sqrt 2 }}{{4/5}} = 2R\) \( \Rightarrow R = \frac{{5\sqrt 2 }}{2}\)