問題の解答
検索用コード（LaTeX）
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% 例題I4.2.5：正弦定理と余弦定理の利用（One More）★★★
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$\triangle\mathrm{ABC}$において，$\sin A:\sin B:\sin C=5:8:7$が成り立つとき，2番目に大きい角の大きさを求めよ．
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% 例題I4.2.5の解答（One More）★★★
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正弦定理$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$より，
\[
a:b:c=\sin A:\sin B:\sin C
\]
これと$\sin A:\sin B:\sin C=5:8:7$より，
\[
a:b:c=5:8:7
\]
したがって，2番目に長い辺は$c$であるから，$C$が2番目に大きい角である．
$a=5k,b=8k,c=7k(k>0)$とおくと，余弦定理より，
\begin{align*}
\cos C&=\frac{(5k)^2+(8k)^2-(7k)^2}{2\cdot 5k\cdot 8k}\\
&=\frac{25k^2+64k^2-49k^2}{80k^2}\\
&=\frac{40k^2}{80k^2}=\frac{1}{2}
\end{align*}
$0^{\circ}<C<180^{\circ}$より，$C=60^{\circ}$
よって，2番目に大きい角の大きさは，$60^{\circ}$である．
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% 問題I4.2.5：正弦定理と余弦定理の利用（One More）★★★
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$\triangle\mathrm{ABC}$において，$\sin A:\sin B:\sin C=3:5:7$が成り立つとき，最大の角の大きさを求めよ．
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% 問題I4.2.5の解答（One More）★★★
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正弦定理$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$より，
\[
a:b:c=\sin A:\sin B:\sin C
\]
したがって，$\sin A:\sin B:\sin C=3:5:7$であるから，
\[
a:b:c=3:5:7
\]
ゆえに，最も長い辺は$c$であるから，$C$が最大の角である．
$a=3k,b=5k,c=7k(k>0)$とおくと，余弦定理より，
\begin{align*}
\cos C&=\frac{(3k)^2+(5k)^2-(7k)^2}{2\cdot 3k\cdot 5k}\\
&=\frac{9k^2+25k^2-49k^2}{30k^2}\\
&=\frac{-15k^2}{30k^2}=-\frac{1}{2}
\end{align*}
$0^{\circ}<C<180^{\circ}$より，$C=120^{\circ}$
よって，最大の角の大きさは，$120^{\circ}$である．
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