【数学I】例題4.2.3：三角形の辺と角1（One More）★★

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% 例題I4.2.3：三角形の辺と角1（One More）★★
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次の場合について，$\triangle\mathrm{ABC}$の残りの辺の長さと角の大きさを求めよ．

(1) $b=2\sqrt{3},c=3-\sqrt{3},A=120^{\circ}$

(2) $a=\sqrt{2},b=2,c=\sqrt{3}-1$

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% 例題I4.2.3の解答（One More）★★
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(1) 余弦定理より，

\[
a^2=(2\sqrt{3})^2+(3-\sqrt{3})^2-2\cdot 2\sqrt{3}\cdot(3-\sqrt{3})\cdot\cos 120^{\circ}=18
\]

したがって，$a>0$であるから，$a=\sqrt{18}=3\sqrt{2}$

余弦定理より，

\[
\cos B=\frac{(3-\sqrt{3})^2+(3\sqrt{2})^2-(2\sqrt{3})^2}{2\cdot(3-\sqrt{3})\cdot 3\sqrt{2}}=\frac{1}{\sqrt{2}}
\]

ゆえに，$B=45^{\circ}$

よって，$C=180^{\circ}-(120^{\circ}+45^{\circ})=15^{\circ}$

(2) 余弦定理より，

\[
\cos A=\frac{2^2+(\sqrt{3}-1)^2-(\sqrt{2})^2}{2\cdot 2\cdot(\sqrt{3}-1)}=\frac{6-2\sqrt{3}}{2\cdot 2(\sqrt{3}-1)}=\frac{2\sqrt{3}(\sqrt{3}-1)}{2\cdot 2(\sqrt{3}-1)}=\frac{\sqrt{3}}{2}
\]

したがって，$A=30^{\circ}$

余弦定理より，

\begin{align*}
\cos B&=\frac{(\sqrt{3}-1)^2+(\sqrt{2})^2-2^2}{2\cdot(\sqrt{3}-1)\cdot\sqrt{2}}\\
&=\frac{2-2\sqrt{3}}{2\sqrt{2}(\sqrt{3}-1)}\\
&=\frac{-2(\sqrt{3}-1)}{2\sqrt{2}(\sqrt{3}-1)}=-\frac{1}{\sqrt{2}}
\end{align*}

したがって，$B=135^{\circ}$

よって，$C=180^{\circ}-(30^{\circ}+135^{\circ})=15^{\circ}$

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% 問題I4.2.3：三角形の辺と角1（One More）★★
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次の場合について，$\triangle\mathrm{ABC}$の残りの辺の長さと角の大きさを求めよ．

(1) $b=\sqrt{6}-\sqrt{2},c=2,A=135^{\circ}$

(2) $a=2\sqrt{2},b=2,c=\sqrt{6}+\sqrt{2}$

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% 問題I4.2.3の解答（One More）★★
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(1) 余弦定理より，

\[
a^2=(\sqrt{6}-\sqrt{2})^2+2^2-2\cdot(\sqrt{6}-\sqrt{2})\cdot 2\cdot\cos 135^{\circ}=8
\]

したがって，$a>0$であるから，$a=\sqrt{8}=2\sqrt{2}$

正弦定理より，

\[
\frac{2}{\sin C}=\frac{2\sqrt{2}}{\sin 135^\circ}
\]

ゆえに，$\sin C=\frac{1}{2}$

$0^{\circ}<C<180^{\circ}-135^{\circ}$より，$0^{\circ}<C<45^{\circ}$であるから，$C=30^{\circ}$

よって，$B=180^{\circ}-(135^{\circ}+30^{\circ})=15^{\circ}$

(2) 余弦定理より，

\[
\cos A=\frac{2^2+(\sqrt{6}+\sqrt{2})^2-(2\sqrt{2})^2}{2\cdot 2\cdot(\sqrt{6}+\sqrt{2})}=\frac{\sqrt{2}}{2}
\]

したがって，$A=45^{\circ}$

余弦定理より，

\begin{align*}
\cos B&=\frac{(\sqrt{6}+\sqrt{2})^2+(2\sqrt{2})^2-2^2}{2\cdot(\sqrt{6}+\sqrt{2})\cdot 2\sqrt{2}}\\
&=\frac{12+4\sqrt{3}}{4\sqrt{2}(\sqrt{6}+\sqrt{2})}\\
&=\frac{\sqrt{3}}{2}
\end{align*}

したがって，$B=30^{\circ}$

よって，$C=180^{\circ}-(45^{\circ}+30^{\circ})=105^{\circ}$

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% 例題I4.2.3の側注：三角形の辺と角1（One More）★★
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% 側注1
$C=180^\circ-(A+B)$

% 側注2
$\cos B$から考えてもよい．

% 側注3
$C=180^\circ-(A+B)$

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% 問題I4.2.3の側注：三角形の辺と角1（One More）★★
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% 側注1
余弦定理でも求めることができるが，計算に手間が掛かる．そこで，正弦定理を用いるとよい．

% 側注2
$C=150^\circ$は不適である．

% 側注3
$\cos B$から考えてもよい．

% 側注4
$C=180^\circ-(A+B)$

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% 例題I4.2.3の考え方・補足など：三角形の辺と角1（One More）★★
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% 考え方
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与えられた辺や角に応じて，余弦定理（正弦定理）を利用する．正弦定理，余弦定理のいずれかを使うかの判断は，与えられた辺や角を図示して考えるとよい．なお，一般に，2つの角と1辺が与えられているときは正弦定理を使い，3辺や2辺とその間の角が与えられているときは余弦定理を使うとよい．

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例えば，(1)は$a$を求めた後に$B$を求めるのに，次のように正弦定理を用いてもよい．

正弦定理より，$\frac{3\sqrt{2}}{\sin 120^{\circ}}=\frac{2\sqrt{3}}{\sin B}$であるから，$\sin B=\frac{2\sqrt{3}\sin 120^{\circ}}{3\sqrt{2}}=\frac{1}{\sqrt{2}}$

$A=120^{\circ}$より，$0^{\circ}<B<60^{\circ}$であるから，$B=45^{\circ}$となり，$C=180^{\circ}-(120^{\circ}+45^{\circ})=15^{\circ}$

このように，与えられた辺や角によって，自在に正弦定理と余弦定理を扱うことができるようになるとよい．ただし，$\sin B=\frac{1}{\sqrt{2}}$となる$B$は$45^\circ,135^\circ$と2通りの可能性が導かれることもあるので，辺や角度の大小関係についての吟味を忘れないように注意すること．