【数学I】例題4.2.4：三角形の辺と角2（One More）★★

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 例題I4.2.4：三角形の辺と角2（One More）★★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

$\triangle\mathrm{ABC}$において，$b=2\sqrt{3},c=2,C=30^{\circ}$のとき，残りの辺の長さと角の大きさを求めよ．

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 例題I4.2.4の解答（One More）★★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

余弦定理より，$2^2=a^2+(2\sqrt{3})^2-2\cdot a\cdot 2\sqrt{3}\cdot\cos 30^{\circ}$

したがって，$a^2-6a+8=0$

ゆえに，$(a-2)(a-4)=0$

これを解くと，$a=2,4$

(i) $a=4$のとき

余弦定理より，$\cos A=\frac{(2\sqrt{3})^2+2^2-4^2}{2\cdot 2\sqrt{3}\cdot 2}=0$

したがって，$A=90^{\circ}$

ゆえに，$B=180^{\circ}-(90^{\circ}+30^{\circ})=60^{\circ}$

(ii) $a=2$のとき

余弦定理より，$\cos A=\frac{(2\sqrt{3})^2+2^2-2^2}{2\cdot 2\sqrt{3}\cdot 2}=\frac{\sqrt{3}}{2}$

したがって，$A=30^{\circ}$

ゆえに，$B=180^{\circ}-(30^{\circ}+30^{\circ})=120^{\circ}$

よって，(i)，(ii)より，$a=4,A=90^{\circ},B=60^{\circ}$または$a=2,A=30^{\circ},B=120^{\circ}$

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 例題I4.2.4の別解
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

正弦定理より，$\frac{2\sqrt{3}}{\sin B}=\frac{2}{\sin 30^{\circ}}$であるから，$\sin B=\sqrt{3}\cdot\sin 30^\circ=\frac{\sqrt{3}}{2}$

したがって，$B=60^\circ,120^\circ$

(i) $B=60^\circ$のとき，$A=180^\circ-(60^\circ+30^\circ)=90^\circ$であるから，$a=4$

(ii) $B=120^\circ$のとき，$A=180^\circ-(120^\circ+30^\circ)=30^\circ$

$\triangle\mathrm{ABC}$は，$\mathrm{BA}=\mathrm{BC}$の二等辺三角形であるから，$a=2$

よって，(i)，(ii)より，$a=4,A=90^{\circ},B=60^{\circ}$または$a=2,A=30^{\circ},B=120^{\circ}$

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 問題I4.2.4：三角形の辺と角2（One More）★★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

$\triangle\mathrm{ABC}$において，$b=2\sqrt{2},c=2,C=30^{\circ}$のとき，残りの辺の長さと角の大きさを求めよ．

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 問題I4.2.4の解答（One More）★★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

余弦定理より，$2^2=a^2+(2\sqrt{2})^2-2\cdot a\cdot 2\sqrt{2}\cdot\cos 30^{\circ}$

したがって，$a^2-2\sqrt{6}a+4=0$

これを解くと，$a=\sqrt{6}\pm\sqrt{2}$

(i) $a=\sqrt{6}+\sqrt{2}$のとき

余弦定理より，

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

したがって，$B=45^{\circ}$

ゆえに，$A=180^{\circ}-(45^{\circ}+30^{\circ})=105^{\circ}$

(ii) $a=\sqrt{6}-\sqrt{2}$のとき

余弦定理より，

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

したがって，$B=135^{\circ}$

ゆえに，$A=180^{\circ}-(135^{\circ}+30^{\circ})=15^{\circ}$

よって，(i)，(ii)より，$a=\sqrt{6}+\sqrt{2},A=105^{\circ},B=45^{\circ}$または$a=\sqrt{6}-\sqrt{2},A=15^{\circ},B=135^{\circ}$

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 問題I4.2.4の別解
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

正弦定理より，$\frac{2\sqrt{2}}{\sin B}=\frac{2}{\sin 30^{\circ}}$であるから，$\sin B=\sqrt{2}\cdot\sin 30^\circ=\frac{\sqrt{2}}{2}$

したがって，$B=45^\circ,135^\circ$

(i) $B=45^\circ$のとき，$A=180^\circ-(45^\circ+30^\circ)=105^\circ$

このとき，

\begin{align*}
a&=c\cos B+b\cos C\\
&=2\cos 45^{\circ}+2\sqrt{2}\cos 30^{\circ}=\sqrt{6}+\sqrt{2}
\end{align*}

(ii) $B=135^\circ$のとき，$A=180^\circ-(135^\circ+30^\circ)=15^\circ$

このとき，

\begin{align*}
a&=b\cos C+c\cos B\\
&=2\sqrt{2}\cos 30^{\circ}+2\cos 135^{\circ}=\sqrt{6}-\sqrt{2}
\end{align*}

よって，(i)，(ii)より，$a=\sqrt{6}+\sqrt{2},A=105^{\circ},B=45^{\circ}$または$a=\sqrt{6}-\sqrt{2},A=15^{\circ},B=135^{\circ}$

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 例題I4.2.4の側注：三角形の辺と角2（One More）★★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

% 側注1
$B=180^\circ-(A+C)$

% 側注2
$B=180^\circ-(A+C)$

% 側注3
いずれも$a>0$であるから，場合分けをして考える．

% 側注4
有名な角の三角形となる．

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 問題I4.2.4の側注：三角形の辺と角2（One More）★★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

% 側注1
計算に手間が掛かる（別解のように，第1余弦定理を用いると計算が楽になる）．

% 側注2
$A=180^\circ-(B+C)$

% 側注3
いずれも$a>0$であるから，場合分けをして考える．

% 側注4
第1余弦定理より，

\[
a=c\cos B+b\cos C
\]

第1余弦定理を用いると，計算が楽になり，役に立つことがある．

% 側注5
第1余弦定理は，$90^\circ$や鈍角についても成り立つ．

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 例題I4.2.4の考え方・補足など：三角形の辺と角2（One More）★★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 考え方
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

余弦定理を用いるとよい（正弦定理を用いた別解も考えられる）．三角形が1通りに定まらない場合であるので，場合分けをして考える．

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 余談
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

正弦定理を用いた別解において，例題では有名な角の三角形や二等辺三角形であることを利用したが，利用できない問題もある．そのような場合は，余弦定理や三平方の定理を用いた解法も考えられるが，第1余弦定理（既習の余弦定理は第2余弦定理）という定理が役に立つことがあるので下に紹介する．

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 第1余弦定理
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

$\triangle\mathrm{ABC}$において，次のような関係式が成り立つ．

\[
a=b\cos C+c\cos B,b=a\cos C+c\cos A,c=a\cos B+b\cos A
\]