【数学I】例題4.2.1：正弦定理（One More）★

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% 例題I4.2.1：正弦定理（One More）★
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$\triangle\mathrm{ABC}$において，次の値を求めよ．ただし，外接円の半径を$R$とする．

(1) $b=2,A=105^{\circ},C=30^{\circ}$のとき，$c,R$

(2) $R=3,a=3\sqrt{2}$のとき，$A$

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% 例題I4.2.1の解答（One More）★
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(1) $A+B+C=180^{\circ}$であるから，$B=180^{\circ}-(30^{\circ}+105^{\circ})=45^{\circ}$正弦定理より，

\[
\frac{2}{\sin 45^{\circ}}=\frac{c}{\sin 30^{\circ}}
\]

よって，

\[
c=\frac{2}{\sin 45^{\circ}}\cdot\sin 30^{\circ}=2\div\frac{1}{\sqrt{2}}\cdot\frac{1}{2}=\sqrt{2}
\]

また，$\frac{\sqrt{2}}{\sin 30^{\circ}}=2R$より，

\[
R=\frac{1}{2}\cdot\sqrt{2}\div\frac{1}{2}=\sqrt{2}
\]

(2) 正弦定理より，

\[
\frac{3\sqrt{2}}{\sin A}=2\cdot 3
\]

したがって，

\[
\sin A=\frac{3\sqrt{2}}{6}=\frac{\sqrt{2}}{2}
\]

よって，$0^{\circ}<A<180^{\circ}$より，$A=45^{\circ},135^{\circ}$

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% 問題I4.2.1：正弦定理（One More）★
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$\triangle\mathrm{ABC}$において，次の値を求めよ．ただし，外接円の半径を$R$とする．

(1) $a=3,A=60^{\circ},C=45^{\circ}$のとき，$c,R$

(2) $R=1,a=\sqrt{3}$のとき，$A$

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% 問題I4.2.1の解答（One More）★
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(1) 正弦定理より，

\[
\frac{3}{\sin 60^{\circ}}=\frac{c}{\sin 45^{\circ}}
\]

よって，

\[
c=\frac{3}{\sin 60^{\circ}}\cdot\sin 45^{\circ}=3\div\frac{\sqrt{3}}{2}\cdot\frac{1}{\sqrt{2}}={\frac{6}{\sqrt{6}}}=\sqrt{6}
\]

また，$\frac{3}{\sin 60^{\circ}}=2R$より，

\[
R=\frac{1}{2}\cdot 3\div\frac{\sqrt{3}}{2}=\sqrt{3}
\]

(2) 正弦定理より，

\[
\frac{\sqrt{3}}{\sin A}=2\cdot 1
\]

したがって，

\[
\sin A=\frac{\sqrt{3}}{2}
\]

よって，$0^{\circ}<A<180^{\circ}$より，

\[
A=60^{\circ},120^{\circ}
\]

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% 例題I4.2.1の側注：正弦定理（One More）★
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% 側注1
正弦定理$\frac{b}{\sin B}=\frac{c}{\sin C}$を用いる．両辺に$\sin 30^\circ$を掛けると，$c=\frac{2}{\sin 45^\circ}\cdot\sin 30^\circ$

% 側注2
$c=\frac{2}{\frac{1}{\sqrt{2}}}\cdot\frac{1}{2}$

% 側注3
正弦定理$\frac{a}{\sin A}=2R$を用いる．

% 側注4
図のように，鋭角と鈍角の2つの場合がある．

% 側注5
三角形の内角の和は$180^\circ$である．

% 側注6
正弦定理$\frac{c}{\sin C}=2R$を用いる．

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% 問題I4.2.1の側注：正弦定理（One More）★
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% 側注1
正弦定理$\frac{a}{\sin A}=\frac{c}{\sin C}$を用いる．両辺に$\sin 45^\circ$を掛けると，$c=\frac{3}{\sin 60^\circ}\cdot\sin 45^\circ$

% 側注2
正弦定理$\frac{a}{\sin A}=2R$を用いる．

% 側注3
図のように，鋭角と鈍角の2つの場合がある．

% 側注4
正弦定理$\frac{a}{\sin A}=2R$を用いる．

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% 例題I4.2.1の考え方・補足など：正弦定理（One More）★
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% 考え方
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正弦定理を利用する．このとき，与えられた辺や角に応じて，(i)の形や(ii)の形の等式を使うとよい．

\[
\underbrace{\frac{a}{\sin{A}}=\frac{b}{\sin{B}}}_{(\mathrm{i})}=\underbrace{\frac{c}{\sin{C}}=2R}_{(\mathrm{ii})}
\]

(1) $\frac{b}{\sin B}=\frac{c}{\sin C},\frac{c}{\sin C}=2R$の等式を用いる．

(2) $\frac{a}{\sin A}=2R$の等式を用いる．

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% 正弦定理
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\[
\frac{a}{\sin{A}}=\frac{b}{\sin{B}}=\frac{c}{\sin{C}}=2R
\]