【数学I】例題4.3.3：三角形の内接円と外接円の半径（One More）★★

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% 例題I4.3.3：三角形の内接円と外接円の半径（One More）★★
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$\triangle\mathrm{ABC}$において，$a=3,b=\sqrt{5},c=2$とする．このとき，次の値を求めよ．

(1) $\cos B,\sin B$

(2) $\triangle\mathrm{ABC}$の面積$S$

(3) $\triangle\mathrm{ABC}$の内接円の半径$r$

(4) $\triangle\mathrm{ABC}$の外接円の半径$R$

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% 例題I4.3.3の解答（One More）★★
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(1) 余弦定理より，$\cos B=\frac{2^2+3^2-(\sqrt{5})^2}{2\cdot 3\cdot 2}=\frac{2}{3}$

また，$\sin B>0$より，

\[
\sin B=\sqrt{1-(\frac{2}{3})^2}=\frac{\sqrt{5}}{3}
\]

(2) 求める$\triangle\mathrm{ABC}$の面積$S$は，

\[
S=\frac{1}{2}ca\sin B=\frac{1}{2}\cdot 3\cdot 2\cdot\frac{\sqrt{5}}{3}=\sqrt{5}
\]

(3) (2)，$S=\frac{1}{2}r(a+b+c)$より，

\[
\sqrt{5}=\frac{1}{2}r(3+\sqrt{5}+2)
\]

よって，

\[
r=\frac{2\sqrt{5}}{5+\sqrt{5}}=\frac{2\sqrt{5}(5-\sqrt{5})}{20}=\frac{\sqrt{5}-1}{2}
\]

(4) 正弦定理より，

\[
R=\frac{b}{2\sin B}=\sqrt{5}\div(2\cdot\frac{\sqrt{5}}{3})=\frac{3}{2}
\]

よって，外接円の半径は，$R=\frac{3}{2}$

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% 問題I4.3.3：三角形の内接円と外接円の半径（One More）★★
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$\triangle\mathrm{ABC}$において，$a=4,b=5,c=6$とする．このとき，次の値を求めよ．

(1) $\cos A,\sin A$

(2) $\triangle\mathrm{ABC}$の面積$S$

(3) $\triangle\mathrm{ABC}$の内接円の半径$r$

(4) $\triangle\mathrm{ABC}$の外接円の半径$R$

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% 問題I4.3.3の解答（One More）★★
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(1) 余弦定理より，$\cos A=\frac{5^2+6^2-4^2}{2\cdot 5\cdot 6}=\frac{3}{4}$

また，$\sin A>0$より，

\[
\sin A=\sqrt{1-(\frac{3}{4})^2}=\sqrt{\frac{16}{16}-\frac{9}{16}}=\sqrt{\frac{7}{16}}=\frac{\sqrt{7}}{4}
\]

(2) 求める$\triangle\mathrm{ABC}$の面積$S$は，$S=\frac{1}{2}bc\sin A=\frac{1}{2}\cdot 5\cdot 6\cdot\frac{\sqrt{7}}{4}=\frac{15\sqrt{7}}{4}$

(3) (2)，$S=\frac{1}{2}r(a+b+c)$より，$\frac{15\sqrt{7}}{4}=\frac{1}{2}r(4+5+6)$

よって，$r=\frac{2\cdot 15\sqrt{7}}{4\cdot 15}=\frac{\sqrt{7}}{2}$

(4) 正弦定理より，$R=\frac{a}{2\sin A}=\frac{4}{2\cdot\frac{\sqrt{7}}{4}}={\frac{8}{\sqrt{7}}}={\frac{8\sqrt{7}}{7}}$

よって，外接円の半径は，$R=\frac{8\sqrt{7}}{7}$

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% 例題I4.3.3の側注：三角形の内接円と外接円の半径（One More）★★
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% 側注1
$\cos B=\frac{c^2+a^2-b^2}{2ca}$

% 側注2
$\sin^2B+\cos^2B=1$

% 側注3
分母を有理化する．

% 側注4
$2R=\frac{b}{\sin B}$より，
\[
R=\frac{b}{2\sin B}
\]

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% 問題I4.3.3の側注：三角形の内接円と外接円の半径（One More）★★
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% 側注1
$\cos A=\frac{b^2+c^2-a^2}{2bc}$

% 側注2
$\sin^2A+\cos^2A=1$

% 側注3
$2R=\frac{a}{\sin A}$より，
\[
R=\frac{a}{2\sin A}
\]

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% 例題I4.3.3の考え方・補足など：三角形の内接円と外接円の半径（One More）★★
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(1) 3辺が与えられているので，余弦定理によって$\cos B$を求める．

(2) 2辺とその間の角が与えられているので，$S=\frac{1}{2}ca\sin B$を用いる．

(3) $\triangle\mathrm{ABC}$の内接円の中心をI，半径を$r$とするとき，

\[
\triangle\mathrm{ABC}=\triangle\mathrm{IBC}+\triangle\mathrm{ICA}+\triangle\mathrm{IAB}
\]

よって，$S=\frac{1}{2}ar+\frac{1}{2}br+\frac{1}{2}cr=\frac{1}{2}(a+b+c)r$

これと，(2)の結果を用いて半径$r$を求める．

(4) 正弦定理を用いる．

(3)