【数学I】例題4.3.5：円に内接する四角形2（One More）★★★

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% 例題I4.3.5：円に内接する四角形2（One More）★★★
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円に内接する四角形ABCDにおいて，$\mathrm{AB}=2,\mathrm{BC}=4,\mathrm{CD}=3,\mathrm{DA}=2$とする．このとき，次の値を求めよ．

(1) $\cos B$

(2) 四角形ABCDの面積$S$

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% 例題I4.3.5の解答（One More）★★★
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(1) 四角形ABCDは円に内接するから，

\[
D=180^{\circ}-B
\]

$\triangle\mathrm{ABC}$において，余弦定理より，

\begin{align*}
\mathrm{AC}^2&=2^2+4^2-2\cdot 2\cdot 4\cdot\cos B\\
&=20-16\cos B\cdots(\mathrm{i})
\end{align*}

$\triangle\mathrm{ACD}$において，余弦定理より，

\[
\mathrm{AC}^2=3^2+2^2-2\cdot 3\cdot 2\cdot\cos(180^{\circ}-B)=13+12\cos B\cdots(\mathrm{ii})
\]

(i)，(ii)より，$20-16\cos B=13+12\cos B$

よって，$\cos B=\frac{1}{4}$

(2) $\sin B>0$より，

\[
\sin B=\sqrt{1-(\frac{1}{4})^2}=\sqrt{\frac{16-1}{16}}=\frac{\sqrt{15}}{4}
\]

また，$\sin D=\sin(180^{\circ}-B)=\sin B=\frac{\sqrt{15}}{4}$

よって，求める四角形ABCDの面積$S$は，

\begin{align*}
S&=\triangle\mathrm{ABC}+\triangle\mathrm{ACD}\\
&=\frac{1}{2}\cdot 2\cdot 4\cdot\sin B+\frac{1}{2}\cdot 3\cdot 2\cdot\sin D\\
&=\frac{1}{2}\cdot 2\cdot 4\cdot\frac{\sqrt{15}}{4}+\frac{1}{2}\cdot 3\cdot 2\cdot\frac{\sqrt{15}}{4}=\frac{7\sqrt{15}}{4}
\end{align*}

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% 問題I4.3.5：円に内接する四角形2（One More）★★★
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円に内接する四角形ABCDにおいて，$\mathrm{AB}=3,\mathrm{BC}=\sqrt{2},\mathrm{CD}=\sqrt{2},\mathrm{DA}=1$とする．このとき，次の値を求めよ．

(1) $\cos B$

(2) 四角形ABCDの面積$S$

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% 問題I4.3.5の解答（One More）★★★
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(1) 四角形ABCDは円に内接するから，

\[
D=180^{\circ}-B
\]

$\triangle\mathrm{ABC}$において，余弦定理より，

\begin{align*}
{\mathrm{AC}^2}&=3^2+(\sqrt{2})^2-2\cdot 3\cdot\sqrt{2}\cdot\cos B\\
&=11-6\sqrt{2}\cos B\cdots(\mathrm{i})
\end{align*}

$\triangle\mathrm{ACD}$において，余弦定理より，

\[
{\mathrm{AC}^2}=(\sqrt{2})^2+1^2-2\cdot\sqrt{2}\cdot 1\cdot\cos(180^{\circ}-B)=3+2\sqrt{2}\cos B\cdots(\mathrm{ii})
\]

(i)，(ii)より，$11-6\sqrt{2}\cos B=3+2\sqrt{2}\cos B$

よって，$\cos B=\frac{1}{\sqrt{2}}$

(2) $\sin B>0$より，

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

また，$\sin D=\sin(180^{\circ}-B)=\sin B=\frac{1}{\sqrt{2}}$

よって，求める四角形ABCDの面積$S$は，

\begin{align*}
S&=\triangle\mathrm{ABC}+\triangle\mathrm{ACD}\\
&=\frac{1}{2}\cdot 3\cdot\sqrt{2}\cdot\sin B+\frac{1}{2}\cdot\sqrt{2}\cdot 1\cdot\sin D\\
&=\frac{1}{2}\cdot 3\cdot\sqrt{2}\cdot\frac{1}{\sqrt{2}}+\frac{1}{2}\cdot\sqrt{2}\cdot 1\cdot\frac{1}{\sqrt{2}}=2
\end{align*}

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% 例題I4.3.5の側注：円に内接する四角形2（One More）★★★
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% 側注1
$B+D=180^\circ$

% 側注2
$\cos(180^{\circ}-\theta)=-\cos\theta$

% 側注3
$\sin^2B+\cos^2B=1$

% 側注4
$\sin(180^{\circ}-\theta)=\sin\theta$

% 側注5
四角形の面積は，2つの三角形の面積の和と考えられる．

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% 問題I4.3.5の側注：円に内接する四角形2（One More）★★★
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% 側注1
$B+D=180^\circ$

% 側注2
$\cos(180^{\circ}-\theta)=-\cos\theta$

% 側注3
$\sin^2B+\cos^2B=1$

% 側注4
$\sin(180^{\circ}-\theta)=\sin\theta$

% 側注5
四角形の面積は，2つの三角形の面積の和と考えられる．

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% 例題I4.3.5の考え方・補足など：円に内接する四角形2（One More）★★★
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% 考え方
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円に内接する四角形の対角の和は$180^\circ$であることを用いる．$\triangle\mathrm{ABC}$と$\triangle\mathrm{ACD}$に余弦定理をそれぞれ用いて，$\mathrm{AC}^2$を2通りに表す．また，$\sin(180^{\circ}-\theta)=\sin\theta,\cos(180^{\circ}-\theta)=-\cos\theta$を用いて式変形する．