問題の解答
検索用コード（LaTeX）
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% 例題I4.3.1：三角形の面積（One More）★★
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次のような$\triangle\mathrm{ABC}$の面積$S$を求めよ．
(1) $a=4,c=\sqrt{6},B=60^{\circ}$
(2) $a=7,b=6,c=5$
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% 例題I4.3.1の解答（One More）★★
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(1) 求める$\triangle\mathrm{ABC}$の面積$S$は，
\[
S=\frac{1}{2}ca\sin B=\frac{1}{2}\cdot 4\cdot\sqrt{6}\cdot\sin 60^{\circ}=\frac{1}{2}\cdot 4\cdot\sqrt{6}\cdot\frac{\sqrt{3}}{2}=\sqrt{18}=3\sqrt{2}
\]
(2) 余弦定理より，$\cos A=\frac{6^2+5^2-7^2}{2\cdot 6\cdot 5}=\frac{1}{5}$
$\sin^2A+\cos^2A=1,\sin A>0$より，
\[
\sin A=\sqrt{1-\cos^2A}=\sqrt{1-(\frac{1}{5})^2}=\sqrt{\frac{24}{25}}=\frac{2\sqrt{6}}{5}
\]
よって，求める$\triangle\mathrm{ABC}$の面積$S$は，
\[
S=\frac{1}{2}bc\sin A=\frac{1}{2}\cdot 6\cdot 5\cdot\frac{2\sqrt{6}}{5}=6\sqrt{6}
\]
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% 例題I4.3.1の別解
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ヘロンの公式より，$s=\frac{7+6+5}{2}=9$であるから，
\begin{align*}
S&=\sqrt{s(s-a)(s-b)(s-c)}\\
&=\sqrt{9(9-7)(9-6)(9-5)}\\
&=\sqrt{9\cdot 2\cdot 3\cdot 4}\\
&=\sqrt{216}=6\sqrt{6}
\end{align*}
よって，求める面積$S$は，$S=6\sqrt{6}$
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% 問題I4.3.1：三角形の面積（One More）★★
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次のような$\triangle\mathrm{ABC}$の面積$S$を求めよ．
(1) $a=5,c=\sqrt{10},B=45^{\circ}$
(2) $a=7,b=5,c=8$
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% 問題I4.3.1の解答（One More）★★
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(1) 求める$\triangle\mathrm{ABC}$の面積$S$は，
\[
S=\frac{1}{2}ca\sin B=\frac{1}{2}\cdot\sqrt{10}\cdot 5\cdot\sin 45^{\circ}=\frac{1}{2}\cdot\sqrt{10}\cdot 5\cdot\frac{\sqrt{2}}{2}=\frac{5\sqrt{20}}{4}=\frac{5\sqrt{5}}{2}
\]
(2) 余弦定理より，$\cos A=\frac{5^2+8^2-7^2}{2\cdot 5\cdot 8}=\frac{40}{80}=\frac{1}{2}$
$\sin^2A+\cos^2A=1,\sin A>0$より，
\[
\sin A=\sqrt{1-\cos^2A}=\sqrt{1-(\frac{1}{2})^2}=\frac{\sqrt{3}}{2}
\]
よって，求める$\triangle\mathrm{ABC}$の面積$S$は，
\[
S=\frac{1}{2}bc\sin A=\frac{1}{2}\cdot 5\cdot 8\cdot\frac{\sqrt{3}}{2}=10\sqrt{3}
\]
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% 問題I4.3.1の別解
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ヘロンの公式より，$s=\frac{7+5+8}{2}=10$であるから，
\begin{align*}
S&=\sqrt{s(s-a)(s-b)(s-c)}\\
&=\sqrt{10(10-7)(10-5)(10-8)}\\
&=\sqrt{10\cdot 3\cdot 5\cdot 2}\\
&=\sqrt{300}=10\sqrt{3}
\end{align*}
よって，求める面積$S$は，$S=10\sqrt{3}$
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