{"id":3035,"date":"2025-09-06T00:48:50","date_gmt":"2025-09-05T15:48:50","guid":{"rendered":"https:\/\/math-friend.com\/?p=3035"},"modified":"2025-09-06T01:03:37","modified_gmt":"2025-09-05T16:03:37","slug":"%e3%80%90%e5%8d%83%e8%91%89%e5%a4%a7%e5%ad%a6%e5%85%a5%e8%a9%a6%e3%80%912014","status":"publish","type":"post","link":"https:\/\/math-friend.com\/?p=3035","title":{"rendered":"\u3010\u5343\u8449\u5927\u5b66\u5165\u8a66\u3011\u6700\u5927\u89d2120\u00b0\u306e\u4e09\u89d2\u5f62\u3067\u4f59\u5f26\u306e\u5024\u3092\u6c42\u3081\u308b(2014)"},"content":{"rendered":"\n

\u4eca\u56de\u306f\u3053\u3061\u3089\u306e\u554f\u984c\u3092\u89e3\u3044\u3066\u3044\u304d\u307e\u3059.<\/p>\n\n\n\n

\\(\\triangle{\\mathrm{ABC}}\\)\u306b\u3064\u3044\u3066, \\(\\mathrm{A}\\), \\(\\mathrm{B}\\), \\(\\mathrm{C}\\)\u306e\u3046\u3061, \u6700\u3082\u5927\u304d\u3044\u89d2\u306f\\(120^\\circ\\)\u3067\u3042\u308a, \u3055\u3089\u306b\\(\\displaystyle \\frac{\\sin{\\mathrm{A}}}{5}=\\frac{\\sin{\\mathrm{B}}}{3}\\)\u3092\u6e80\u305f\u3057\u3066\u3044\u308b. \u3053\u306e\u3068\u304d, \\(\\cos{\\mathrm{A}}\\), \\(\\cos{\\mathrm{B}}\\), \\(\\cos{\\mathrm{C}}\\)\u306e\u5024\u3092\u305d\u308c\u305e\u308c\u6c42\u3081\u3088.
(2014 \u5343\u8449\u5927\u5b66 [2])<\/span><\/p>\n\n\n\n

\u305d\u308c\u3067\u306f\u89e3\u3044\u3066\u3044\u304d\u307e\u3057\u3087\u3046.<\/p>\n\n\n\n

\n

\u56f3\u306e\u3088\u3046\u306b, \\(a=\\mathrm{BC}\\), \\(b=\\mathrm{CA}\\), \\(c=\\mathrm{AB}\\)\u3068\u304a\u304f.<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

\\(\\triangle{\\mathrm{ABC}}\\)\u3067\u6b63\u5f26\u5b9a\u7406\u3088\u308a,$$
\\frac{a}{\\sin{\\mathrm{A}}}=\\frac{b}{\\sin{\\mathrm{B}}}
$$\u3068\u306a\u308a, \\(\\displaystyle \\frac{\\sin{\\mathrm{A}}}{5}=\\frac{\\sin{\\mathrm{B}}}{3}\\)\u3068\u8fba\u3005\u3092\u304b\u3051\u3066,$$
\\begin{align}
\\frac{a}{\\sin{\\mathrm{A}}}\\cdot \\frac{\\sin{\\mathrm{A}}}{5}&=\\frac{b}{\\sin{\\mathrm{B}}}\\cdot\\frac{\\sin{\\mathrm{B}}}{3}\\\\[1.5ex]
\\frac{a}{5}&=\\frac{b}{3}
\\end{align}
$$\u304c\u308f\u304b\u308b. \\(\\displaystyle \\frac{a}{5}=\\frac{b}{3}=k\\)\u3068\u304a\u304f\u3068, \\(k\\)\u306f\u6b63\u306e\u5b9f\u6570\u3067\u3042\u308a, $$
\\begin{align}
a&=5k\\\\[1.5ex]
b&=3k
\\end{align}
$$\u3068\u306a\u308b.<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

\u3053\u3053\u3067,\\(\\sin{\\mathrm{B}}>0\\)\u3067\u3042\u308b\u3053\u3068\u304b\u3089,$$
\\sin{\\mathrm{A}}=\\frac{5}{3}\\sin{\\mathrm{B}}>\\sin{\\mathrm{B}}
$$\u3067\u3042\u308a, \u4e00\u822c\u306b, \\(\\triangle{\\mathrm{ABC}}\\)\u306b\u304a\u3044\u3066\u306f, \u89d2\\(\\mathrm{A}\\), \\(\\mathrm{B}\\)\u306e\u5927\u5c0f\u95a2\u4fc2\u3068, \\(\\sin{\\mathrm{A}}\\), \\(\\sin{\\mathrm{B}}\\)\u306e\u5927\u5c0f\u95a2\u4fc2\u306f\u4e00\u81f4\u3059\u308b\u306e\u3067, \\(\\mathrm{A}>\\mathrm{B}\\)\u304c\u308f\u304b\u308b. \u3053\u308c\u304b\u3089, \\(\\triangle{\\mathrm{ABC}}\\)\u306e\u6700\u5927\u306e\u89d2\u306f, \\(\\mathrm{A}\\), \\(\\mathrm{C}\\)\u306e\u3044\u305a\u308c\u304b\u3067\u3042\u308b\u3053\u3068\u304c\u308f\u304b\u308b.

\u2460 \\(\\mathrm{A}=120^\\circ\\)\u306e\u3068\u304d<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

\u4f59\u5f26\u5b9a\u7406\u304b\u3089,$$
\\begin{align}
&a^2=b^2+c^2-2bc\\cos{\\mathrm{A}}\\\\[1.5ex]
&(5k)^2=(3k)^2+c^2-2\\cdot 3kc\\cdot\\left(-\\frac{1}{2}\\right)\\\\[1.5ex]
&c^2+3kc-16k^2=0\\\\[1.5ex]
&c=\\frac{-3k\\pm\\sqrt{9k^2+64k^2}}{2}=\\frac{-3\\pm\\sqrt{73}}{2}k
\\end{align}
$$\u3068\u306a\u308b\u304c, \\(c>0\\)\u3088\u308a, \\(\\displaystyle c=\\frac{-3+\\sqrt{73}}{2}k\\)\u304c\u308f\u304b\u308b.<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

\u3088\u3063\u3066\u4f59\u5f26\u5b9a\u7406\u3088\u308a,$$
\\begin{align}
\\cos{\\mathrm{B}}&=\\frac{c^2+a^2-b^2}{2ca}\\\\[1.5ex]
&=\\frac{\\left(\\frac{-3+\\sqrt{73}}{2}k\\right)^2+(5k)^2-(3k)^2}{2\\cdot\\frac{-3+\\sqrt{73}}{2}k\\cdot 5k}\\\\[1.5ex]
&=\\frac{9-6\\sqrt{73}+73+4\\cdot 16}{20(-3+\\sqrt{73}}\\\\[1.5ex]
&=\\frac{2\\cdot 73-6\\sqrt{73}}{20(-3+\\sqrt{73})}\\\\[1.5ex]
&=\\frac{2\\sqrt{73}(\\sqrt{73}-3)}{20(-3+\\sqrt{73})}\\\\[1.5ex]
&=\\frac{\\sqrt{73}}{10}
\\end{align}
$$
$$
\\begin{align}
\\cos{\\mathrm{C}}&=\\frac{a^2+b^2-c^2}{2ab}\\\\[1.5ex]
&=\\frac{25k^2+9k^2-\\frac{(-3+\\sqrt{73})^2}{4}k^2}{2\\cdot 5k\\cdot 3k}\\\\[1.5ex]
&=\\frac{4\\cdot 34-9+6\\sqrt{73}-73}{4\\cdot 2\\cdot 5\\cdot 3}\\\\[1.5ex]
&=\\frac{54+6\\sqrt{73}}{4\\cdot 2\\cdot 5\\cdot 3}\\\\[1.5ex]
&=\\frac{9+\\sqrt{73}}{20}
\\end{align}
$$\u3068\u306a\u308b. \u307e\u305f, \\(\\mathrm{A}=120^\\circ\\)\u3088\u308a, \\(\\displaystyle \\cos{\\mathrm{A}}=-\\frac{1}{2}\\)\u3067\u3042\u308b.

\u2461 \\(\\mathrm{C}=120^\\circ\\)\u306e\u3068\u304d<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

\u4f59\u5f26\u5b9a\u7406\u304b\u3089,$$
\\begin{align}
c^2&=a^2+b^2-2ab\\cos{\\mathrm{C}}\\\\[1.5ex]
&=(5k)^2+(3k)^2-2\\cdot 5k \\cdot 3k\\left(-\\frac{1}{2}\\right)\\\\[1.5ex]
&=25k^2+9k^2+15k^2\\\\[1.5ex]
&=49k^2
\\end{align}
$$\u3068\u306a\u308a, \\(c=7k\\)\u304c\u308f\u304b\u308b.<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

\u3088\u3063\u3066\u4f59\u5f26\u5b9a\u7406\u3088\u308a,$$
\\begin{align}
\\cos{\\mathrm{A}}&=\\frac{b^2+c^2-a^2}{2bc}\\\\[1.5ex]
&=\\frac{9k^2+49k^2-25k^2}{2\\cdot 3k \\cdot 7k}\\\\[1.5ex]
&=\\frac{33}{2\\cdot 3\\cdot 7}\\\\[1.5ex]
&=\\frac{11}{14}
\\end{align}
$$
$$
\\begin{align}
\\cos{\\mathrm{B}}&=\\frac{c^2+a^2-b^2}{2ca}\\\\[1.5ex]
&=\\frac{49k^2+25k^2-9k^2}{2\\cdot 7k \\cdot 5k}\\\\[1.5ex]
&=\\frac{65}{2\\cdot 7\\cdot 5}\\\\[1.5ex]
&=\\frac{13}{14}
\\end{align}
$$\u3068\u306a\u308b. \u307e\u305f, \\(\\mathrm{C}=120^\\circ\\)\u3088\u308a, \\(\\displaystyle \\cos{\\mathrm{C}}=-\\frac{1}{2}\\)\u3067\u3042\u308b.

\u4ee5\u4e0a\u304b\u3089, $$
(\\cos{\\mathrm{A}}, \\cos{\\mathrm{B}}, \\cos{\\mathrm{C}})=\\left(-\\frac{1}{2}, \\frac{\\sqrt{73}}{10}, \\frac{9+\\sqrt{73}}{20}\\right), \\left(\\frac{11}{14}, \\frac{13}{14}, -\\frac{1}{2}\\right)
$$\u3068\u306a\u308b.<\/p>\n<\/div><\/div>\n\n\n\n

youtube\u3067\u3082\u89e3\u8aac\u3057\u3066\u3044\u307e\u3059.<\/p>\n\n\n\n