已知a=根号3+1.b=2,c=根号2,则角C等于cosC=(a^2+b^2-c^2)/2ab=[(√3+1)^2+2^2-(√2)^2]/[2*(√3+1)*2]=[4+2√3+4-2]/[4(√3+1)]=[6+2√3]/[4(√3+1)]=2√3(√3+1)/[4(√3+1)]=√3/2C=30答案是这样的,我有一个问题是最后第二步,
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/08 06:59:14
![已知a=根号3+1.b=2,c=根号2,则角C等于cosC=(a^2+b^2-c^2)/2ab=[(√3+1)^2+2^2-(√2)^2]/[2*(√3+1)*2]=[4+2√3+4-2]/[4(√3+1)]=[6+2√3]/[4(√3+1)]=2√3(√3+1)/[4(√3+1)]=√3/2C=30答案是这样的,我有一个问题是最后第二步,](/uploads/image/z/2765830-22-0.jpg?t=%E5%B7%B2%E7%9F%A5a%3D%E6%A0%B9%E5%8F%B73%2B1.b%3D2%2Cc%3D%E6%A0%B9%E5%8F%B72%2C%E5%88%99%E8%A7%92C%E7%AD%89%E4%BA%8EcosC%3D%28a%5E2%2Bb%5E2-c%5E2%29%2F2ab%3D%5B%28%E2%88%9A3%2B1%29%5E2%2B2%5E2-%28%E2%88%9A2%29%5E2%5D%2F%5B2%2A%28%E2%88%9A3%2B1%29%2A2%5D%3D%5B4%2B2%E2%88%9A3%2B4-2%5D%2F%5B4%28%E2%88%9A3%2B1%29%5D%3D%5B6%2B2%E2%88%9A3%5D%2F%5B4%28%E2%88%9A3%2B1%29%5D%3D2%E2%88%9A3%28%E2%88%9A3%2B1%29%2F%5B4%28%E2%88%9A3%2B1%29%5D%3D%E2%88%9A3%2F2C%3D30%E7%AD%94%E6%A1%88%E6%98%AF%E8%BF%99%E6%A0%B7%E7%9A%84%2C%E6%88%91%E6%9C%89%E4%B8%80%E4%B8%AA%E9%97%AE%E9%A2%98%E6%98%AF%E6%9C%80%E5%90%8E%E7%AC%AC%E4%BA%8C%E6%AD%A5%2C)
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