Re: [中學] 請教一題題目
※ 引述《yuchuen ()》之銘言:
: 三角形ABC中,a^2-c^2=bc,角B=51度,求角C=?
: 請各位前輩解答一下
不知道原PO題目是不是有打錯?
如果改成 三角形ABC中,a^2-b^2=bc,角B=51度,求角C=?
解法如下:
a^2-b^2=bc
=> (sinA)^2 - (sinB)^2 = sinB sinC
=> 2(sinA)^2 - 2(sinB)^2 = 2sinB sinC
=> 1-2(sinA)^2 - (1-2(sinB)^2) = -2sinB sinC
=> cos 2A - cos2B = -2sinB sinC
=> -2sin(A+B)sin(A-B) = -2sinB sinC
=> sin(180-C)sin(180-2B-C) = sinB sinC
=> sin(2B+C)sinC = sinBsinC [sinC≠0]
=> sin(2B+C) = sinB
=> 3B+C=180
B=51 , C=180-51*3=27度
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04/25 12:38, , 1F
04/25 12:38, 1F
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04/25 12:39, , 2F
04/25 12:39, 2F
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04/25 12:42, , 3F
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04/25 12:51, , 4F
04/25 12:51, 4F
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04/25 12:51, 5F
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04/25 12:57, , 6F
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