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若三角形ABC的面积为(a²+b²+c²)/4,则内角C=
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三角形ABC的面积为(a²+b²+c²)/4
1/2*ab*sinC=(a²+b²+c²)/4
a²+b²+c²=2ab*sinC.....(1)
由余弦定理:c²=a²+b²-2abcosC
a²+b²-c²=2ab*cosC......(2)
(1)-(2),得;
c²=ab(sinC-cosC)
c²/ab=sinC-cosC
由正弦定理:a/sinA=b/sinB=c/sinC
c/a=sinC/sinA
c/b=sinC/sinB
(sinC)²/sinA*sinB=sinC-cosC
(sinC)²=sinA*sinB*(sinC-cosC)
1/2*ab*sinC=(a²+b²+c²)/4
a²+b²+c²=2ab*sinC.....(1)
由余弦定理:c²=a²+b²-2abcosC
a²+b²-c²=2ab*cosC......(2)
(1)-(2),得;
c²=ab(sinC-cosC)
c²/ab=sinC-cosC
由正弦定理:a/sinA=b/sinB=c/sinC
c/a=sinC/sinA
c/b=sinC/sinB
(sinC)²/sinA*sinB=sinC-cosC
(sinC)²=sinA*sinB*(sinC-cosC)
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