在三角形ABC中,角A,B,C所对的边为a,b,c,且(b^2-a^2-c^2)/ac=cos(A+C)/sinAcosA,若sinB/cosC>根号2,求C的范围
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![在三角形ABC中,角A,B,C所对的边为a,b,c,且(b^2-a^2-c^2)/ac=cos(A+C)/sinAcosA,若sinB/cosC>根号2,求C的范围](/uploads/image/z/968636-20-6.jpg?t=%E5%9C%A8%E4%B8%89%E8%A7%92%E5%BD%A2ABC%E4%B8%AD%2C%E8%A7%92A%2CB%2CC%E6%89%80%E5%AF%B9%E7%9A%84%E8%BE%B9%E4%B8%BAa%2Cb%2Cc%2C%E4%B8%94%EF%BC%88b%5E2-a%5E2-c%5E2%EF%BC%89%2Fac%3Dcos%28A%2BC%29%2FsinAcosA%2C%E8%8B%A5sinB%2FcosC%3E%E6%A0%B9%E5%8F%B72%2C%E6%B1%82C%E7%9A%84%E8%8C%83%E5%9B%B4)
在三角形ABC中,角A,B,C所对的边为a,b,c,且(b^2-a^2-c^2)/ac=cos(A+C)/sinAcosA,若sinB/cosC>根号2,求C的范围
在三角形ABC中,角A,B,C所对的边为a,b,c,且(b^2-a^2-c^2)/ac=cos(A+C)/sinAcosA,若sinB/cosC>根号2,
求C的范围
在三角形ABC中,角A,B,C所对的边为a,b,c,且(b^2-a^2-c^2)/ac=cos(A+C)/sinAcosA,若sinB/cosC>根号2,求C的范围
(b^2-a^2-c^2)/ac=-2cosB
cos(A+C)/sinAcosA=-cosB/(sinAcosA)
2sinAcosA=1
sin2A=1
A=π/4
sinB/cosC=sin(A+C)/cosC=sinA+cosAsinC/cosC>√2
sinC/cosC>1
C的范围是(π/4,π/2)
由余弦定理得,
b²-a²-c²=-2ac·cosB
∴(b²-a²-c²)/(ac)=-2cosB
∴-2cosB=cos(A+C)/(sinAcosA)
由于A+B+C=π
∴cosB=-cos(A+C)
∴-2cosB=2cos(A+C)
∴2cos(A+C)=cos(A+C)...
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由余弦定理得,
b²-a²-c²=-2ac·cosB
∴(b²-a²-c²)/(ac)=-2cosB
∴-2cosB=cos(A+C)/(sinAcosA)
由于A+B+C=π
∴cosB=-cos(A+C)
∴-2cosB=2cos(A+C)
∴2cos(A+C)=cos(A+C)/(sinAcosA)
∴1/2sin(2A)=1/2
即sin(2A)=1
∴A=π/4
∴ B+C=π-A=3π/4
∴ B=3π/4-C
sinB/cosC>√2
sin(3π/4-C)/cosC>√2
sin(3π/4-C)>√2 cosC
sin(C+π/4)>√2 cosC
√2 sin(C+π/4)>2 cosC
sinC+cosC>2cosC
sinC>cosC
tanC>1
所以
π/4
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