已知三角形ABC的周长为根号2+1且sinA+sinB=根号2sinC.求:1求边AB的长.2若sinAsinB=1/3
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已知三角形ABC的周长为根号2+1且sinA+sinB=根号2sinC.求:1求边AB的长.2若sinAsinB=1/3sin^2C求角C的大小
![已知三角形ABC的周长为根号2+1且sinA+sinB=根号2sinC.求:1求边AB的长.2若sinAsinB=1/3](/uploads/image/z/11739703-31-3.jpg?t=%E5%B7%B2%E7%9F%A5%E4%B8%89%E8%A7%92%E5%BD%A2ABC%E7%9A%84%E5%91%A8%E9%95%BF%E4%B8%BA%E6%A0%B9%E5%8F%B72%EF%BC%8B1%E4%B8%94sinA%2BsinB%3D%E6%A0%B9%E5%8F%B72sinC.%E6%B1%82%EF%BC%9A1%E6%B1%82%E8%BE%B9AB%E7%9A%84%E9%95%BF.2%E8%8B%A5sinAsinB%3D1%2F3)
1、R(sinA+sinB+sinC)=2^0.5+1
因为sinA+sinB=2^0.5sinC
故R[2^0.5+1]sinC=2^0.5+1
则RsinC=1,即AB=1
2 cosC=(a*a+b*b-c*c)/2a*b=(sinA*sinA+sinB*sinB-sinC*sinC)/2sinA*sinB
=[(sinA+sinB)^2-2sinA*sinB-sinC*sinC]/2*sinA*sinB
=[2sinC^2-2/3sinC^2-sinC^2]/2/3sinC^2
=1/3/(2/3)
=1/2
因为0
因为sinA+sinB=2^0.5sinC
故R[2^0.5+1]sinC=2^0.5+1
则RsinC=1,即AB=1
2 cosC=(a*a+b*b-c*c)/2a*b=(sinA*sinA+sinB*sinB-sinC*sinC)/2sinA*sinB
=[(sinA+sinB)^2-2sinA*sinB-sinC*sinC]/2*sinA*sinB
=[2sinC^2-2/3sinC^2-sinC^2]/2/3sinC^2
=1/3/(2/3)
=1/2
因为0
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