在三角形ABc中2√2sin^2A-sin^2c)=(a-b)sinB 三角形ABC的外接圆的半径为√2 1)求角c 2）求三角形SABC最值

2√2(sin^2A-sin^2C)=(a-b)sinB a/sinA=b/sinB=c/sinC=2R=2√2原式可化为(a^2-c^2)/(2√2)=(a-b)b/(2√2)即a^2+b^2-c^2=ab∴cosC=(a^2+b^2-c^2)/(2ab)=1/2∴C=60°∴c=2R*sinC=√6∵a^2+b^2-c^2=ab,c=√6∴a^2+b^2=c...