题目内容：

△ABC的三个内角A、B、C所对的边分别为a、b、c,asinAsinB+bcos^2A=根号2a． 若C^2=b2^+根号3a^2,求B．

优质解答

bsinA=asinBasinAsinB+bcos^2A=√2absin²a+bcos²A=√2ab=√2ab^2=2a^2c^2=b^2+√3a^2=2a^2+√3a^2c=a√[2+√3]=a√[(4+2√3)/2]=a(√3+1)/√2cosB=(a^2+c^2-b^2)/(2ac)=(a^2+2a^2+√3a^2-2a^2)/[2a*(√3+1...