△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．

Question

△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．

Answer 1

bsinA=asinB
asinAsinB+bcos^2A=√2a
bsin²a+bcos²A=√2a
b=√2a
b^2=2a^2
c^2=b^2+√3a^2
=2a^2+√3a^2
c=a√[2+√3]
=a√[(4+2√3)/2]
=a(√3+1)/√2
cosB=(a^2+c^2-b^2)/(2ac)
=(a^2+2a^2+√3a^2-2a^2)/[2a*(√3+1)a/√2]
=(1+√3)a/√2(√3+1)a
=√2/2
所以 B=45°