解析:∵向量m=(b,3a),向量n=(c,b),向量m//向量n
∴b/c=3a/b,即b^2=3ac
由正弦定理知,sin²B=3sinAsinC
∵C-A=90°,∴2A+B=90°,A=(90°-B)/2
则,sin²B=3sin((90°-B)/2)sin(90°+A)= 3sin((90°-B)/2)sin((270°-B)/2)
= 3/2根(1-cos(90°-B))根(1-cos(270°-B)) = 3/2根[(1-sinB)(1+sinB)] (∵B为锐角)
= 3/2cosB
即,1-cos²B=3/2cosB,2cos²B+3cosB-2=0
解得cosB=1/2,cosB=-2(舍),则B=60°