△ABC中,A,B为锐角,角A,B,C所对的边分别为a,b,c,cos2A=3/5,sinB=√10/10,a-b=√2-1,求a,b,c的值
问题描述：
△ABC中,A,B为锐角,角A,B,C所对的边分别为a,b,c,cos2A=3/5,sinB=√10/10,a-b=√2-1,求a,b,c的值
答
cos2A=1-2(sinA)^2=3/5sinA=√5/5cosA=2√5/5sinB=√10/10cosB=3√10/10cosC=-cos(A+B)=-cosAcosB+sinAsinB=-√2/2sinC=√2/2c/sinC=(a-b)/(sinA-sinB)c=√55=c^2=a^2+b^2-2abcosC=a^2+b^2+√2ab=(a-b)^2+2ab+√2aba...