△ABC的三个内角,A,B,C的对边分别为a,b,c,且
|
∵
=a2-(b-c)2 bc
=1,a2-b2+2bc-c2 bc
∴a2-b2-c2=-bc,即b2+c2-a2=bc,
∴cosA=
=b2+c2-a2 2bc
=bc 2bc
,1 2
又A为三角形的内角,
则A=60°.
故选B
△ABC的三个内角,A,B,C的对边分别为a,b,c,且
|
∵
=a2-(b-c)2 bc
=1,a2-b2+2bc-c2 bc
∴a2-b2-c2=-bc,即b2+c2-a2=bc,
∴cosA=
=b2+c2-a2 2bc
=bc 2bc
,1 2
又A为三角形的内角,
则A=60°.
故选B