Question

在三角形ABC中abc分别为角ABC对边若c×cosB=b×cosC且cosA=2/3则sinB=

Answer 1

c·cosB=b·cosC

b·cosC-c·cosB=0

由正弦定理得sinBcosC-cosCsinB=0

sin(B-C)=0

B、C为三角形内角,B=C

cosA=-cos(B+C)=-cos(2B)=-(1-2sin²B)=2sin²B-1

sin²B=(1+cosA)/2

cosA=2/3代入,sin²B=(1+2/3)/2=5/6

B为三角形内角,sinB>0

sinB=√(5/6)=√30/6