Question

在三角形abc中，角a，b，c所对的边分别为a，b，c，且acosc，bcosb，ccosa成等差数，求角B的大小

Answer 1

cosA=(b²+c²-a²)/2bc
cosC=(a²+b²-c²)/2ab
2bcosB=acosC+ccosA
=a(a²+b²-c²)/2ab+c(b²+c²-a²)/2bc
=(a²+b²-c²)/2b+(b²+c²-a²)/2b
=2b²/2b
=b
cosB=b/2b
=1/2

所以B=60度，楼上的错了，cos120度变负的了

Answer 2

cosA=(b²+c²-a²)/2bc
cosC=(a²+b²-c²)/2ab
2bcosB=acosC+ccosA
=a(a²+b²-c²)/2ab+c(b²+c²-a²)/2bc
=(a²+b²-c²)/2b+(b²+c²-a²)/2b
=2b²/2b
=b
cosB=b/2b
=1/2
∴∠B=60°或120°

Answer 3

2bcosB=acosC+ccosA
由正弦定理可得：
2sinBcosB=sinAcosC+cosAsinC=sin(A+C)=sinB
cosB=1/2
B=60º

Answer 4

60度，用余弦定理做