由正弦定理得
c/sinC=a/sinA
csinA=asinC
又csinA=acosC,因此asinC=acosC
sinC=cosC
C=π/4
√3sinA-cos(B+C)
=√3sinA+cosA
=2[(√3/2)sinA+(1/2)cosA]
=2sin(A+π/6)
B>0 A=π-C-B0
π/6
由正弦定理得
c/sinC=a/sinA
csinA=asinC
又csinA=acosC,因此asinC=acosC
sinC=cosC
C=π/4
√3sinA-cos(B+C)
=√3sinA+cosA
=2[(√3/2)sinA+(1/2)cosA]
=2sin(A+π/6)
B>0 A=π-C-B0
π/6