一道解三角形的题,
1个回答

acosC+√3asinC-b-c=0

sinAcosC+√3sinAsinC-sinB-sinC=0

sinAcosC+√3sinAsinC-sinAcosC-sinCcosA-sinC=0

√3sinAsinC-sinCcosA-sinC=0

√3sinA-cosA-1=0

2sin(A-π/6)=1

A-π/6=π/6 a=π/3 A-π/6=5π/6 A=π(不符题意)

S=√3=1/2bcsinA bc=4

cosA=(b^2+c^2-a^2)/2bc

(b+c)^2-2bc=2*4*1/2-2^2=0

(b+c)^2=8 无解