在△ABC中已知sinA=3/5,B=2π/3,c=4-√3,则a=?
A=arcsin(3/5),故C=π-[2π/3+arcsin(3/5)]=π/3-arcsin(3/5)；
故a=(csinA)/sinC=(4-√3)(3/5)/sin[π/3-arcsin(3/5)]
=[3(4-√3)/5]/[sin(π/3)cosarcsin(3/5)-cos(π/3)sinarcsin(3/5)]
=[3(4-√3)/5]/[(√3/2)(4/5)-(1/2)(3/5)]
=[3(4-√3)/5]/(2√3/5-3/10)=[3(4-√3)/5]/[(4√3-3)/10]
=6(4-√3)/(4√3-3)=6(4-√3)(4√3+3)/39=2√3