Question

三角形ABC中,内角A,B,C的对边分别为a,b,c,2根号3Sin平方 A+B/2=sinC+根号3+1,求角c大小

Answer 1

√3【1-cos（A+B)]=sinC+√3+1

√3(1+cosC)=sinC+√3+1

√3cosC-sinC=1

√3/2cosC-1/2sinC=1/2

sin(60°-C)=1/2

∵0﹤C﹤180°

∴-120°﹤60°-C﹤60°

∴60°-C=30°

∴C=30°