在三角形ABC中,a=根号3,b=根号2,角A=60°,求角B和边c

Question

如题所述

Answer 1

用正弦公式sinA/a=SinB/b=SinC/cSinB=b/a*sinA=√2/√3*√3/2=√2/2解得B=45SinC=sin(A+B)=sinAcosB+cosAsinB=1/2*√2/2+√3/2*√2/2=(√3+1)√2/4c=b*SinC/sinB=√2*(√3+1)/2=(√6+√2)/2