Question

在三角形ABC中，sinA(sinB+cosB)-sinC=0,sinB+2cosC=0,求角A,B,C的大小

如题

Answer 1

解:
sinC=sin(π-(A+B))=sin(A+B)
cos2C=cos2(π-(A+B))=cos2(A+B)
∴sinA(sinB+cosB)- sin(A+B)=0
sinAsinB+sinAcosB)- sinAcosB-cosAsinB=0
sinB(sinA-cosA)=0,又sinB≠0
∴sinA=cosA
∵A∈(0,π)
∴A= π/4
再由sinB+ cos2(A+B)=0
sinB+cos( π/2 +2B)=0
sinB-sin2B=0
cosB=- 1/2
∴B= π/3
又C=π-(A+B)=π-( π/3 + π/4 )= 5π/12
∴A= π/4 ,B= π/3 ,C= 5π/12 .