在任一三角形ABC中，求證:a(sinB-sinC)+b(sinC-sinA)+c(sinA-sinB)=0注a=BC b=AC c=AB

熱心網(wǎng)友

證明:a/sinA=b/sinB=c/sinC---asinB=bsinA;asinC=csinA;bsinC=csinB左邊=asinB-asinC+bsinC-bsinA+csinA-csinB=(asinB-bsinA)+(bsinC-csinB)+(csinA-bsinC)=0+0+0=0

熱心網(wǎng)友

0=sinA(sinB-sinC)+sinB(sinC-sinA)+sinC(sinA-sinB)=a(sinB-sinC)/(2R)+b(sinC-sinA)/(2R)+c(sinA-sinB)/(2R)a(sinB-sinC)+b(sinC-sinA)+c(sinA-sinB)=0