Question

在三角形abc中asina+csinc-根号2asinc=bsinb

Answer 1

asinA+csinC-√2asinC=bsinB
两边同除以sinB
asinA/sinB+csinC/sinB-√2asinC/sinB=b
由正弦定理sinA/sinB=a/b，sinC/sinB=c/b
a�0�5/b+c�0�5/b-√2ac/b=b
化简得a�0�5+c�0�5-b�0�5=√2ac
由余弦定理cosB=(a�0�5+c�0�5-b�0�5)/(2ac)=√2/2
故∠B=45°
（2）由∠B=45°，∠A=75°
∠C=180°-∠B-∠A=60°
由正弦定理b/sinB=c/sinC
c=bsinC/sinB=2*（√3/2）/（√2/2）=√6
由余弦定理a�0�5=b�0�5+c�0�5-2bccosA=4+6-2*2*√6(√6-√2)/4=4+2√3
得a=√(4+2√3)=√3+1