Question

△abc中,已知(b-a)(sinA+sinB)=bsinA,且sinA/sinC=sinC/sinB,判断△abc形状

Answer 1

(b-a)(sinA+sinB)=bsinA,

(b-a)/b=sinA/(sinA+sinB),(1)

根据正弦定理,sinA/a=sinB/b,

(sinA+sinB)/(a+b)=sinA/a,(等比）,

sinA/(sinA+sinB)=a/(a+b),(2),

对比（1）和（2）式,

a/(a+b)=(b-a)/b,

b^2-a^2=ab,(3)

sinA/sinC=sinC/sinB,(已知）,

根据正弦定理,sinA/sinC=a/c,sinC/sinB=c/b,

∴a/c=c/b,

c^2=ab,(4),

对比（3）和（4）式,

b^2-a^2=c^2,

a^2+c^2=b^2,

∴△ABC是RT△,〈B=90°.