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

题目

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

答案

(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°.