在△ABC中,若sinA+sinB=sinC(cosA+cosB).
(1)判断△ABC的形状;
(2)在上述△ABC中,若角C的对边c=1,求该三角形内切圆半径的取值范围.
人气:493 ℃ 时间:2020-04-15 10:40:16
解答
(1)根据正弦定理,原式可变形为:c(cosA+cosB)=a+b①,∵根据任意三角形射影定理得:a=b•cosC+c•cosB,b=c•cosA+a•cosC,∴a+b=c(cosA+cosB)+cosC(a+b)②,由于a+b≠0,故由①式、②式得:cosC=0,∴在...
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