作业帮在三角形ABC中,sinC(cosA+cosB)=sinA+sinB.求∠C的度数；若角C所对的边c=1,求内切圆半径r的取值范围.
来源：学生作业帮助网 编辑：作业帮 时间：2024/05/05 16:47:10
在三角形ABC中,sinC(cosA+cosB)=sinA+sinB.求∠C的度数；若角C所对的边c=1,求内切圆半径r的取值范围.
在三角形ABC中,sinC(cosA+cosB)=sinA+sinB.求∠C的度数；若角C所对的边c=1,求内切圆半径r的取值范围.
在三角形ABC中,sinC(cosA+cosB)=sinA+sinB.求∠C的度数；若角C所对的边c=1,求内切圆半径r的取值范围.
根据正弦定理,原式可变形为：
c(cosA+cosB)=a+b.①
∵ 根据任意三角形射影定理（又称“第一余弦定理”）：
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
因此,在△ABC中,∠C=90°.
参考资料：射影定理
方法二:
即a+b=c(cosA+cosB)=(c^2+b^2-a^2)/(2b)+(c^2+a^2-b^2)/(2a)
也就是2a+2b=c^2(1/b+1/a)+a+b-(a^2/b+b^2/a)
两边同时约去a+b得
2=c^2/ab+1-(a^2+b^2-ab)/ab
即c^2=a^2+b^2
C为90°
（2）r=ab/(a+b+c)
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