方程x^2-(bcosA)x+acosB=0的两根之积等于两跟之和,a,b是三角形ABC的两边A,B为a,b的对角,判断三角形形状

方程x^2-(bcosA)x+acosB=0的两根之积等于两跟之和,a,b是三角形ABC的两边A,B为a,b的对角,判断三角形形状
方程x^2-(bcosA)x+acosB=0的两根之积等于两跟之和,a,b是三角形ABC的两边A,B为a,b的对角,判断三角形形状

方程x^2-(bcosA)x+acosB=0的两根之积等于两跟之和,a,b是三角形ABC的两边A,B为a,b的对角,判断三角形形状
依题可得bcosA=acosB,由正弦定理得bsinA=asinB
得cosB＊sinA／cosA＝sinB
又得tanA＝tanB
所以A＝B
所以此三角形为等腰三角形

三角形为等腰三角形。
x1+x2=bcosA x1*x2=acosB
bcosA=acosB 即a/cosA=b/cosB.
可以说明是等腰三角形了

由伟大定理有a cosB=bcosA，由正弦定理得sinA cosB=sinB cosA即 sinA cosB-sinB cosA=0则sin（A-B）=0而A,B为三角形ABC的两边a,b的对角，所以A=B故该三角形为等腰三角形

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由题义得bcosA=acosB,由正弦定理得bsinA=asinB
得cosB＊sinA／cosA＝sinB
又得tanA＝tanB
所以A＝B
所以此三角形为等腰三角形