设 3x^2-27x+32=0 的两根为 b.c(b>c)
b+c=9 bc=32/3 b^2+c^2=(b+c)^2-2bc=81-64/3=179/3
BC^2=a^2=b^2+c^2-2bc cosA=179/3-64/3*(1/2)
=179/3-32*3=147/3=49
BC=7
在三角形ABC中,A=60°,最大边与最小边是方程3x2-27+32=0的两个实根,那么BC是?
在三角形ABC中,A=60°,最大边与最小边是方程3x2-27+32=0的两个实根,那么BC是?
数学人气:934 ℃时间:2020-04-12 23:00:17
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