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半径为R的圆外接与三角形ABC 且2R(sin^2A-sin^2c)=(根号3*a-b)sinB求角C和△abc的面积最大值
人气:116 ℃ 时间:2019-10-19 08:11:10
解答
用弦化边来做
∵2R(sin^2A-sin^2c)=(根号3*a-b)sinB
asinA-csinC=(根号3*a-b)sinB
a²-c²=根号3*ab-b²
∴cosC=(a²+b²-c²)/2ab=根号3*ab/2ab=根号3/2
C=π/6
S△abc=absinC/2=根号3ab/4≤根号3(a²+b²)/8,当且仅当a=b时等号成立.
此时,A=7π/12
由正弦定理,得
a/sinA=c/sinC=2R
a/sin(7π/12)=2R
a=(根号2+根号6)R
S△abc最大值=根号3a²/4=(2*根号3+3)R²
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