三角形ABC中,a、b、c分别是A、B、C的对边,且有sin2C+根号3*cos(A+B)=0(1)若a=4,c=根号13,求三角形ABC的

三角形ABC中,a、b、c分别是A、B、C的对边,且有sin2C+根号3*cos(A+B)=0(1)若a=4,c=根号13,求三角形ABC的面积... 三角形ABC中,a、b、c分别是A、B、C的对边,且有sin2C+根号3*cos(A+B)=0(1)若a=4,c=根号13,求三角形ABC的面积 展开
匿名用户
2011-06-30
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sin2C+√3cos(A+B)=0
sin2C+√3cos(pi-C)=0
2sinCcosC-√3cosC=0
cosC(sinC-(√3)/2)=0
所以:cosC=0, 或sinC=(√3)/2
而:a=4,c=√13
a>c 如cosC=0,则C=90度,A<C, a<c,与条件矛盾,所以cosC=0不成立
所以:sinC=(√3)/2, C=pi/3
a/sinA=c/sinC
sinA=(a/c)sinC=(2/13)√39
cosA=(1-(12/13))^(1/2)=(√13)/13
sinB=sin(pi-(A+C))=sin(A+C)=sinAcosC+cosAsinC
=(3/26)√39
△ABC的面积=(1/2)ac*sinB=3√3
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a(3a-2b)=3(c+b)(c-b),
变形得(a²+b²-c²)/2ab=1/3
就是cosC=1/3
则sinC=2√2/3
sin2C+cos(2A+2B)
=sin2C+cos2C
=2sinCcosC+cos²C-sin²C
=(4√2-7)/9
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