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在三角形ABC中sin^2A+Cos^2B%Cos^2C+sinAsinC=0 B=

sinA+1-sinB-(1-sinC)+sinAsinC=0正弦定理令a/sinA=b/sinB=c/sinC=1/ksinA=ka,sinB=kb,sinC=kc代入a-b+c+ac=0a+c-b=-ac余弦定理cosB=(a+c触鞭溉庄防彪狮波饯-b)/2ac=(-ac)/(2ac)=-1/2B=120度

已知sin^2A+sin^2C-sin^2B=sinAsinC由正弦定理知a^2+c^2-b^2=ac ∴又由余弦定理知 cosB=(a^2+c^2-b^2)/2ac=1/2∴B=60°A+C=120度,故A-C=A-(120-A)=2A-120度2cos^2A+cos(A-C)=cos2A+1+cos(2A-120)=cos2A+1+(cos2Acos1

sin^2a+sinasinb=sin^2c-sin^2b (sina)^2+(sinb)^2-(sinc)^2=-sinasinb (2rsina)^2+(2rsinb)^2-(2rsinc)^2=-(2rsina)(2rsinb) a^2+b^2-c^2=-ab (a^2+b^2-c^2)^2/(2ab)=-1/2 cosc=-1/2 c=120

由正弦定理有sina/a=sinb/b=sinc/c=2r所以sina=2ar,sinb=2br,sinc=2cr因为sina+sinb=sinc所以(2ar)+(2br)=(2cr)即a+b=c所以三角形是直角三角形如果不懂,请追问,祝学习愉快!

在△ABC中,sin^2A+sin^2C=sin^B+sinAsinC则sin^2A+sin^2C-sin^2B=sinAsinC由正弦定理知a^2+c^2-b^2=ac ∴又由余弦定理知 cosB=(a^2+c^2-b^2)/2ac=1/2 ∴B=60°H为△ABC的垂心,延长AH角BC边于D点,则AD⊥BC,因为B=60°,则|

cos^2A+cos^2B>1+cos^2C1-sin^2A+1-sin^2B>1+1-sin^2Csin^2C>sin^2A+sin^2Ba/sinA=b/sinB=c/sinCc^2>a^2+b^2cosC=(a^2+b^2-c^2)/2ab

解:由sin^2A+sin^2B=6sin^2C得:a^2+b^2=6*c^2(1/tanA+1/tanB)*tanC=(cosA/sinA+cosB/sinB)sinC/cosc=(cosAsinB=sinAcosB)sinC/sinAsinBcosC=sin(A+B)sinC/sonAsinBcosC=(sinC)^2/sinAsinBscosC=c^2*2ab/ab(a^2+b^2-c^2)=2c^2/(6c^2-c^2)=2/5

a^2+b^2=2c^2 (a^2+b^2)/c^2=2 a/sina=b/sinb=c/sinc 令1/k=a/sina=b/sinb=c/sinc sina=ak,sinb=bk,sinc=ck sin^2a+sin^2b/sin^2c =(a^2k^2+b^2k^2)/c^2k^2 =(a^2+b^2)/c^2 =2

在三角形ABC中sin^2A+sin^2B+sin^2C-2cosAcosBcosC=2∵sina^2+sinb^2+sinc^2-2cosacosbcosc =3-(cosa^2+cosb^2+cosc^2+2cosacosbcosc) =3-{cosa*[cosa+2cosb*cosc]+(1/2)*[cos(2b)+cos(2c)+2]} =3-{-cos(b+c)*[-cos(b+c)+2cosb*cosc]+

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