已知O是三角形ABC的外心,AB=2 AC=1,角BAC=120°.若向量AO=m*向量AB+n*向量AC 则m+n=由余弦定理:BC=(AB^2+AC^2-2*AB*AC*cos120°)^(1/2)= (4+1+2) ^(1/2)=7^(1/2)则 AO=(BC/2)/cos30°=(7/3)^(1/2)过O作AC的垂线与AC交于D,再过O作A
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![已知O是三角形ABC的外心,AB=2 AC=1,角BAC=120°.若向量AO=m*向量AB+n*向量AC 则m+n=由余弦定理:BC=(AB^2+AC^2-2*AB*AC*cos120°)^(1/2)= (4+1+2) ^(1/2)=7^(1/2)则 AO=(BC/2)/cos30°=(7/3)^(1/2)过O作AC的垂线与AC交于D,再过O作A](/uploads/image/z/6872167-55-7.jpg?t=%E5%B7%B2%E7%9F%A5O%E6%98%AF%E4%B8%89%E8%A7%92%E5%BD%A2ABC%E7%9A%84%E5%A4%96%E5%BF%83%2CAB%3D2+AC%3D1%2C%E8%A7%92BAC%3D120%C2%B0.%E8%8B%A5%E5%90%91%E9%87%8FAO%3Dm%2A%E5%90%91%E9%87%8FAB%2Bn%2A%E5%90%91%E9%87%8FAC+%E5%88%99m%2Bn%3D%E7%94%B1%E4%BD%99%E5%BC%A6%E5%AE%9A%E7%90%86%EF%BC%9ABC%3D%28AB%5E2%2BAC%5E2-2%2AAB%2AAC%2Acos120%C2%B0%29%5E%281%2F2%29%3D+%284%2B1%2B2%29+%5E%281%2F2%29%3D7%5E%281%2F2%29%E5%88%99+AO%3D%28BC%2F2%29%2Fcos30%C2%B0%3D%287%2F3%29%5E%281%2F2%29%E8%BF%87O%E4%BD%9CAC%E7%9A%84%E5%9E%82%E7%BA%BF%E4%B8%8EAC%E4%BA%A4%E4%BA%8ED%2C%E5%86%8D%E8%BF%87O%E4%BD%9CA)
已知O是三角形ABC的外心,AB=2 AC=1,角BAC=120°.若向量AO=m*向量AB+n*向量AC 则m+n=由余弦定理:BC=(AB^2+AC^2-2*AB*AC*cos120°)^(1/2)= (4+1+2) ^(1/2)=7^(1/2)则 AO=(BC/2)/cos30°=(7/3)^(1/2)过O作AC的垂线与AC交于D,再过O作A
已知O是三角形ABC的外心,AB=2 AC=1,角BAC=120°.若向量AO=m*向量AB+n*向量AC 则m+n=
由余弦定理:BC=(AB^2+AC^2-2*AB*AC*cos120°)^(1/2)
= (4+1+2) ^(1/2)=7^(1/2)
则 AO=(BC/2)/cos30°=(7/3)^(1/2)
过O作AC的垂线与AC交于D,再过O作AB的平行线与AC的延长线交于E,
则 DO=(AO^2-(AC/2)^2)^(1/2)=(7/3-1/4)^(1/2)=(25/12)^(1/2)
∵∠DEO=60°
∴DO/EO=cos30°
∴EO=DO/cos30°=(25/12)^(1/2)*(2/3^(1/2))=5/3
∴DE=EO/2=5/6
∴AE=DE+AC/2=5/6+1/2=4/3
过O作AC的平行线与AB交于F,则四边形FAEO是平行四边形,
向量AO=向量AF+向量AE=m*向量a+n*向量b
∴|向量AF|=m*|向量a|,|向量AE|=n*|向量b|
∵|向量AF|=EO=5/3,|向量a|=2,|向量AE|=4/3,|向量b|=1
∴5/3=2m,4/3=n
∴m + n = 5/6 + 4/3 = 13/6
请问AO=(BC/2)/cos30°=(7/3)^(1/2)怎么来的?
已知O是三角形ABC的外心,AB=2 AC=1,角BAC=120°.若向量AO=m*向量AB+n*向量AC 则m+n=由余弦定理:BC=(AB^2+AC^2-2*AB*AC*cos120°)^(1/2)= (4+1+2) ^(1/2)=7^(1/2)则 AO=(BC/2)/cos30°=(7/3)^(1/2)过O作AC的垂线与AC交于D,再过O作A
解析:设外接圆半径为R=AO=BO=CO,不妨连接OC,OB,过O做
OD⊥BC于D,由∠BAC=120°,得∠BOC=120°,
又∵OC=OB,
∴∠ OCB=∠OBC=(180°-120°)/2=30°,
在Rt△OCD中,CD=BC/2=√7/2,∠OCD=30°,
∴ OC=CD/cos30°=(BC/2)/cos30°
即OA==(BC/2)/cos30°=(7/3)^(1/2)
明白了吗?
自己画个图,
点O在△ABC外,∠BOC=360°-2∠A=120°,
因为BO=CO,所以∠OBC=∠OCB=30°,则
BO=(BC/2)/cos30°=AO
6
先用余弦定理求BC边,在用正弦定理求外接球的半径