已知o为锐角△ABC的外心,且A=6/π,若cosB/sinC向量AB+cosC/sinB向量AC=m向量OA,求
已知o为锐角△ABC的外心,且A=6/π,若cosB/sinC向量AB+cosC/sinB向量AC=m向量OA,求
已知o为锐角△ABC的外心,且A=6/π,若cosB/sinC向量AB+cosC/sinB向量AC=m向量OA,求
已知o为锐角△ABC的外心,且A=6/π,若cosB/sinC向量AB+cosC/sinB向量AC=m向量OA,求
取AB中点D,则有 AO = AD + DO ,
代入cosB /sinC AB +cosC/ sinB AC =2m AO 得:
cosB/ sinC AB +cosC/ sinB AC =m( AD + DO ),
由 OD ⊥ AB ,得 DO • AB =0,
∴两边同乘 AB ,化简得:
cosB /sinC AB • AB +cosC/ sinB AC • AB =m( AD + DO )• AB =1/2m AB • AB ,
即cosB /sinC c2+cosC/ sinB bc•cosA=1/2mc2,
由正弦定理a/ sinA =b/ sinB =c/ sinC 化简得:
cosB/ sinC* sin^2C+cosC /sinB* sinBsinCcosA=1/2msin^2C,
由sinC≠0,两边同时除以sinC得:cosB+cosAcosC=1/2msinC,
∴m=2[cosB+cosAcosC]/ sinC =2[-cos(A+C)+cosAcosC]/ sinC=2[-cosAcosC+sinAsinC+cosAcosC ]/sinC =2sinA,
又∠A=Pai/6
则m=1
故答案为:m=1
由|OA|=|OB|=|OC|知O是△ABC的外心,
∴AB*AO=(1/2)AB^2,AC*AO=(1/2)AC^2,
由(cosB/sinC)AB+(cosC/sinB)AC=2mOA得
(cosB/sinC)AB^2+(cosC/sinB)AC^2=4mOA^2,
由正弦定理得
m=sinCcosB+sinBcosC=sin(B+C)=sinA=sin30°=1/2.