已知O为锐角三角形ABC的外心,角B=30°,若(向量)BA*cosA/sinC+(向量)BC*cosC/sinA=2m(向量)OB,则实数m的值为?
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![已知O为锐角三角形ABC的外心,角B=30°,若(向量)BA*cosA/sinC+(向量)BC*cosC/sinA=2m(向量)OB,则实数m的值为?](/uploads/image/z/10155414-30-4.jpg?t=%E5%B7%B2%E7%9F%A5O%E4%B8%BA%E9%94%90%E8%A7%92%E4%B8%89%E8%A7%92%E5%BD%A2ABC%E7%9A%84%E5%A4%96%E5%BF%83%2C%E8%A7%92B%3D30%C2%B0%2C%E8%8B%A5%28%E5%90%91%E9%87%8F%29BA%2AcosA%2FsinC%2B%28%E5%90%91%E9%87%8F%29BC%2AcosC%2FsinA%3D2m%28%E5%90%91%E9%87%8F%29OB%2C%E5%88%99%E5%AE%9E%E6%95%B0m%E7%9A%84%E5%80%BC%E4%B8%BA%3F)
已知O为锐角三角形ABC的外心,角B=30°,若(向量)BA*cosA/sinC+(向量)BC*cosC/sinA=2m(向量)OB,则实数m的值为?
已知O为锐角三角形ABC的外心,角B=30°,若(向量)BA*cosA/sinC+(向量)BC*cosC/sinA=2m(向量)OB,则实数m的值为?
已知O为锐角三角形ABC的外心,角B=30°,若(向量)BA*cosA/sinC+(向量)BC*cosC/sinA=2m(向量)OB,则实数m的值为?
取AB中点D,则有 OB = OD + DB ,
代入cosA /sinC BA +cosC/ sinA BC =2m OB 得:
cosB/ sinC AB +cosC/ sinB AC =2m( 0D + DB ),
由 OD ⊥ AB ,得 OD • AB =0,
∴两边同乘BA ,化简得:
cosA /sinC BA • BA +cosC/ sinA BC • BA =2m( OD + DB )• BA =m AB • AB ,
即cosA /sinC c^2+cosC/ sinA ac•cosB=mc^2,
由正弦定理a/ sinA =b/ sinB =c/ sinC 化简得:
cosA/ sinC* sin^2C+cosC /sinA* sinAsinCcosB=msin^2C,
由sinC≠0,两边同时除以sinC得:cosA+cosBcosC=msinC,
∴m=[cosA+cosBcosC]/ sinC =[-cos(B+C)+cosBcosC]/ sinC=[-cosBcosC+sinBsinC+cosBcosC ]/sinC =sinB,
又∠B=30,所以有m=sin30=1/2.
前面这位老兄在两边同乘BA后,等号右边应该是负的。所以最后结果应该是—1/2。这方法真不错!