Question

已知O是锐角△ABC的外接圆的圆心，且∠A=θ，若cosB/sinC*向量AB+cosC/sinB*向量AC=2m*向量AO，则m=

Answer 1

cosB/sinC*向量AB+cosC/sinB*向量AC=2m*向量AO, 向量AO=cosB/(2msinC)*向量AB+cosC/(2msinB)*向量AC, 向量BO=向量BA+AO=(cosB-2msinC)/(2msinC)*向量AB+cosC/(2msinB)*向量AC, O是锐角△ABC的外接圆的圆心, ∴OA^2=BO^2， ∴[cosB/(2msinC)]^2*AB^2+2cosBcosC/(4m^2*sinBsinC)*AB*AC =[(cosB-2msinC)/(2msinC)]^2*AB^2+2(cosB-2msinC)cosC/(4m^2*sinBsinC)*AB*AC, ∴[4m^2*(sinC)^2-4mcosBsinC]/(sinC)^2*AB^2 -4msinCcosC/(sinBsinC)*AB*AC=0, 由正弦定理，msinC-cosB-cosCcosA=0,