Question

在斜三角形abc中,sinA=-根号2*cosB*cosC,且tanB*tanC=1-根号2,则角a的值为多少.要有过程

Answer 1

sinA==-根号2*cosB*cosC => cosB*cosC= -根号2/2sinA

tanB*tanC=1-根号2 => sinB/cosB*sinC/cosC=1-根号2 => （sinB*sinC)/(cosB*cosC)=1-根号2

=> (sinB*sinC)/(-根号2/2sinA)=1-根号2

=> (sinB*sinC)/sinA=1

sinA=sinB*sinC

sin(180-(B+C))=sinB*sinC

后面运用一下公式吧,公式我记不得了.