在三角形ABC中,sin(C-A)=1,sinB=1/3,求sinA,若AC=根号下6,求三角型面积

网友回答

C=90+A
sinB=sin(A+C) = sinAcosC+cosAsinC = -sin^2A + cos^2A = 1-2sin^2A
sinA = sqrt(3)/3
AC=b = sqrt(6)
a /sinA = b/sinB
a = sinA/sinB*b
S=1/2 ab sinC = 1/2 b^2 sinAsinC/sinB = sqrt(3)*sqrt(6)/3*3 = 3sqrt(2)