Question

在三角形ABC中,B=30度,AB=2倍根号3,面积S=根号3,求AC?

Answer 1

S=0.5*a*c*sinB

=0.5*BC*AB*sinB

=BC*2√3*0.5

=√3

得BC=2

cosB=(a^2+c^2-b^2)/(2a*c)

=(BC^2+AB^2-AC^2)/(2*BC*AB)

=(4+12-AC^2)/(2*2*2√3)

=(√3)/2

得AC=2