Question

在△ABC中,∠B=π/3,AB=1,AC=根号13,求BC边中线AD=?

Answer 1

AC^2=AB^2+BC^2-2AB*BC*cosB

13=1+BC^2-2*1*BC*0.5

BC^2-BC-12=0

BC=-3(舍）或BC=4

BD=2

AD^2=AB^2+BD^2-2×AB×BD×cosB

AD^2=1+4-2×1×2×0.5=3

AD=根号3