在任意△ABC中
作AD⊥BC.
∠C所對(duì)的邊為c,∠B所對(duì)的邊為b,∠A所對(duì)的邊為a
則有BD=cosB*c,AD=sinB*c,DC=BC-BD=a-cosB*c
根據(jù)勾股定理可得：
AC^2=AD^2+DC^2
b^2=(sinB*c)^2+(a-cosB*c)^2
b^2=sin^2B*c^2+a^2+cos^2B*c^2-2ac*cosB
b^2=(sin^2B+cos^2B)*c^2-2ac*cosB+a^2
b^2=c^2+a^2-2ac*cosB
cosB=(c^2+a^2-b^2)/2ac