- 解三角形的问题3三角形ABC中边长a,b,c满足(1/b+c)+
- 三角形AB中边长a,b,c满足(1/b+c)+(1/c+a)=3/a+b+c,求角C
- (1/b+c)+(1/c+a)=3/a+b+c
==> (a+b+c)/(b+c) + (a+b+c)/(c+a) = 3
==> 1+ a/(b+c) + 1 + b/(a+c) =3
==> a/(b+c) + b/(a+c) =1
去分母,化简得
c^2 = a^2 + b^2 - ab
根据余弦定理有 c^2 = a^2 + b^2 -2abcosC
比较上下两个式子,得到 2cosC=1 ==> C=60°