三角形ABC中,三边满足于1/(a+b)+1/(b+c)=3/(a+b+c),则∠B=?如题

Question

三角形ABC中,三边满足于1/(a+b)+1/(b+c)=3/(a+b+c),则∠B=?
如题

Answer 1

1/(a+b)+1/(b+c)=3/(a+b+c)
(b+c)/[(a+b)(b+c)]+(a+b)/[(a+b)(b+c)]=3/(a+b+c)
(a+b+b+c)/[(a+b)(b+c)]=3/(a+b+c)
(a+2b+c)(a+b+c)=3(a+b)(b+c)
a^2+ab+ac+2ab+2b^2+2bc+ac+bc+c^2
=3ab+3ac+3b^2+3bc
a^2+c^2-b^2=ac
cosB=(a^2+c^2-b^2)/2ac=ac/2ac=1/2>0
B为锐角,∠B=60度