三角形ABC中边长a,b,c满足(1/b+c)+(1/c+a)=3/a+b+c,求角C
热心网友
(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°
三角形ABC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°