# 1 a + b </mrow> </mfrac> + 1 a

$\frac{1}{a+b}+\frac{1}{a+c}=\frac{3}{a+b+c}$ in a triangle
Find the angle $\alpha$ of a triangle with sides a, b and c for which the equality
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Valeria Wolfe
Step 1
$\mathrm{cos}\alpha =\frac{{b}^{2}+{c}^{2}-{a}^{2}}{2bc}$
after simplifying the given equality and plug in something, but this does not seem to be easy.
$\frac{1}{a+b}+\frac{1}{a+c}=\frac{3}{a+b+c}\phantom{\rule{0ex}{0ex}}\frac{2a+b+c}{\left(a+b\right)\left(a+c\right)}=\frac{3}{a+b+c}\phantom{\rule{0ex}{0ex}}\left(2a+b+c\right)\left(a+b+c\right)=3\left(a+b\right)\left(a+c\right)$
Step 2
The last line, you get ${b}^{2}+{c}^{2}={a}^{2}+bc$ , so the $\mathrm{cos}\alpha =1/2$cosα=1/2 .

Logan Wyatt
Step 1
Multiplying both sides of the equation by
$\left(a+b\right)\left(a+c\right)\left(a+b+c\right)$ , we get
$\begin{array}{rl}\left(a+c\right)\left(a+b+c\right)+\left(a+b\right)\left(a+b+c\right)& =3\left(a+c\right)\left(a+b\right)\\ {a}^{2}+ab+2ac+bc+{c}^{2}+{a}^{2}+2ab+ac+{b}^{2}+bc& =3{a}^{2}+3ab+3ac+3bc\\ 2{a}^{2}+3ab+3ac+{b}^{2}+2bc+{c}^{2}& =3{a}^{2}+3ab+3ac+3bc\\ bc& =-{a}^{2}+{b}^{2}+{c}^{2}\end{array}$
Hence, we have $\mathrm{cos}\alpha =\frac{1}{2}$ . Since we know $0<\alpha <\pi$, we have $\alpha =\overline{)\frac{\pi }{3}}$