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If in a triangle $A B C, \frac{b+c}{11}=\frac{c+a}{12}=\frac{a+b}{13}$, then $\cos \mathrm{A}$ is equal to
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$1 / 5$
$1 / 5$
Let $\frac{b+c}{11}=\frac{c+a}{12}=\frac{a+b}{13}=K$ in a triangle $A B C$. $\Rightarrow b+c=11 K, c+a=12 K, a+b=13 K$ On solving these, we get $a=7 K, b=6 K, c=5 K$
Now we know,
$$
\begin{aligned}
& \cos A=\frac{b^2+c^2-a^2}{2 b c} \\
& =\frac{36 K^2+25 K^2-49 K^2}{2(6 K) 5 K}=\frac{1}{5}
\end{aligned}
$$
Now we know,
$$
\begin{aligned}
& \cos A=\frac{b^2+c^2-a^2}{2 b c} \\
& =\frac{36 K^2+25 K^2-49 K^2}{2(6 K) 5 K}=\frac{1}{5}
\end{aligned}
$$
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