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If $\frac{b+c}{11}=\frac{c+a}{12}=\frac{a+b}{13}$, then $\cos A$ is equal to
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Verified Answer
The correct answer is:
$\frac{1}{5}$
Let $\frac{b+c}{11}=\frac{c+a}{12}=\frac{a+b}{13}=k$ $\Rightarrow b+c=11 k, c+a=12 k, a+b=13 k$
Now, $2(a+b+c)=36 k$
$a+b+c=18 k$
So, $a=7 k$
$b=6 k$
and $\quad c=5 k$
Now, $\quad \cos A=\frac{b^{2}+c^{2}-a^{2}}{2 b c}$
$$
=\frac{36 k^{2}+25 k^{2}-49 k^{2}}{2\left(30 k^{2}\right)}=\frac{12 k^{2}}{60 k^{2}}=\frac{1}{5}
$$
Now, $2(a+b+c)=36 k$
$a+b+c=18 k$
So, $a=7 k$
$b=6 k$
and $\quad c=5 k$
Now, $\quad \cos A=\frac{b^{2}+c^{2}-a^{2}}{2 b c}$
$$
=\frac{36 k^{2}+25 k^{2}-49 k^{2}}{2\left(30 k^{2}\right)}=\frac{12 k^{2}}{60 k^{2}}=\frac{1}{5}
$$
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