Question

If $\frac{b+c}{11}=\frac{c+a}{12}=\frac{a+b}{13}$ then $\cos A=$

• A. $\frac{5}{7}$
• B. $\frac{1}{5}$
• C. $\frac{2}{5}$
• D. $\frac{1}{7}$

Answer 1

The correct option is B $\frac{1}{5}$

$\frac{b+c}{11}=\frac{c+a}{12}=\frac{a+b}{13}=k$
$\Rightarrow b+c=11k,c+a=12k,a+b=13k$
adding them, $2(a+b+c)=36k$
$\Rightarrow a+b+c=18k\Rightarrow a=7k,b=6k,c=5k$
$\therefore \cos A=\frac{b^2+c^2-a^2}{2bc}=\frac{36k^2+25k^2-49k^2}{2\left(30k^2\right)}=\frac{12k^2}{60k^2}\Rightarrow \cos A=\frac{1}{5}$