Question

With usual notation, if in a $△ABC,\frac{b+c}{11}=\frac{c+a}{12}=\frac{a+b}{13}$,

then prove that $\frac{\cos A}{7}=\frac{\cos B}{19}=\frac{\cos C}{25}$

Answer 1

$\frac{b+c}{11}=\frac{c+a}{12}\Rightarrow c=11a-12b$

Similarly, $a=12b-13c$ and $2b=11a-13c$

So, substracting or adding them one another gives,

$c-2b=13c-12b\Rightarrow \frac{b}{c}=\frac{12}{10}=\frac{6}{5}$

$c+a=11a-13c\Rightarrow \frac{a}{c}=\frac{14}{10}=\frac{7}{5}$

$\Rightarrow \frac{a}{b}=\frac{7}{6}$

So we can say that, $a=7k,b=6k$ and $c=5k$

$\cos A=\frac{b^2+c^2-a^2}{2bc}=\frac{36+25-49}{2\times 6\times 5}=\frac{1}{5}=\frac{7}{7\times 5}$

Similarly, $\cos B=\frac{19}{7\times 5}$ and $\cos C=\frac{5}{7}=\frac{25}{7\times 5}$

ie, $\frac{\cos A}{7}=\frac{\cos B}{19}=\frac{\cos C}{25}$