Question

If $\frac{b+c}{11}=\frac{c+a}{12}=\frac{a+b}{13},$ prove that $\frac{\cos A}{7}=\frac{\cos B}{19}=\frac{\cos C}{25}$

Answer 1

$\frac{b+c}{11}=\frac{c+a}{12}=\frac{a+b}{13}=\frac{b+c+a+b+c+a}{11+12+13}$

$=\frac{2(a+b+c)}{36}$

$=\frac{a+b+c}{18}$

$\Rightarrow \frac{b+c}{11}=\frac{a}{7},\frac{c+a}{12}=\frac{b}{6},\frac{a+b}{13}=\frac{c}{5}$ then

Let $\frac{a}{7}=\frac{b}{6}=\frac{c}{5}=k$

$\Rightarrow \cos A=\frac{k^2(6^2+5^2-7^2)}{k^2\times 2\times 6\times 5}=\frac{36+25-49}{60}=\frac{12}{60}=\frac{1}{5}=\frac{7}{35}$

$\Rightarrow \cos B=\frac{k^2(5^2+7^2-6^2)}{k^2\times 2\times 5\times 7}=\frac{25+49-36}{70}=\frac{38}{70}=\frac{19}{35}$

$\Rightarrow \cos C=\frac{k^2(6^2+7^2-5^2)}{k^2\times 2\times 6\times 7}=\frac{36+49-25}{84}=\frac{60}{84}=\frac{25}{35}$

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

$\Rightarrow \frac{\frac{7}{35}}{7}=\frac{\frac{19}{35}}{19}=\frac{\frac{25}{35}}{25}=\frac{1}{35}=k$

Hence $\frac{\cos A}{7}=\frac{\cos B}{19}=\frac{\cos C}{25}$ is proved.