Question

In $△ABC$, if $\frac{b+c}{11}=\frac{c+a}{12}=\frac{a+b}{13}$, then prove $\frac{\cos B}{19}=\frac{\cos C}{25}$

Answer 1

$\begin{array}{c}Given,\\ \frac{b+c}{11}=\frac{c+a}{12}=\frac{a+b}{13}\\ Now,\\ \Rightarrow \frac{b+c}{11}=\frac{c+a}{12}=\frac{a+b}{13}=\frac{2\left(a+b+c\right)}{36}=\frac{\left(a+b+c\right)}{18}\\ \Rightarrow \frac{b+c}{11}=\frac{\left(a+b+c\right)}{18}\\ \Rightarrow \frac{a}{7}=\frac{\left(a+b+c\right)}{18}\\ Similarly\frac{b}{6}=\frac{\left(a+b+c\right)}{18}\\ \frac{c}{5}=\frac{\left(a+b+c\right)}{18}\\ Hence,\frac{\sin A}{7}=\frac{\sin B}{6}=sinC\\ \Rightarrow \frac{\cos A}{7}=\frac{\cos B}{19}=\frac{\cos C}{25}\end{array}$