Question

In $∆ABC,\frac{b+c}{12}=\frac{c+a}{13}=\frac{a+b}{15}$. Prove that $\frac{\cos A}{2}=\frac{\cos B}{7}=\frac{\cos C}{11}$.

Answer 1

$$\begin{gathered} \mathrm{Let}\frac{b+c}{12}=\frac{c+a}{13}=\frac{a+b}{15}=k \\ \Rightarrow b+c=12k,c+a=13k,a+b=15k \\ \Rightarrow b+c+c+a+a+b=12k+13k+15k \\ \Rightarrow 2\left(a+b+c\right)=40k \\ \Rightarrow a+c+b=20k \\ \Rightarrow a+12k=20k\left(\because b+c=12k\right) \\ \Rightarrow a=8k \\ \mathrm{Also}, \\ c+a=13k \\ \Rightarrow c=13k-a=13k-8k=5k \\ \mathrm{and} \\ a+b=15k \\ \Rightarrow b=15k-a=15k-8k=7k \\ \mathrm{Now}, \\ \cos A=\frac{b^2+c^2-a^2}{2bc}=\frac{k^2\left(49+25-64\right)}{k^2\left(2\times 35\right)}=\frac{1}{7} \\ \cos B=\frac{a^2+c^2-b^2}{2ac}=\frac{k^2\left(64+25-49\right)}{\left(2\times 40\right)k^2}=\frac{1}{2} \\ \cos C=\frac{a^2+b^2-c^2}{2ab}=\frac{k^2\left(64+49-25\right)}{\left(2\times 56\right)k^2}=\frac{11}{14} \\ \therefore \cos A:\cos B:\cos C=\frac{1}{7}:\frac{1}{2}:\frac{11}{14}=2:7:11 \\ \Rightarrow \frac{\cos A}{2}=\frac{\cos B}{7}=\frac{\cos C}{11} \end{gathered}$$

Hence proved.