Question

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

Answer 1

Let $\frac{b+c}{12}=\frac{c+a}{13}=\frac{a+b}{15}=k$

$\Rightarrow b+c=12k,c+a=13k,a+b=15k$

On solving, $a=8k,b=7k,c=5k$

Now, $\cos A=\frac{b^2+c^2-a^2}{2bc}=\frac{49+25-64}{\mathrm{2.7.5}}=\frac{1}{7}$

$\cos B=\frac{a^2+c^2-b^2}{2ac}=\frac{64+25-49}{\mathrm{2.8.5}}=\frac{1}{2}$

$\cos C=\frac{a^2+b^2-c^2}{2ab}=\frac{64+49-25}{\mathrm{2.8.7}}=\frac{11}{14}$

$\cos A:\cos B:\cos C=\frac{1}{7}:\frac{1}{2}:\frac{11}{14}=2:7:11$

Hence, proved