Question

If in a triangle $ABC,\frac{b+c}{11}=\frac{c+a}{12}=\frac{a+b}{13}$, then $\cos A$ is equal to:

• A. $\frac{19}{35}$
• B. $\frac{5}{7}$
• C. $\frac{1}{5}$
• D. $\frac{35}{19}$

Answer 1

The correct option is C $\frac{1}{5}$

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

$b+c=11k,c+a=12k,a+b=13k$.

So $2(a+b+c)=36k$

$a+b+c=18k$

$a=7k,b=6k,c=5k$

$\cos A=\frac{b^2+c^2-a^2}{2bc}$

$=\frac{(6k{)}^2+(5k{)}^2-(7k{)}^2}{2\times 6k\times 5k}$

$=\frac{61k^2-49k^2}{60k^2}=\frac{12k^2}{60k^2}$

$\cos A=\frac{1}{5}$