Question

With usual notations, if in a triangle $ABC,\frac{b+c}{11}=\frac{c+a}{12}=\frac{a+b}{13}$, then $\cos A:\cos B:\cos C$ is equal to

• A. $7:19:25$
• B. $19:7:25$
• C. $12:14:20$
• D. $19:25:20$

Answer 1

Answer: (A)

$7:19:25$

$\frac{b+c}{11}=\frac{c+a}{12}=\frac{a+b}{13}=\frac{2(a+b+c)}{36}$
or $\frac{b+c}{11}=\frac{c+a}{12}=\frac{a+b}{13}=\frac{a+b+c}{18}=k$
$\therefore b+c=11k,c+a=12k,a+b=13k,a+b+c=18k$
Substituting $b+c=11k$ in $a+b+c=18k$ we get
$a+11k=18k$ or $a=7k$
Substituting $a=7k$ in $c+a=12k$ we get
$c+7k=12k$ or $c=5k$
Substituting $c=5k$ in $b+c=11k$ we get
$b+5k=11k$ or $b=6k$
$\therefore a=7k,b=6k$ and $c=5k$
Using cosine rule, we get
$\cos A=\frac{b^2+c^2-a^2}{2bc}=\frac{{\left(6k\right)}^2+{\left(5k\right)}^2-{\left(7k\right)}^2}{2\times 6k\times 5k}=\frac{36k^2+25k^2-49k^2}{60k^2}=\frac{1}{5}$(On simplification)
$\cos B=\frac{c^2+a^2-b^2}{2ca}=\frac{{\left(5k\right)}^2+{\left(7k\right)}^2-{\left(6k\right)}^2}{2\times 5k\times 7k}=\frac{25k^2+49k^2-36k^2}{70k^2}=\frac{19}{35}$(On simplification)
$\cos C=\frac{a^2+b^2-c^2}{2ab}=\frac{{\left(7k\right)}^2+{\left(6k\right)}^2-{\left(5k\right)}^2}{2\times 7k\times 6k}=\frac{49k^2+36k^2-25k^2}{84k^2}=\frac{5}{7}$(On simplification)
$\cos A:\cos B:\cos C=\frac{1}{5}:\frac{19}{35}:\frac{5}{7}$
On simplification, we get
$\cos A:\cos B:\cos C=\frac{7}{35}:\frac{19}{35}:\frac{25}{35}=7:19:25$