# The question is below?

## If in a triangle $A B C$ ,$\frac{b + c}{11} = \frac{c + a}{12} = \frac{a + b}{13}$ then prove that $\cos \frac{A}{7} = \cos \frac{B}{19} = \cos \frac{C}{25}$

Jun 5, 2018

Given
$I n \Delta A B C , \frac{b + c}{11} = \frac{c + a}{12} = \frac{a + b}{13}$

Let

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

=>(b+c)/11=(c+a)/12=(a+b)/13=(2(a+b+c))/((11+12+13)

$\implies \frac{b + c}{11} = \frac{c + a}{12} = \frac{a + b}{13} = \frac{a + b + c}{18}$

$\implies \frac{a}{7} = \frac{b}{6} = \frac{c}{5} = k \left(s a y\right)$

So

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

$\implies \cos A = \frac{36 {k}^{2} + 25 {k}^{2} - 49 {k}^{2}}{2 \cdot 6 \cdot 5 {k}^{2}} = \frac{1}{5}$

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

$\implies \cos B = \frac{25 {k}^{2} + 49 {k}^{2} - 36 {k}^{2}}{2 \cdot 5 \cdot 7 {k}^{2}} = \frac{19}{35}$

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

$\implies \cos C = \frac{49 {k}^{2} + 36 {k}^{2} - 25 {k}^{2}}{2 \cdot 7 \cdot 6 {k}^{2}} = \frac{5}{7}$

Now

$\cos A : \cos B : \cos C = \frac{1}{5} : \frac{19}{35} : \frac{5}{7} = 7 : 19 : 25$

$\implies \cos \frac{A}{7} = \cos \frac{B}{19} = \cos \frac{C}{25}$