Trigonometry! #70

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

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

$a-b=b-c=\frac{a-c}{2}$

$\therefore \frac{b+c}{11}=b-c\rightarrow5b=6c$

$\therefore\frac{a+c}{12}=\frac{a-c}{2}\rightarrow5a=7c$

$\therefore\frac{a+b}{13}=a-b\rightarrow6a=7b$

$\therefore \frac{a}{7}=\frac{b}{6}=\frac{c}{5}=p$

$\Rightarrow a=7p,~b=6p,~c=5p$

$\cos A=\frac{b^{2}+c^{2}-a^{2}}{2bc}=\frac{36p^{2}+25p^{2}-49p^{2}}{2(6p)(5p)}=\frac{12p^{2}}{30p^{2}}=\frac{1}{5} \\ \Rightarrow5\cos A=1$

$\cos B=\frac{a^{2}+c^{2}-b^{2}}{2ac}=\frac{49p^{2}+25p^{2}-36p^{2}}{2(7p)(5p)}=\frac{38p^{2}}{70p^{2}}=\frac{19}{35} \\ \Rightarrow \frac{35\cos B}{19}=1$

$\cos C==\frac{a^{2}+b^{2}-c^{2}}{2ab}=\frac{49p^{2}+36p^{2}-25p^{2}}{2(7p)(6p)}=\frac{60p^{2}}{84p^{2}}=\frac{5}{7} \\ \Rightarrow\frac{7\cos C}{5}=1$

$\therefore 5\cos A=\frac{35\cos B}{19}=\frac{7\cos C}{5}$

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

(b + c)/ 11 = (c + a)/ 12 = (a + b)/ 13 = k = (a - b)/ (12 - 11) = (a - c)/ (13 - 11)

13 (a - b) = a + b and 12 (a - c) = 2 (a + c)

a/ 7 = b/ 6 = c/ 5

Cos A = (36 + 25 - 49)/ (2)(6)(5) = 1/ 5

Cos B = (25 + 49 - 36)/ (2)(5)(7) = 19/ 35

Cos C =(49 + 36 - 25)/ (2)(7)(6) = 5/ 7

For ratios but not exact or absolute, same concept of coprime is applicable to describe 3 elements x, y and z wanted. Therefore,

Cos A/ 7 = Cos B/ 19 = Cos C/ 25 = 1/ 35 <> 1

Sum of integers = 7 + 19 + 25 = 51