With usual notion, if in triangle `A B C ,` `(b+c)/(11)=(c+a)/(12)=(a+b)/(13),t h e np rov et h a t` `(cosA)/7=(cosB)/(19)=(cosC)/(25)`

Explanation

Let `(b + c)/(11) = (c + a)/(12) = (a +b)/(13) = k`
`:. B + c = 11 k`
`c + a = 12 k`
`a + b = 13 k`
Adding the above three equations, we get
`2(a + b + c) = 36k`
or `a + b + c = 18 k`
From Eqs. (i) and (iv), `a = 7 k`
From Eqs. (ii) and (iv), `b = 6k`
From Eqs. (iii) and (iv), `c = 5k`
`cos A = (b^(2) + c^(2) - a^(2))/(2bc) = (36k^(2) + 25k^(2) - 49k^(2))/(2 xx 6k xx 5k)`
`= (12k^(2))/(60k^(2)) = (1)/(5)`
`cos B = (c^(2) + a^(2) -b^(2))/(2ca) = (25 k^(2) + 49k^(2) - 36k^(2))/(2 xx 5k xx 7k)`
`= (38k^(2))/(70k^(2)) = (19)/(35)`
`cos C = (a^(2) + b^(2) -c^(2))/(2ab) = (49k^(2) + 36 k^(2) - 25 k^(2))/(2 xx 7k xx 6k)`
`= (60k^(2))/(84 k^(2)) = (5)/(7)`
`:. (cos A)/((1//5)) = (cos B)/((19 //35)) = (cos C)/((5//7))`
or `(cos A)/((7//35)) = (cos B)/((19//35)) = (cos C)/((25//35))`
or `(cos A)/(7) = (cos B)/(19) = (cos C)/(25)`