Question

The question is below?

If in a triangle `ABC` ,`(b+c)/11=(c+a)/12=(a+b)/13` then prove that `cosA/7=cosB/19=cosC/25`

Answer 1

Given
`InDeltaABC ,(b+c)/11=(c+a)/12=(a+b)/13`

Let

`(b+c)/11=(c+a)/12=(a+b)/13`

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

`=>(b+c)/11=(c+a)/12=(a+b)/13=(a+b+c)/18`

`=>a/7=b/6=c/5=k(say)`

So

`cosA=(b^2+c^2-a^2)/(2bc)`

`=>cosA=(36k^2+25k^2-49k^2)/(2*6*5k^2)=1/5`

`cosB=(c^2+a^2-b^2)/(2ca)`

`=>cosB=(25k^2+49k^2-36k^2)/(2*5*7k^2)=19/35`

`cosC=(a^2+b^2-c^2)/(2ab)`

`=>cosC=(49k^2+36k^2-25k^2)/(2*7*6k^2)=5/7`

Now

`cosA:cosB:cosC=1/5:19/35:5/7=7:19:25`

`=>cosA/7=cosB/19=cosC/25`