In `triangleABC`, If `(1)/(b+c)+(1)/(c+a)=(3)/(a+b+c)`, then `cosC=`

In triangleABC , If (b+c)/(11)=(c+a)/(12)=(a+b)/(13) , then cosC=

In triangleABC , If (b+c)/(11)=(c+a)/(12)=(a+b)/(13) , then cosC=

In a triangleABC , if 1/(b+c)+1/(c+a)=3/(a+b+c) , then angleC =

IN triangleABC, (cosC+cosA)/(c+a)+(cosB)/(b)=

IN triangleABC, (cosC+cosA)/(c+a)+(cosB)/(b)=

In any triangle ABC, If b^(2) = a(c+a), c^(2) = b(a+b) , then prove that, cosAcosB cosC =-1/8 .

If in triangleABC, tan^(-1)(a/(b+c))+ tan^(-1)(c/(a+b)) = pi/4 , then prove that, the triangle is right angled.

If, in triangleABC, B=3C , then: (b-c)/(2c)=