# Thread: Relation between cosines and lenghts of triangle

1. ## Relation between cosines and lenghts of triangle

Prove that in any triangle we have the following relation $\frac{a+b+c}{R}=8\cos{\frac{A}{2}}\cos{\frac{B}{2} }\cos{\frac{C}{2}}$

2. ## Re: Relation between cosines and lenghts of triangle

What's $R$? ...

It can be useful to know that in a triangle:
A+B+C=180°

3. ## Re: Relation between cosines and lenghts of triangle

R is radius of the circumscribed circle.

4. ## Re: Relation between cosines and lenghts of triangle

RHS = 8cos(A/2) cos(B/2) cos(C/2) = 8 sqrt {[s^3(s-a)(s-b)(s-c)]/(abc)^2 }

= 8s(area of triangle)/abc

now using s = (a+b+c)/2 and 4(area of triangle)/abc = 1/R you get required result

5. ## Re: Relation between cosines and lenghts of triangle

could you be more speciffic?

6. ## Re: Relation between cosines and lenghts of triangle

about what, if you point out the problem

7. ## Re: Relation between cosines and lenghts of triangle

Originally Posted by waqarhaider
RHS = 8cos(A/2) cos(B/2) cos(C/2) = 8 sqrt {[s^3(s-a)(s-b)(s-c)]/(abc)^2 }