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

Printable View

- Jul 18th 2011, 10:41 AMTheodorMunteanuRelation between cosines and lenghts of triangle
Prove that in any triangle we have the following relation $\displaystyle \frac{a+b+c}{R}=8\cos{\frac{A}{2}}\cos{\frac{B}{2} }\cos{\frac{C}{2}}$

- Jul 18th 2011, 10:46 AMSironRe: Relation between cosines and lenghts of triangle
What's $\displaystyle R$? ...

It can be useful to know that in a triangle:

A+B+C=180° - Jul 18th 2011, 10:47 AMTheodorMunteanuRe: Relation between cosines and lenghts of triangle
R is radius of the circumscribed circle.

- Jul 18th 2011, 11:17 AMwaqarhaiderRe: 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 - Jul 18th 2011, 11:23 AMTheodorMunteanuRe: Relation between cosines and lenghts of triangle
could you be more speciffic?

- Jul 18th 2011, 11:36 AMwaqarhaiderRe: Relation between cosines and lenghts of triangle
about what, if you point out the problem

- Jul 18th 2011, 11:53 AMTheodorMunteanuRe: Relation between cosines and lenghts of triangle
- Jul 19th 2011, 10:12 AMwaqarhaiderRe: Relation between cosines and lenghts of triangle
cos(A/2) = sqrt[ s(s-a)/bc] , cos(B/2) = sqrt[ s(s-b)/ac] and cos(C/2) = sqrt[s(s-c)/ab]