T TheodorMunteanu Jul 2011 19 0 Jul 18, 2011 #1 Prove that in any triangle we have the following relation \(\displaystyle \frac{a+b+c}{R}=8\cos{\frac{A}{2}}\cos{\frac{B}{2}}\cos{\frac{C}{2}}\)

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

Siron Jul 2011 1,254 389 Belgium Jul 18, 2011 #2 What's \(\displaystyle R\)? ... It can be useful to know that in a triangle: A+B+C=180°

W waqarhaider May 2011 99 32 Islamabad Jul 18, 2011 #4 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

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

T TheodorMunteanu Jul 2011 19 0 Jul 18, 2011 #7 waqarhaider said: RHS = 8cos(A/2) cos(B/2) cos(C/2) = 8 sqrt {[s^3(s-a)(s-b)(s-c)]/(abc)^2 } Click to expand... about this

waqarhaider said: RHS = 8cos(A/2) cos(B/2) cos(C/2) = 8 sqrt {[s^3(s-a)(s-b)(s-c)]/(abc)^2 } Click to expand... about this

W waqarhaider May 2011 99 32 Islamabad Jul 19, 2011 #8 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]