Prove that in any triangle we have the following relation $\displaystyle \frac{a+b+c}{R}=8\cos{\frac{A}{2}}\cos{\frac{B}{2} }\cos{\frac{C}{2}}$
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What's $\displaystyle R$? ... It can be useful to know that in a triangle: A+B+C=180°
R is radius of the circumscribed circle.
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
could you be more speciffic?
about what, if you point out the problem
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 } about this
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]
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