A hexagon with sides of length 2, 7, 2, 11, 7, 11 is inscribed in a circle. Find the radius of the circle.
As suggested, *if* it matters, you may assume that the sides listed are given in order
(In reply to
No Subject by Gamer)
I also got the equation: arccos(1-(2/x)²/2)+ arccos(1-(7/x)²/2)+ arccos(1-(11/x)²/2) = pi
To solve it, first take sin()'s of each side and use trig identities to get the equivalent equation: (2/x)*(sqrt(1-4/(4x^2)))*(1-49/(2x^2))*)))*(1-121/(2x^2)) + (7/x)*(sqrt(1-49/(4x^2)))*(1-4/(2x^2))*)))*(1-121/(2x^2)) + (11/x)*(sqrt(1-121/(4x^2)))*(1-4/(2x^2))*)))*(1-49/(2x^2)) - (154/x^3)*(sqrt(1-4/(4x^2)))*(sqrt(1-49/(4x^2)))*(sqrt(1-121/(4x^2))) = 0
A little bit of algebra reduces the equation to: (8x^4 - 680x^2 + 11858)*sqrt(4x^2-4) + (28x^4 - 1750x^2 + 3388)*sqrt(4x^2-49) + (44x^4 - 1166x^2 + 2156)*sqrt(4x^2-121) - 154*sqrt(4x^2-4)*(sqrt(4x^2-49)*sqrt(4x^2-121).
To evaluate this monster equation, I used the polynomial abilities of UBASIC to simplify the polynomials generated by isolating the roots and squaring. The equation left over after eliminating common factors is: 3840x^8 - 310364x^6 + 5202948x^4 + 39184761x^2 - 35153041 = 0. 7 is indeed a root of the equation.
Solving the arccos equation.
