In a circle, two chords of lengths 4 and 11 respectively subtend central angles whose degree-measures are in the ratio 1 to 3 respectively. Determine the length of the radius of the circle.
By the Law of Cosines, we have that
4^2 = 2R^2 - 2R^2 ( cos A)
16 -2R^2
_______ = (cos A)
-2R^2
R^2 - 8
_______ = (cos A)
R^2
11^2 = 2R^2 - 2R^2 (cos 3A)
121 - 2R^2
__________ = cos(3A)
- 2R^2
2R^2 - 121
_________ = cos(3A)
2R^2
Using the identity (cos3A) = 4 (cos A)^3 - 3cosA
2R^2 - 121 4 [ R^2 - 8 ] ^3 3 [ R^2 - 8 ]
__________ = _____________ - ____________
2R^2 R^6 R^2
2R^2 - 121 6 [ R^2 - 8] 4 [ R^2 - 8]^3
_________ + ___________ = ____________
2R^2 2R^2 R^6
8R^2 - 169 4 [ R2 - 8 ] ^3
__________ = ____________
2R^2 R^6
R^4 [8R^2 - 169] = 8 [ R^2 - 8 ] ^3
8R^6 - 169R^4 = 8 [ R^6 - 24R^4 + 192R^2 - 512 ]
8R^6 - 169R^4 = 8R^6 - 192R^4 + 1536R^2 - 4096
23R^4 - 1536R^2 + 4096 = 0
Factor as
(23R^2 - 64) (R^2 - 64) = 0
So either
23R^2 - 64 = 0
R^2 = 64 / 23
R = ± √[64/ 23] ≈ ±1.66 units ( reject.....the diameter would be shorter than either chord....which is impossible )
Or
R^2 - 64 = 0
R = 8 units = the radius