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Topic: CAML (Read 1156 times) |
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LZJ
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Re: CAML
« Reply #25 on: Jun 21st, 2003, 2:25am » |
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Sorry for not posting my reply before...
Let cos-1x be a, cos-12x be b, cos-13x be c, and let a, b and c be the angles of a triangle.
Let the length of the triangle side opposite angle a be A, and similarly for the other 2 angles.
Using cosine rule,
C2 = A2 + B2 - 6ABx --(1)
(since cos(c) = 3x)
B2 = A2 + C2 - 4ACx --(2)
A2 = B2 + C2 - 2BCx --(3)
Simplify:
Example of simplification...sub (3) into (2)
B2 = (B2 + C2 - 2BCx) + C2 - 4ACx
2C2 = 2C(2Ax + Bx)
Since A, B, C > 0,
C = 2Ax + Bx
B = 3Ax + Cx
A = 2Cx + 3Bx
C = 4Cx2 + 6Bx2 + Bx,
C(1 - 4x2) = B(6x2 + x)
B = 9Bx2 + 6Cx2 + Cx,
B(1 - 9x2) = C(6x2 + x)
1 + 36x4 - 13x2
= 36x4 + x2 + 12x3
12x3 + 14x2 - 1 = 0
See? Readily solvable in less than 10 minutes 
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Sir Col
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impudens simia et macrologus profundus fabulae
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Re: CAML
« Reply #26 on: Jun 21st, 2003, 3:24am » |
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Very impressive, LZJ, and thanks for taking the time to share your solution. When I tried it, I missed the rather obvious fact that A,B,C>0! 
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