Let a and B be the two sides of a triangle, and let a square plus b square minus 8A minus 10B plus 41 = 0?

If the sum of the two squares of (A-4) ^ 2 + (B-5) ^ 2 = 0 is zero, then they are all zero, so a = 4; b = 5; these are identical transformations, so as long as a = 4, B = 5 must satisfy the formula given in the question. If we tell the angle between a and B, then we can get the third side C with the cosine theorem

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