If the three sides of △ ABC are three consecutive positive integers, and a > b > C, 3b = 20acos & nbsp; a, then Sin & nbsp; a: Sin & nbsp; B: Sin & nbsp; C=______ ．

If the three sides of △ ABC are three consecutive positive integers, and a > b > C, 3b = 20acos & nbsp; a, then Sin & nbsp; a: Sin & nbsp; B: Sin & nbsp; C=______ ．


Since the lengths of a, B, C are three consecutive positive integers, and a ＞ B ＞ C, we can assume that the lengths of three sides are a, A-1, A-2 respectively. From the cosine theorem, we can get: cosa = B2 + C2 − a22bc = (a − 1) 2 + (a − 2) 2 − A22 (a − 1) (a − 2) = a − 52 (a − 2), and 3b = 20acosa, we can get: cosa = 3b20a = 3A − 320A

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