Find missing angles/sides with (a−b)/(a+b) = tan((A−B)/2)/tan((A+B)/2).
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Why it works
The law of tangents links side differences to angle differences. Given a, b and C, half-sum S = (180°−C)/2 is known, so tan((A−B)/2) = ((a−b)/(a+b)) tan S unlocks the half-difference — then A = S+D, B = S−D. It was the pre-calculator workhorse for SAS triangles.
