JEE Main 2002 Mathematics question on Trigonometry

From the sine rule, a/sin A = 2R. If a and sin A are both given, then 2R is already determined — meaning R provides no new information. So the set {a, sin A, R} gives only a/sin A = 2R (one equation), which fixes the shape's sine ratios but not a unique triangle; infinitely many similar triangles (scaled differently, or with different angle combinations satisfying the same ratio) remain possible since knowing a and sin A already forces 2R, and no additional independent constraint on B or C is introduced. In contrast, option 3 gives sin B from which B is known, then R gives 2R = b/sin B fixing b, and a is also known — enough to uniquely construct the triangle. Thus option 4 is redundant and does not uniquely determine the triangle.