For a triangle with sides aaa, bbb, ccc, prove that asin(B)=bsin(A)a\sin(B) = b\sin(A)asin(B)=bsin(A).
False; this only holds for right triangles
True; this follows directly from the Law of Sines
False; the correct statement is asin(C)=csin(A)a\sin(C) = c\sin(A)asin(C)=csin(A)
True; but requires A≠BA \neq BA=B