Is it Pythagoras related?

Relevant wiki: Sine Rule (Law of Sines)

$\text{Using the sine rule }$ $\large{\frac { a }{ \sin { A } } =\frac { b }{ \sin { B } } =\frac { c }{ \sin { C } } }= \small 2R$ $\text{we get }$ $\frac { 3 }{ \frac { 5 }{ 13 } } =\frac { 8 }{ \sin { B } }$ $\sin { B=\frac { 40 }{ 39 } } \Rightarrow \sin { B>1 } \\$ $\text{As we know that the maximum value of }sin B \text{ can be 1, hence this triangle is not possible.}$

We are given that $a=3, b=8,$ and $\sin A=\dfrac{5}{13}$ so considering the possibility of triangle $ABC$ being a right triangle, we see that $\dfrac{5}{13}=\dfrac{3}{c},$ because $\sin A= \dfrac{\text{Opposite(A)}}{\text{Hypotenuse}},$ where $c$ is the length of the hypotenuse of such a triangle. This leaves us with $\dfrac{5}{39}=\dfrac{1}{c} \\ \implies c=\dfrac{39}{5} \Longrightarrow c<8.$

For any triangle $ABC,$ such that $a=3$ and $\sin A=\dfrac{5}{13},$ the maximum possible length of any side of such a triangle should be $\dfrac{39}{5}.$ Here, we have a side of length $8$ which is larger than this maximal value hence, there are no triangles that can be formed from this data.

Shortest value ' a' can have is the perpendicular from B to AC .
The length of the perpendicular =b * Sin A =3.08.
But a=3<3.08.
So the triangle is not possible.

I used law of cosin and found the side opposite angle A to be approximatly 10.83. Then i used law of sin and found the two other angles. Logically the longest side should be opposite the largest angle and that didn't happen. Thus 0 triangles. This required use of calculator so not best solution.