We begin by finding the third side of the triangle. Using the Law of Cosines, we have

\begin{align*}
b^{2} \amp = a^{2} + c^{2} - 2ac \cos (B) \amp\amp \blert{\text{Substitute the known values.}}\\
3^{2} \amp = 8^{2} + c^{2} - 2(8)c \cos (14.4\degree) \amp\amp \blert{\text{Simplify.}}\\
9 \amp = 64 + c^{2} - 16c (0.9686) \amp\amp \blert{\text{Write the quadratic equation in standard form.}}\\
0 \amp = c^{2}-15.497c + 55 \amp\amp \blert{\text{Apply the quadratic formula.}}
\end{align*}

\begin{align*}
c \amp = \dfrac{15.497 \pm \sqrt{(-15.497)^{2} - 4(1)(55)}}{2(1)} \amp\amp \blert{\text{Simplify.}}\\
\amp = \dfrac{15.497 \pm 4.490}{2} = 5.503~ \text{ or }~ 9.994
\end{align*}

Because there are two positive solutions for side \(c\text{,}\) either \(c = 5.503\) or \(c = 9.994\text{,}\) there are two triangles with the given properties. We apply the Law of Cosines again to find angle \(C\) in each triangle.

For the triangle with \(c = 5.503\text{,}\) we have

\begin{align*}
c^{2} \amp = a^{2} + b^{2} - 2ab \cos (C) \amp\amp \blert{\text{Substitute the known values.}}\\
5.503^{2} \amp = 8^{2} + 3^{2} - 2(8)(3) \cos (C) \amp\amp \blert{\text{Solve for} \cos (C).}\\
\cos (C) \amp = \dfrac{5.503^{2} - 64 - 9}{-48} = 0.889876\\
C \amp = \cos^{-1} 0.889876) = 27.1\degree
\end{align*}

and \(A = 180\degree - (14.4\degree + 27.1\degree) = 138.5\degree\text{.}\)

For the triangle with \(c = 9.994\text{,}\) we have

\begin{align*}
c^{2} \amp = a^{2} + b^{2} - 2ab \cos (C) \amp\amp \blert{\text{Substitute the known values.}}\\
9.994^{2} \amp = 8^{2} + 3^{2} - 2(8)(3) \cos (C) \amp\amp \blert{\text{Solve for} \cos (C).}\\
\cos (C) \amp = \dfrac{9.994^{2} - 64 - 9}{-48} = -0.560027\\
C \amp = \cos^{-1} (-0.560027) = 124.1\degree
\end{align*}

and \(A = 180\degree - (14.4\degree + 124.1\degree) = 41.5\degree\text{.}\) Both triangles are shown below.