\(\angle ABC=180-(\angle BAC+\angle BCA)=34^\circ\)
using \(\sin\) rule of triangles:
\(\Large\frac{AC}{\sin{\angle{ABC}}}=\frac{BC}{\sin{\angle{BAC}}}\\ \Large\Rightarrow BC=\frac{5\sin{34^\circ}}{\sin{108^\circ}}\approx3\)
Exact value is \(2.93985\)