Trigonometry

Just a slight error by EnormousBighead....

BC has to be >  5 because angle A > angle B

AC/ sin 34  = BC / sin 108

5 / sin 34 = BC / sin 108

BC  = 5*sin 108 / sin 34  ≈ 8.50

$\angle ABC=180-(\angle BAC+\angle BCA)=34^\circ$

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 $2.93985$