We note that both sine and cosine of $A$ and $B$ appear in the given relation and also the coefficients at symmetric. Hence, let's square both equations: \begin{align*} 9\sin^2 A + 24\sin A\cos B + 16\cos^2 B &= 36\\ 16\sin^2 B + 24\sin B\cos A + 9\cos^2 B &= 1 \end{align*} Adding these two equations and rearranging leads to: $$9(\sin^2 A +\cos^A) + 16(\cos^2B+\sin^2B) + 24(\sin A\cos B+\cos A\sin B)=37$$ $$\implies 9+16 + 24\sin(A+B)=37$$ $$\therefore\quad 24\sin(A+B) =12\implies \sin(A+B)=\frac{1}{2}$$ This means that $A+B=30^\circ$ or $A+B=150^\circ$. However, if $A+B=30^\circ$, then $A<30^\circ$ which will imply $$3\sin A + 4\cos B < 3\sin 30^\circ + 4\cos B \le 3\times\frac{1}{2} + 4 <6$$ This contradicts to the given condition. Therefore $A+B=150^\circ$ which means $C=\boxed{30^\circ}$.