In triangle $ ABC$, $ 3\sin A \plus{} 4\cos B \equal{} 6$ and $ 4\sin B \plus{} 3\cos A \equal{} 1$. Then $ \angle C$ in degrees is

$ \textbf{(A)}\ 30\qquad 
\textbf{(B)}\ 60\qquad 
\textbf{(C)}\ 90\qquad 
\textbf{(D)}\ 120\qquad 
\textbf{(E)}\ 150$
1. Given the equations:
   \[
   3\sin A + 4\cos B = 6
   \]
   \[
   4\sin B + 3\cos A = 1
   \]

2. Square both sides of each equation:
   \[
   (3\sin A + 4\cos B)^2 = 6^2
   \]
   \[
   (4\sin B + 3\cos A)^2 = 1^2
   \]

3. Expand the squared equations:
   \[
   9\sin^2 A + 24\sin A \cos B + 16\cos^2 B = 36
   \]
   \[
   16\sin^2 B + 24\sin B \cos A + 9\cos^2 A = 1
   \]

4. Add the two expanded equations:
   \[
   9\sin^2 A + 16\cos^2 B + 16\sin^2 B + 9\cos^2 A + 24\sin A \cos B + 24\sin B \cos A = 37
   \]

5. Use the Pythagorean identity \(\sin^2 x + \cos^2 x = 1\):
   \[
   9(\sin^2 A + \cos^2 A) + 16(\sin^2 B + \cos^2 B) + 24(\sin A \cos B + \sin B \cos A) = 37
   \]
   \[
   9(1) + 16(1) + 24(\sin A \cos B + \sin B \cos A) = 37
   \]
   \[
   9 + 16 + 24(\sin A \cos B + \sin B \cos A) = 37
   \]
   \[
   25 + 24(\sin A \cos B + \sin B \cos A) = 37
   \]

6. Solve for \(\sin A \cos B + \sin B \cos A\):
   \[
   24(\sin A \cos B + \sin B \cos A) = 12
   \]
   \[
   \sin A \cos B + \sin B \cos A = \frac{1}{2}
   \]

7. Recognize that \(\sin A \cos B + \sin B \cos A = \sin(A + B)\):
   \[
   \sin(A + B) = \frac{1}{2}
   \]

8. Since \(\sin(A + B) = \sin C\) and \(\sin C = \frac{1}{2}\), we have:
   \[
   C = 30^\circ \text{ or } 150^\circ
   \]

9. Check the feasibility of \(C = 150^\circ\):
   - If \(C = 150^\circ\), then \(A + B = 30^\circ\).
   - This implies \(A < 30^\circ\) and \(B < 30^\circ\).
   - However, substituting \(A < 30^\circ\) into the original equation \(3\sin A + 4\cos B = 6\) would yield:
     \[
     3 \sin A + 4 \cos B < 3 \times \frac{1}{2} + 4 = 1.5 + 4 = 5.5 < 6
     \]
   - This is a contradiction.

10. Therefore, the only feasible solution is \(C = 30^\circ\).

The final answer is \(\boxed{30^\circ}\)