Finding $C$ if $ 3\sin A + 4\cos B = 6 $ and $ 3\cos A + 4\sin B = 1 $ in a triangle $ABC$ 
In a triangle $ABC$, it's given that the following two equations are satisfied:
$$ 3\sin A + 4\cos B = 6 $$
$$ 3\cos A + 4\sin B = 1 $$
Source: ISI B-math UGA 2017 

We have to find the angle $ C$. Now, it's easy to see that $ \sin C = 0.5 $ (by squaring and adding both the equations). Now we have to decide whether $ C = \pi/6 $ or $5\pi/6$. The solution given in the book goes something like this:
Assume $ C = 5\pi/6$, then even if $B = 0$ and $A = \pi/6$, then the quantity $3\sin A + 4\cos B = 5.5 < 6$ and hence $ C \not= 5\pi/6 $.
But then if we do the same thing by setting $C = \pi/6$, again we face the same problem. So how to find the value of $C$?
EDIT:
Squaring and adding the two equations, we get:
$$ 9(\sin^2 A + \cos^2 A) + 16 (\sin^2 B + \cos^2 B) + 24\sin(A+B) = 37  $$
$$ 24\sin(A+B) = 24\sin C = 12 \implies \sin C = 0.5 $$

when $C=\dfrac{\pi}{6}, A $ can be large enough,it is different from the case $C=\dfrac{5\pi}{6}$as $A\le \dfrac{\pi}{6}$ that  you can't find solution for $A,B$
Now in case $C=\dfrac{\pi}{6}$, it is possible to find solution but you can't use same  method to check as $\sin{A} $ can large than $\dfrac{1}{2}$.
