In a triangle, $\Delta ABC$ we are given that $3 \sin A + 4 \cos B = 6$ and $4 \sin B + 3\cos A = 1$. Then what is the measure of angle $C$? 
In a triangle, $\Delta ABC$ we are given that $3 \sin A + 4 \cos B = 6$ and $4 \sin B + 3\cos A = 1$. Then what is the measure of angle $C$?

On squaring and adding both the equations,we get
$$9 + 16 + 24 \sin (A + B) = 37\implies$$
$$\sin (A + B) =
\frac{1}{2}\implies A+B=\frac{\pi}{6}\implies C=5\pi/6$$
But,in answer key it is given that $C=\pi/6$
I wanted to know where i'm wrong.
I've also gone through Finding $C$ if $ 3\sin A + 4\cos B = 6 $ and $ 3\cos A + 4\sin B = 1 $ in a triangle $ABC$ ,but here also,i didn't get my answer.

Note that $\sin(A+B) = \frac{1}{2}$ has two solutions in $(0,\pi)$, that are $A+B \in \{\frac{\pi}{6},\frac{5\pi}{6}\}$. 
Now check which of them satisfies original equations. Lets suppose $A+B = \frac{\pi}{6}$. In triangle, all angles are positive, so both $0 < A,B \lt \frac{\pi}{6}$. This implies that 


*

*$0 < \sin A< \frac{1}{2}$

*$\frac{\sqrt{3}}{2} < \cos B < 1$
From first equation,
$$3\sin(A) +4\cos(B) = 6$$
Maximum value of LHS is $4+\frac{3}{2}$, that too not achievable, and it is less than $6$. So this equation is never satisfied for $A+B = \frac{\pi}{6}$.
Similarly analyse the equations for second solution, that is $A+B = \frac{5\pi}{6}$
