Finding $C$ if $ 3\sin A + 4\cos B = 6 $ and $ 3\cos A + 4\sin B = 1 $ in a triangle $ABC$

Question

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: $$ 9(\sin^2 A + \cos^2 A) + 16 (\sin^2 B + \cos^2 B) + 24\sin(A+B) = 37 $$ $$ 24\sin(A+B) = 24\sin C = 12 \implies \sin C = 0.5 $$

Answer 1

$ 3\sin A + 4\cos B = 6 $ ......(i)

$ 3\cos A + 4\sin B = 1 $.... (ii)

Squaring (i) we get :

$9 \sin^2 A +16\cos^2 B +24 \sin A \cos B = 36 $ ....(iii)

Now Squaring (ii) we get :

$9 \cos^2 A +16\sin^2 B +24 \sin B \cos A =1 $...(iv)

Now adding (iii) and (iv) we get :

$9(\sin^2 A +\cos^2 A) +16(\cos^2 B + \sin^2 B) +24 ( \sin A \cos B +\cos A \sin B ) = 37 $

$\Rightarrow 9 +16 + 24 \sin(A+B) = 37 $

$\Rightarrow 24 \sin C = 37 -25 = 12 $

$\Rightarrow \sin C =\frac{1}{2} = \sin \frac{\pi}{6}$

Now general solution of the equation $\sin\theta =\sin\alpha$ is given by :

$$\theta =n\pi +(-1)^n \alpha, n \in Z$$

$$\therefore C = n\pi +(-1)^n \frac{\pi}{6}$$ is the solution.

Answer 2

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 $\dfrac{1}{2}$.