If $A$, $B$, and & $C$ are angles of a triangle. Find angle $C$. System of equations involving trigonometric functions.

Question

If $A$, $B$, and & $C$ are angles of a triangle and

$5\sin(A) + 12\cos(B) = 15$
$12\sin(B) + 5\cos(A) = 2$

Find angle $C$.

I ended up with $\sin(A+B) = \frac{1}{2}$ and there seems to be two values of $C$ which are $30°$ and $150°$. How would I know which is correct? Or are both answers considered correct?

Answer 1

Assuming that $sin(A+B) = \dfrac{1}{2}$ is correct, you can use the first equation: $$5\cdot sin(A) + 12 \cdot cos(B) = 15$$ $$\Rightarrow 12 \cdot cos(B) = 15 - 5\cdot sin(A)$$ using $0 < sin(A) < 1$ for $A \in (0, \pi)$ $$\Rightarrow 12 \cdot cos(B) \ge 10$$ $$\Rightarrow cos(B) \ge \dfrac{10}{12}$$

From this, we can safely conclude that $B \le 57°$.

Second, from the second equation, we get: $$sin(A) = \dfrac{2-5\cdot cos(B)}{12}$$ $$sin(A) \le \dfrac{2}{12}$$

And therefor: $A\le 10°$.

Therefor, C cannot be $30°$ and must be $150°$!

Answer 2

Squaring and summing give:$$25+144+120\sin(A+B)=225+4$$ or $$\sin(A+B)=\frac{1}{2},$$ which gives $C=30^{\circ}$, which contradicts with the first equality , or $C=150^{\circ}.$

Indeed, if $A+B=30^{\circ}$, so $A<30^{\circ}$ and $$12\cos B=15-5\sin A>15-5\cdot\frac{1}{2}=12.5>12,$$ which is impossible.