I'm stuck with a problem. Given is a triangle $\Delta ABC$ with $\angle A = 35°, BC=3$ and $AC=5$.
I need to find the two possible values for $\angle C$. I only managed to found one angle. I did the following:
$\LARGE\frac{BC}{\sin(\angle A)}=\frac{AC}{\sin(\angle B)}=\frac{AB}{\sin(\angle C)}$
This makes:
$\LARGE\frac{3}{\sin(35°)}=\frac{5}{\sin(\angle B)}=\frac{AB}{\sin(\angle C)}$
Conclusion: $\sin(\angle B)=\frac{5\sin(35°)}{3} \rightarrow \angle B=72.93°$ and $\angle C=180-\angle A-\angle B = 72.07°$. However, this is only one of the two solutions. I need to find the other solution, but I have no idea what to do.
use the equation $3^3=5^2+c^2-10c\cos(35^{\circ})$ and solve this equation for $c$