Okay, here we have this:
Considering the provided information, we are going to solve the triangle, so we obtain the following:
[tex]\begin{gathered} b=\sqrt[]{a^2+c^2-2ac\cdot\cos\mleft(B\mright)} \\ b=\sqrt[]{631^2+199^2-2(631)(199)\cdot\cos (104.8)} \\ b\approx708ft \end{gathered}[/tex][tex]\begin{gathered} \angle A=arccos(\frac{b^2+c^2-a^2}{2bc}) \\ \angle A=arccos(\frac{708.46^2+199^2-631^2}{2(708.46)(199)}) \\ \angle A\approx59.4degrees \end{gathered}[/tex][tex]\begin{gathered} \angle C=arccos(\frac{b^2+a^2-c^2}{2ba}) \\ \angle C=arccos(\frac{708.46^2+631^2-199^2}{2(708.46)(631)}) \\ \angle C\approx15.8degrees \end{gathered}[/tex]
