so check the picture below
let's find the missing side first, either b or c, say side "b"
[tex]\bf \textit{Law of Cosines}\\ \quad \\
c^2 = {{ a}}^2+{{ b}}^2-(2{{ a}}{{ b}})cos(C)\implies
c = \sqrt{{{ a}}^2+{{ b}}^2-(2{{ a}}{{ b}})cos(C)}\\\\
-----------------------------\\\\
b = \sqrt{{{ a}}^2+{{ c}}^2-(2{{ a}}{{ c}})cos(B)}
\\\\\\
b = \sqrt{{{ 79.9}}^2+{{ 46.8}}^2-2(79.9)(46.8)cos(16.3^o)}
\\\\\\
b\approx 37.3659169728602838312\implies b\approx 37.37[/tex]
so, b is about 37.37
now, let us use the law of cosines, to find either angle A or C, hmmm say A
[tex]\bf \textit{Law of Cosines}\\ \quad \\
c^2 = {{ a}}^2+{{ b}}^2-(2{{ a}}{{ b}})cos(C)\implies
c = \sqrt{{{ a}}^2+{{ b}}^2-(2{{ a}}{{ b}})cos(C)}\\ \quad \\
\cfrac{{{ a}}^2+{{ b}}^2-c^2}{2{{ a}}{{ b}}}=cos(C)\implies cos^{-1}\left(\cfrac{{{ a}}^2+{{ b}}^2-c^2}{2{{ a}}{{ b}}}\right)=\measuredangle C\\\\
-----------------------------\\\\
[/tex]
[tex]\bf cos^{-1}\left(\cfrac{{{ b}}^2+{{ c}}^2-a^2}{2{{ b}}{{ c}}}\right)=\measuredangle A
\\\\\\
cos^{-1}\left(\cfrac{{{ 37.37}}^2+{{ 46.8}}^2-79.9^2}{2(37.37)(46.8)}\right)=\measuredangle A[/tex]
now, what's ∡C? well, all internal angles in a triangle are 180, thus ∡C picks up the slack from A and B, or 180 - A - B
