[tex]\bf \textit{Law of sines}
\\ \quad \\
\cfrac{sin(\measuredangle A)}{a}=\cfrac{sin(\measuredangle B)}{b}=\cfrac{sin(\measuredangle C)}{c}\\\\
-----------------------------\\\\
[/tex]
[tex]\bf \cfrac{sin(48^o)}{32}=\cfrac{sin(B)}{27}\implies\cfrac{27sin(48^o)}{32}=sin(B)
\\\\\\
sin^{-1}\left[ \cfrac{27sin(48^o)}{32} \right]=sin^{-1}[sin(B)]\implies sin^{-1}\left[ \cfrac{27sin(48^o)}{32} \right]=\measuredangle B
\\\\\\
38.83\approx \measuredangle B
\\\\\\
\textit{now, we know angles A and B, angle is C is just 180-A-B}
\\\\\\
\measuredangle C = 180-48-38.83\implies C\approx 93.17^o[/tex]
so, now we know the ∡C, so, let's use the law of sines to get side "c"
[tex]\bf \cfrac{sin(48^o)}{32}=\cfrac{sin(93.17^o)}{c}\implies c=\cfrac{32sin(93.17^o)}{sin(48^o)}[/tex]
and surely you know how much that is
