[tex]\bf \begin{cases}
\measuredangle A=66^o\\
\measuredangle B=53^o\\
\textit{what about the last angle? C?}\\
\textit{all internal angles in a triangle add up to 180, thus}\\
\measuredangle C = 180-(A+B)\to 61^o
\\\\
\textit{side "c" is } 23
\end{cases}\qquad thus\\\\
-----------------------------\\\\
\textit{Law of sines}
\\ \quad \\
\cfrac{sin(\measuredangle A)}{a}=\cfrac{sin(\measuredangle B)}{b}=\cfrac{sin(\measuredangle C)}{c}\\\\
-----------------------------\\\\
[/tex][tex]\bf thus
\\\\
\cfrac{sin(\measuredangle C)}{c}=\cfrac{sin(B)}{b}\implies \cfrac{sin(61^o)}{23}=\cfrac{sin(53^o)}{b}[/tex]
solve for "b",
when taking the sines, make sure your calculator is in Degree mode, since the angles are in degrees
