We know that the sum of angles in a triangle equals 180 degrees. Therefore,

\(\displaystyle\angle{B}{A}{C}+\angle{c}{b}{a}+\angle{A}{C}{B}={180}^{\circ}\)

\(\displaystyle\angle{C}{B}{A}={18}^{\circ}-\angle{B}{A}{C}-\angle{A}{C}{B}\)

\(\displaystyle\angle{C}{B}{A}={180}^{\circ}-{44}^{\circ}-{103}^{\circ}\)

\(\displaystyle\angle{C}{B}{A}={33}^{\circ}\)

By using law of sines in the given triangle,

\(\displaystyle{\left(\frac{{{\sin}\angle{B}{A}{C}}}{{x}}\right)}={\left(\frac{{{\sin}\angle{C}{B}{A}}}{{y}}\right)}\)

\(\displaystyle{\left(\frac{{{\sin{{44}}}^{\circ}}}{{x}}\right)}={\left(\frac{{{\sin{{33}}}^{\circ}}}{{50.5}}\right)}\)

\(\displaystyle{x}=\frac{{{\sin{{44}}}^{\circ}}}{{{\sin{{33}}}^{\circ}}}\cdot{50.5}\)

\(\displaystyle{x}=\frac{{{0.6946}}}{{{0.5446}}}\cdot{50.5}\)

\(\displaystyle{x}={64.41}\)

\(\displaystyle\angle{B}{A}{C}+\angle{c}{b}{a}+\angle{A}{C}{B}={180}^{\circ}\)

\(\displaystyle\angle{C}{B}{A}={18}^{\circ}-\angle{B}{A}{C}-\angle{A}{C}{B}\)

\(\displaystyle\angle{C}{B}{A}={180}^{\circ}-{44}^{\circ}-{103}^{\circ}\)

\(\displaystyle\angle{C}{B}{A}={33}^{\circ}\)

By using law of sines in the given triangle,

\(\displaystyle{\left(\frac{{{\sin}\angle{B}{A}{C}}}{{x}}\right)}={\left(\frac{{{\sin}\angle{C}{B}{A}}}{{y}}\right)}\)

\(\displaystyle{\left(\frac{{{\sin{{44}}}^{\circ}}}{{x}}\right)}={\left(\frac{{{\sin{{33}}}^{\circ}}}{{50.5}}\right)}\)

\(\displaystyle{x}=\frac{{{\sin{{44}}}^{\circ}}}{{{\sin{{33}}}^{\circ}}}\cdot{50.5}\)

\(\displaystyle{x}=\frac{{{0.6946}}}{{{0.5446}}}\cdot{50.5}\)

\(\displaystyle{x}={64.41}\)