using Cosine rule, \(\displaystyle{\cos{{A}}}={\frac{{{b}^{{{2}}}+{c}^{{{2}}}-{a}^{{{2}}}}}{{{2}{b}{c}}}}\)

\(\displaystyle\Rightarrow{\cos{{\left({22}\right)}}}={\frac{{{\left({6}\right)}^{{{2}}}+{\left({15}\right)}^{{{2}}}-{a}^{{{2}}}}}{{{2}{\left({6}\right)}{\left({15}\right)}}}}\)

\(\displaystyle\Rightarrow\frac{{0}}{{9272}}={\frac{{{36}+{225}-{a}^{{{2}}}}}{{{180}}}}\)

\(\displaystyle\Rightarrow{166.893}={261}-{a}^{{{2}}}\)

\(\displaystyle\Rightarrow{a}^{{{2}}}={94.1069}\)

\(\displaystyle\Rightarrow{a}={9.7009}\)

\(\displaystyle\Rightarrow{a}\approx{10}\)

using Sine rule, \(\displaystyle{\frac{{{\sin{{A}}}}}{{{a}}}}={\frac{{{\sin{{B}}}}}{{{b}}}}\)

\(\displaystyle\Rightarrow{\frac{{{\sin{{\left({22}\right)}}}}}{{{10}}}}={\frac{{{\sin{{B}}}}}{{{6}}}}\)

\(\displaystyle\Rightarrow{\sin{{B}}}={\frac{{{6}{\sin{{\left({22}\right)}}}}}{{{10}}}}\)

\(\displaystyle\Rightarrow{B}\approx{13}°\)

using angle sum property of a triangle,

\(\displaystyle{C}={180}-{\left({A}+{B}\right)}\)

\(\displaystyle={180}-{\left({22}+{13}\right)}\)

\(\displaystyle={145}^{{\circ}}\)

Therefore, \(\displaystyle{a}={10},{B}={13}^{{\circ}},{C}={145}^{{\circ}}\)

\(\displaystyle\Rightarrow{\cos{{\left({22}\right)}}}={\frac{{{\left({6}\right)}^{{{2}}}+{\left({15}\right)}^{{{2}}}-{a}^{{{2}}}}}{{{2}{\left({6}\right)}{\left({15}\right)}}}}\)

\(\displaystyle\Rightarrow\frac{{0}}{{9272}}={\frac{{{36}+{225}-{a}^{{{2}}}}}{{{180}}}}\)

\(\displaystyle\Rightarrow{166.893}={261}-{a}^{{{2}}}\)

\(\displaystyle\Rightarrow{a}^{{{2}}}={94.1069}\)

\(\displaystyle\Rightarrow{a}={9.7009}\)

\(\displaystyle\Rightarrow{a}\approx{10}\)

using Sine rule, \(\displaystyle{\frac{{{\sin{{A}}}}}{{{a}}}}={\frac{{{\sin{{B}}}}}{{{b}}}}\)

\(\displaystyle\Rightarrow{\frac{{{\sin{{\left({22}\right)}}}}}{{{10}}}}={\frac{{{\sin{{B}}}}}{{{6}}}}\)

\(\displaystyle\Rightarrow{\sin{{B}}}={\frac{{{6}{\sin{{\left({22}\right)}}}}}{{{10}}}}\)

\(\displaystyle\Rightarrow{B}\approx{13}°\)

using angle sum property of a triangle,

\(\displaystyle{C}={180}-{\left({A}+{B}\right)}\)

\(\displaystyle={180}-{\left({22}+{13}\right)}\)

\(\displaystyle={145}^{{\circ}}\)

Therefore, \(\displaystyle{a}={10},{B}={13}^{{\circ}},{C}={145}^{{\circ}}\)