If A = 10°, what is the value of: $$\frac{{12\sin 3A + 5\cos \left( {5A - {5^ \circ }} \right)}}{{9\sin \frac{{9A}}{2} - 4\cos \left( {5A + {{10}^ \circ }} \right)}}?$$

$$\eqalign{ & A = {10^ \circ } \cr & \frac{{12\sin 3A + 5\cos \left( {5A - {5^ \circ }} \right)}}{{9\sin \frac{{9A}}{2} - 4\cos \left( {5A + {{10}^ \circ }} \right)}} \cr & = \frac{{12\sin 3 \times {{10}^ \circ } + 5\cos \left( {5 \times {{10}^ \circ } - {5^ \circ }} \right)}}{{9\sin \frac{{9 \times {{10}^ \circ }}}{2} - 4\cos \left( {5 \times {{10}^ \circ } + {{10}^ \circ }} \right)}} \cr & = \frac{{12\sin {{30}^ \circ } + 5\cos \left( {{{50}^ \circ } - {5^ \circ }} \right)}}{{9\sin \frac{{{{90}^ \circ }}}{2} - 4\cos \left( {{{50}^ \circ } + {{10}^ \circ }} \right)}} \cr & = \frac{{12\sin {{30}^ \circ } + 5\cos {{45}^ \circ }}}{{9\sin {{45}^ \circ } - 4\cos {{60}^ \circ }}} \cr & = \frac{{12 \times \frac{1}{2} + 5 \times \frac{1}{{\sqrt 2 }}}}{{9 \times \frac{1}{{\sqrt 2 }} - 4 \times \frac{1}{2}}} \cr & = \frac{{6 + \frac{5}{{\sqrt 2 }}}}{{\frac{9}{{\sqrt 2 }} - 2}} \cr & = \frac{{6\sqrt 2 + 5}}{{9 - 2\sqrt 2 }} \cr} $$