Bài giải:

Ta có: $\sin {{50}^{o}}=\cos {{40}^{o}};\sin {{60}^{o}}=\cos {{30}^{o}}$

$A={{\sin }^{2}}{{30}^{o}}+{{\sin }^{2}}{{40}^{o}}+{{\sin }^{2}}{{50}^{o}}+{{\sin }^{2}}{{60}^{o}}$

 $=({{\sin }^{2}}{{30}^{o}}+{{\sin }^{2}}{{60}^{o}})+({{\sin }^{2}}{{40}^{o}}+{{\sin }^{2}}{{50}^{o}})$

$=({{\sin }^{2}}{{30}^{o}}+{{\cos }^{2}}{{30}^{o}})+({{\sin }^{2}}{{40}^{o}}+{{\cos }^{2}}{{40}^{o}})$

$=1+1=2$