Từ giản đồ, ta có: 

\(\frac{U}{\sin \left( \frac{\pi }{2}-{{\varphi }_{AN}} \right)}=\frac{{{U}_{AN}}}{\sin \left( \frac{\pi }{2}-\varphi  \right)}=\frac{{{U}_{NB}}}{\sin \left( {{\varphi }_{AN}}+\varphi  \right)}=\frac{3{{U}_{NB}}}{3\sin \left( {{\varphi }_{AN}}+\varphi  \right)}\)

\(\Leftrightarrow \frac{U}{\cos {{\varphi }_{AN}}}=\frac{{{U}_{AN}}+3{{U}_{NB}}}{\cos \varphi +3\sin \left( {{\varphi }_{AN}}+\varphi  \right)}\) \(\Rightarrow {{U}_{AN}}+3{{U}_{NB}}=\frac{U}{\cos {{\varphi }_{AN}}}\left( \cos \varphi +3\sin \left( {{\varphi }_{AN}}+\varphi  \right) \right)\)

\({{\left( {{U}_{AN}}+3{{U}_{NB}} \right)}_{\max }}\) khi \({{\left( \frac{\cos \varphi +3\sin \left( {{\varphi }_{AN}}+\varphi  \right)}{\cos {{\varphi }_{AN}}} \right)}_{\max }}\)

Ta có: \(\frac{\cos \varphi +3\sin \left( {{\varphi }_{AN}}+\varphi  \right)}{\cos {{\varphi }_{AN}}}=\frac{\cos \varphi +3\sin \varphi \cos {{\varphi }_{AN}}+3\cos \varphi \sin {{\varphi }_{AN}}}{\cos {{\varphi }_{AN}}}\)

\(=\frac{\cos \varphi \left( 1+3\sin {{\varphi }_{AN}} \right)+3\sin \varphi .\cos {{\varphi }_{AN}}}{\cos {{\varphi }_{AN}}}\text{    }(*)\)

Áp dụng bất đẳng thức Bunhia ta có: 

\((*)\le \frac{\left( {{\cos }^{2}}\varphi +{{\sin }^{2}}\varphi  \right)\left( {{\left( 1+3\sin {{\varphi }_{AN}} \right)}^{2}}+{{\left( 3\cos {{\varphi }_{AN}} \right)}^{2}} \right)}{\cos {{\varphi }_{AN}}}\)

Dấu = xảy ra khi: \(\frac{1+3\sin {{\varphi }_{AN}}}{\cos \varphi }=\frac{3\cos {{\varphi }_{AN}}}{\sin \varphi }\)

Lại có: \(\cos \varphi =\sin \varphi =\frac{1}{\sqrt{2}}\) (đề bài cho) 

\(\Rightarrow \frac{1+3\sin {{\varphi }_{AN}}}{\frac{1}{\sqrt{2}}}=\frac{3\cos {{\varphi }_{AN}}}{\frac{1}{\sqrt{2}}}\)

\(\cos {{\varphi }_{AN}}-\sin {{\varphi }_{AN}}=\frac{1}{3}\Rightarrow {{\varphi }_{AN}}=0,547rad\Rightarrow \cos {{\varphi }_{AN}}=0,8538\)