The principle amplitude of $(\sin 40^{o}+i \cos 40^{o})^{5}$ is

$40°=2\cfrac { \pi }{ 9 } ,50°=\cfrac { 5\pi }{ 18 }$

So, $\left( \sin { 40° } +i\cos { 40° } \right) =\cos { 50° } +i\sin { 50° }$

Let $T={ \left( \sin { 40° } +i\cos { 40° } \right) }^{ 5 }={ \left( \cos { 50° } +i\sin { 50° } \right) }^{ 5 }$

$T={ \left( { e }^{ i50° } \right) }^{ 5 }=\left( { e }^{ i\cfrac { 5\pi }{ 18 } \times 5 } \right) ={ e }^{ i\cfrac { 25\pi }{ 18 } }$

So, $25\cfrac { \pi }{ 18 } =250°$

Thus, principal argument$=-180°+250°=70°$