# \(\sqrt{A^2 + B^2}\)

**Geometry**Level 5

If A =\(\sin { \dfrac { \pi }{ 7 } } +\sin { \dfrac { 2\pi }{ 7 } } +\sin { \dfrac { 3\pi }{ 7 } } \)and B= \( \cos {\dfrac { \pi }{ 7 } } +\cos { \dfrac { 2\pi }{ 7 } } +\cos{ \dfrac { 3\pi }{ 7 } }\) , then \(\sqrt{A^2 + B^2}\) can be written as \(p^{1/6}\). Then p is