# $$\sqrt{A^2 + B^2}$$

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

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