# Solution to the [[Trisected Hexagon]] Puzzle +-- {.image} [[TrisectedHexagon.png:pic]] > This regular hexagon has been split into thirds. What’s the angle? =-- ## Solution by [[Trigonometry]], Regions in a [[Regular Hexagon]], and [[Area of a Triangle]] +-- {.image} [[TrisectedHexagonAnnotated.png:pic]] =-- With the points labelled as above, the regions $A H G$ and $A B C$ are each one sixth of the hexagon. Therefore, the triangle $G F A$ is one sixth of the hexagon. Triangle $F E A$ is half of the blue region so is also one sixth of the hexagon. Therefore, by the [[area of a triangle]], triangles $F E A$ and $G F A$ must have the same "base", so $G F$ and $F E$ have the same length. Angle $F \hat{A} G$ can then be computed using [[trigonometry]], since $G A$ has length $\sqrt{3}$ times that of $G E$, so angle $F \hat{A} G$ is given by $\tan^{-1}\left(\frac{1}{2\sqrt{3}}\right)$. Then angle $C \hat{A} D$ is the same, and so since angle $C \hat{A} G = 60^\circ$, we have that angle $D \hat{A} F$ is (to $3$ decimal places): $$ 60^\circ - 2 \tan^{-1}\left(\frac{1}{2\sqrt{3}}\right) = 27.796^\circ $$