As shown in the figure, in a right triangle ABC with $\overline { \mathrm { AB } } = 2$ and $\angle \mathrm {~B} = \frac { \pi } { 2 }$, let D and E be the points where the circle with center A and radius 1 meets the two segments $\mathrm { AB }$ and $\mathrm { AC }$ respectively. Let F be the trisection point of arc DE closer to point D, and let G be the point where line AF meets segment BC. Let $\angle \mathrm { BAG } = \theta$. Let $f ( \theta )$ be the area of the common part of the interior of triangle ABG and the exterior of sector ADF, and let $g ( \theta )$ be the area of sector AFE. Find the value of $40 \times \lim _ { \theta \rightarrow 0 + } \frac { f ( \theta ) } { g ( \theta ) }$. (where $0 < \theta < \frac { \pi } { 6 }$) [3 points]
As shown in the figure, in a right triangle ABC with $\overline { \mathrm { AB } } = 2$ and $\angle \mathrm {~B} = \frac { \pi } { 2 }$, let D and E be the points where the circle with center A and radius 1 meets the two segments $\mathrm { AB }$ and $\mathrm { AC }$ respectively.\\
Let F be the trisection point of arc DE closer to point D, and let G be the point where line AF meets segment BC.\\
Let $\angle \mathrm { BAG } = \theta$. Let $f ( \theta )$ be the area of the common part of the interior of triangle ABG and the exterior of sector ADF, and let $g ( \theta )$ be the area of sector AFE. Find the value of $40 \times \lim _ { \theta \rightarrow 0 + } \frac { f ( \theta ) } { g ( \theta ) }$. (where $0 < \theta < \frac { \pi } { 6 }$) [3 points]