Official solution
\[\cos \theta = {{2\sqrt 3 } \over 4} = {{\sqrt 3 } \over 2} \Rightarrow \theta = 30^\circ \]
Area of the required region
\[ = {2 \over 3}\left( {4 \times {1 \over 2}} \right) + {4^2} \times {\pi \over 6} - {1 \over 2} \times 4 \times 2\sqrt 3 \]
\[ = {4 \over 3} + {{8\pi } \over 3} - 4\sqrt 3 = {1 \over 3}\left\{ {4 - 12\sqrt 3 + 8\pi } \right\}\]