Question: Let $T_1$ be the equilateral triangle inscribed by the unit circle centered at the origin of $\mathbb{R^2}.$ Now let $T_2$ be the triangle induced by clockwise rotating each vertex of $T_1$ by $1^\circ.$ What is the area of $T_1 \cap T_2$ ?
The not-to-scale-drawing below illustrates $T_1 \cap T_2.$ It is a convex hexagon. Careful $T_1 \cap T_2$ is not cyclic - none of its vertices are incident to the boundary of the unit circle. For short handedness I write $\mu\left(T_i\right)$ for the area of $T_i;$ where $1\leq i\leq 2.$ I know that $\mu\left(T_i\right)={3\sqrt{3} \above 1.5pt 4}$ and so I am certain $\mu(T_1 \cap T_2) < {3\sqrt{3} \above 1.5pt 4}.$
In a figure like the one in the question, in my experience people would say the unit circle is circumscribed about each of the triangles $T_1$ and $T_2,$ and they would say $T_1$ and $T_2$ are inscribed triangles, specifically, triangles inscribed in the unit circle.
The entire figure has lines of reflection symmetry through the center of the circle and any of the vertices of the resulting figure. There are three such lines of symmetry; each of the lines goes through two vertices. So the segments from the center to any pair of adjacent vertices make an angle of $60$ degrees.
If you find the distance from the center to each of the vertices, you can use the angle between these radial segments to find the area of the figure.