 Solution

Let $A$ be the top vertex, $O$ the center, and $B,C$ the two other points on the circle. The diameter bisects the angle at $A$, so the two right triangles obtained by drawing the diameter to the opposite point show that $B,C$ have coordinates $(\mp \sin \theta ,-\cos \theta )$ when $O=(0,0)$ and $A=(0,1)$. The chord $BC$ cuts off a lower circular segment of central angle $2\theta$. The triangle $ABC$ has area $\sin \theta (1+\cos \theta )$, while the segment has area $\theta -\sin \theta \cos \theta$, by subtracting the central triangle from its circular sector. Adding giveswith $\theta$ measured in radians. This is the unit-radius case of the area enclosed by two chords and a circular arc.