27. Find a circular segment such that its chord is equal to the length of the largest circle inscribed in it.
27. Let $AB$ be a chord, $r$ be the radius of the largest inscribed circle, $\angle AOB = \alpha$, $N$ be the point where the circle touches the chord, and $M$ be the point of tangency with the larger circle. According to the problem, $AN = \pi r$, and on the other hand,

$$
AN^2 = MN(2R - MN), \quad \text{then} \quad \pi^2 r^2 = 2r \left(\frac{2\pi r}{\sin \frac{\alpha}{2}} - 2r\right)
$$

From this, $\sin \frac{\alpha}{2} = \frac{4\pi}{4 + \pi^2} ; \alpha \approx 122^\circ 56'$.

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