12.175. The radius of the arc of sector $A O B$ is $R$, the central angle $A O B$ is $\alpha$. A regular triangle is inscribed in this sector such that one of its vertices coincides with the midpoint of the arc $A B$, and the other two vertices lie on the radii $O A$ and $O B$ respectively. Find the sides of the triangle.
Solution.

Let $M$ be the midpoint of the arc $AB$, and $\Delta LMK$ be the equilateral triangle mentioned in the problem (Fig. 12.45).

Then $\angle LOM = \frac{\alpha}{2}, \angle MLK = 60^{\circ}, \angle OLK = 90^{\circ} - \frac{\alpha}{2}, \angle OLM = $

$$
= \angle OLK + \angle MLK = 150^{\circ} - \frac{\alpha}{2}
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0928.jpg?height=355&width=523&top_left_y=337&top_left_x=110)

Fig. 12.45

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0928.jpg?height=642&width=519&top_left_y=50&top_left_x=742)

Fig. 12.46

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0928.jpg?height=222&width=1100&top_left_y=806&top_left_x=144)

$$
\begin{aligned}
& = \frac{R \sin \frac{\alpha}{2}}{\sin \left(30^{\circ} + \frac{\alpha}{2}\right)} \\
& \text{Therefore: } \frac{R \sin \frac{\alpha}{2}}{\sin \left(30^{\circ} + \frac{\alpha}{2}\right)}
\end{aligned}
$$