The gray area in the figure is bounded by the sides of a $30^{\circ}$ angle with vertex $A$ and an arc of a circle with center $O$. What is the area of the region if

$$
A O=A B=1
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_6edf624d30325b700134g-1.jpg?height=217&width=336&top_left_y=325&top_left_x=884)
Solution. In the isosceles triangle $OAB$, $\angle OAB = 150^\circ$ and $\angle BOA = \angle ABO = 15^\circ$, let the radius of the circle be denoted by $r$.

![](https://cdn.mathpix.com/cropped/2024_05_02_b23a1986b330f51a1981g-1.jpg?height=203&width=466&top_left_y=259&top_left_x=819)

In triangle $OAB$, apply the cosine rule to side $OB$:

$$
r^{2}=1^{2}+1^{2}-2 \cdot 1 \cdot 1 \cdot \cos 150^{\circ}
$$

From this,

$$
\cos 150^{\circ}=-\cos 30^{\circ}=-\frac{\sqrt{3}}{2}
$$

After substitution, we get $r^{2}=2+\sqrt{3}$. The desired area $T$ is obtained by subtracting the area of triangle $OAB$ from the area of the sector $BOC$:

$$
\begin{aligned}
t_{\text {sector }} & =\frac{r^{2} \pi \cdot \alpha}{360^{\circ}}=\frac{(2+\sqrt{3}) \pi \cdot 15^{\circ}}{360^{\circ}}=\frac{(2+\sqrt{3}) \pi}{24} \\
t_{\triangle} & =\frac{1 \cdot 1 \cdot \sin 150^{\circ}}{2}=\frac{1}{4} \\
T & =\frac{(2+\sqrt{3}) \pi-6}{24} \approx 0.2385 \text { square units. }
\end{aligned}
$$