2. We have a pencil, a ruler, and a certain special device that, for any angle drawn on a plane, constructs two rays that divide this angle into three equal angles. Using these tools, construct an angle of $10^{0}$ on the plane. (Let's recall that a pencil can mark a point on the plane, in particular, the intersection point of two lines. The ruler only allows drawing a straight line through two marked points, and it has no "parallel or perpendicular edges".)
Solution. On the line passing through two distinct points $A$ and $D$, mark points $B$ and $C$ as shown in the figure. Divide the straight angles $ABC$ and $DCB$ (which should be imagined as initially laid out in the upper, and then in the lower half-plane relative to the line $AD$) into three equal parts. As a result, we obtain lines forming a $60^{\circ}$ angle with the line $AD$ and intersecting at points $K$ and $L$. Let $O$ be the point of intersection of $AD$ and $KL$. The angle $BKO$ is obviously $30^{\circ}$. Dividing it into three parts, we obtain the required angle.

Construction completed.

![](https://cdn.mathpix.com/cropped/2024_05_06_216317514e8812a39ee1g-1.jpg?height=500&width=489&top_left_y=1857&top_left_x=1497)