A triangle and two points inside it are marked. It is known that one of the triangle’s angles is equal to $58^{\circ}$, one of two remaining angles is equal to $59^{\circ}$, one of two given points is the incenter of the triangle and the second one is its circumcenter. Using only the ruler without partitions determine where is each of the angles and where is each of the centers.
1. **Identify the angles of the triangle:**
   Given that one angle is \(58^\circ\) and another is \(59^\circ\), the third angle can be calculated using the fact that the sum of the angles in a triangle is \(180^\circ\).
   \[
   \angle A + \angle B + \angle C = 180^\circ
   \]
   Let \(\angle A = 58^\circ\), \(\angle B = 59^\circ\), and \(\angle C\) be the unknown angle. Then,
   \[
   \angle C = 180^\circ - 58^\circ - 59^\circ = 63^\circ
   \]
   Therefore, the angles are \(58^\circ\), \(59^\circ\), and \(63^\circ\).

2. **Determine the positions of the incenter and circumcenter:**
   - The incenter \(I\) is the point where the angle bisectors of the triangle intersect.
   - The circumcenter \(O\) is the point where the perpendicular bisectors of the sides of the triangle intersect.

3. **Label the vertices and angles:**
   Without loss of generality, let:
   \[
   \angle A = 63^\circ, \quad \angle B = 59^\circ, \quad \angle C = 58^\circ
   \]

4. **Analyze the given points \(P_1\) and \(P_2\):**
   - Suppose \(P_1\) is the incenter \(I\) and \(P_2\) is the circumcenter \(O\).
   - The angle \(\angle AIB\) formed by the incenter is given by:
     \[
     \angle AIB = 90^\circ + \frac{1}{2} \angle C = 90^\circ + \frac{1}{2} \times 58^\circ = 119^\circ
     \]
   - The angle \(\angle AOB\) formed by the circumcenter is given by:
     \[
     \angle AOB = 2 \times \angle C = 2 \times 58^\circ = 116^\circ
     \]

5. **Compare the angles \(\angle AIB\) and \(\angle AOB\):**
   \[
   \angle AIB = 119^\circ > \angle AOB = 116^\circ
   \]
   This confirms that \(P_1\) (the incenter) forms a larger angle than \(P_2\) (the circumcenter).

6. **Conclusion:**
   Since \(\angle AIB > \angle AOB\), we can conclude that:
   \[
   P_1 = I \quad \text{(incenter)}, \quad P_2 = O \quad \text{(circumcenter)}
   \]

The final answer is \( \boxed{ P_1 = I } \) (incenter) and \(P_2 = O\) (circumcenter).