Vertices $A, B, C$ of a equilateral triangle of side $1$ are in the surface of a sphere with radius $1$ and center $O$. Let $D$ be the orthogonal projection of $A$ on the plane $\alpha$ determined by points $B, C, O$. Let $N$ be one of the intersections of the line perpendicular to $\alpha$ passing through $O$ with the sphere. Find the angle $\angle DNO$.
1. **Identify the geometric configuration:**
   - Given an equilateral triangle \( \triangle ABC \) with side length 1, inscribed on the surface of a sphere with radius 1 and center \( O \).
   - \( D \) is the orthogonal projection of \( A \) onto the plane \( \alpha \) determined by points \( B, C, \) and \( O \).
   - \( N \) is one of the intersections of the line perpendicular to \( \alpha \) passing through \( O \) with the sphere.

2. **Recognize the tetrahedron structure:**
   - Since \( A, B, C \) are vertices of an equilateral triangle on the sphere, and \( O \) is the center of the sphere, \( A, B, C, O \) form a regular tetrahedron with side length 1.

3. **Determine the position of \( D \):**
   - \( D \) is the orthogonal projection of \( A \) onto the plane \( \alpha \). In a regular tetrahedron, the orthogonal projection of a vertex onto the opposite face is the centroid of that face.
   - The centroid of \( \triangle BCO \) divides each median in the ratio 2:1. The distance from the centroid to any vertex of \( \triangle BCO \) is \( \frac{\sqrt{3}}{3} \) times the side length of the triangle.

4. **Calculate \( DO \):**
   - Since \( \triangle BCO \) is an equilateral triangle with side length 1, the distance from the centroid \( D \) to the center \( O \) is:
     \[
     DO = \frac{\sqrt{3}}{3}
     \]

5. **Determine the distance \( ON \):**
   - \( N \) is one of the intersections of the line perpendicular to \( \alpha \) passing through \( O \) with the sphere. Since \( O \) is the center of the sphere and the radius is 1, \( ON = 1 \).

6. **Calculate the angle \( \angle DNO \):**
   - Using the tangent function:
     \[
     \tan{\angle DNO} = \frac{DO}{ON} = \frac{\frac{\sqrt{3}}{3}}{1} = \frac{\sqrt{3}}{3}
     \]
   - Therefore:
     \[
     \angle DNO = \tan^{-1}\left(\frac{\sqrt{3}}{3}\right) = 30^\circ
     \]

The final answer is \( \boxed{30^\circ} \).