Let $ABCD$ be a tetrahedron and $O$ its incenter, and let the line $OD$ be perpendicular to $AD$. Find the angle between the planes $DOB$ and $DOC.$
1. **Define the planes and the spherical surface:**
   Let $\alpha, \beta, \gamma$ denote the planes $DBC, DAC, DAB$ respectively. For convenience, we cut the trihedron formed by $\alpha, \beta, \gamma$ with the spherical surface $\mathcal{E}$ centered at $D$ with radius $DO$. The rays $DA, DB, DC$ intersect $\mathcal{E}$ at points $A', B', C'$ respectively. 

2. **Identify the internal angle bisectors:**
   Since $O$ is the incenter of the tetrahedron $ABCD$, the segments $A'O, B'O, C'O$ are the internal angle bisectors of the spherical triangle $A'B'C'$.

3. **Projection and perpendicularity:**
   The great circle $(D, DO)$ intersects $A'C'$ and $A'B'$ at points $M$ and $N$ such that $\angle OMA' = \angle ONA' = 90^\circ$. This implies that $M$ and $N$ are the projections of $O$ onto $A'C'$ and $A'B'$ respectively.

4. **Projection onto $B'C'$:**
   Let $L$ be the projection of $O$ onto $B'C'$. By the properties of spherical triangles and angle bisectors, we have:
   \[
   \angle LOB' = \frac{1}{2} \angle NOL \quad \text{and} \quad \angle LOC' = \frac{1}{2} \angle MOL
   \]

5. **Calculate the angle between the planes:**
   Since $M$ and $N$ are projections and $O$ is the incenter, the angle between the planes $DOB$ and $DOC$ is given by:
   \[
   \angle B'OC' = \frac{1}{2} \times 180^\circ = 90^\circ
   \]

Thus, the angle between the planes $DOB$ and $DOC$ is $90^\circ$.

The final answer is $\boxed{90^\circ}$.