All angles of $ABC$ are in $(30,90)$. Circumcenter of $ABC$ is $O$ and circumradius is $R$. Point $K$ is projection of $O$ to angle bisector of $\angle B$, point $M$ is midpoint $AC$. It is known, that $2KM=R$. Find $\angle B$
1. **Given Information and Definitions:**
   - All angles of $\triangle ABC$ are in $(30^\circ, 90^\circ)$.
   - $O$ is the circumcenter of $\triangle ABC$ and $R$ is the circumradius.
   - $K$ is the projection of $O$ onto the angle bisector of $\angle B$.
   - $M$ is the midpoint of $AC$.
   - It is known that $2KM = R$.
   - We need to find $\angle B$.

2. **Introduce Orthocenter and Reflection:**
   - Let $H$ be the orthocenter of $\triangle ABC$.
   - Let $H'$ be the point where $OK$ intersects the altitude from $B$.
   - We aim to show that $H = H'$.
   - Let $O'$ be the reflection of $O$ through $AC$.

3. **Using Reflection and Distance Properties:**
   - Since $O'$ is the reflection of $O$ through $AC$, $O'$ lies on the circumcircle of $\triangle ABC$.
   - We know $2KM = O'H' = O'A = O'C = R$.
   - This implies that $O'H' = R$.

4. **Angle Relationships:**
   - Consider the angle $\angle AH'C$:
     \[
     2\angle AH'C = 360^\circ - \angle AO'C = 360^\circ - 2\angle AOC
     \]
   - Since $O$ is the circumcenter, $\angle AOC = 2\angle B$.
     \[
     2\angle AH'C = 360^\circ - 2(2\angle B) = 360^\circ - 4\angle B
     \]
   - Therefore,
     \[
     \angle AH'C = 180^\circ - 2\angle B
     \]

5. **Comparing with $\angle AHC$:**
   - We know $\angle AHC = 180^\circ - \angle B$ (since $H$ is the orthocenter).
   - Therefore,
     \[
     \angle AH'C = 180^\circ - \angle B
     \]
   - This implies that $H' = H$.

6. **Using Euler Line and Equilateral Triangle:**
   - Since $H = H'$, we have $R = BO = BH = OO'$.
   - By Euler's line property, $2OM = BH$.
   - This implies that $\triangle OO'C$ is equilateral, meaning $\angle B = 60^\circ$.

The final answer is $\boxed{60^\circ}$.