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 $\angle B$
Let $H$ be an orthocenter of triangle $ABC$
Let $H'$ be a point where $OK$ intersects the altitude from $B$
First, We will show that $H=H'$
•$O'$ be reflection of $O$ through $AC$
We will get $2KM=O'H'=O'A=O'C=R$
$\rightarrow 2\angle AH'C=360-\angle AO'C=360-2\angle AOC$
$\rightarrow \angle AH'C=180-\angle B=\angle AHC$
Therefore, $H'=H$ and then ,$R=BO=BH=OO'$
(By Euler line that $2OM=BH$
$\implies OO'C$ is equilateral or$\angle B=60$