For reference (exact copy of the question):
In the acute triangle $ABC$, the distance between the feet of the relative heights to sides $AB$ and $BC$ is $24$. Calculate the measure of the circumradius of triangle $ABC$. $\angle B = 37^\circ$
My figure and the relationships I found
I tried to draw $DH\perp AC$ in $H$ $\implies \triangle DCH$ is notable ($37^\circ:53^\circ$) therefore sides = $3k:4k:5k$
$FE$ is a right triangle cevian...but I can't see where it will go into the solution.