In triangle $ABC,$ $\angle C = 90^\circ.$ A semicircle is constructed along side $\overline{AC}$ that is tangent to $\overline{BC}$ and $\overline{AB}.$ If the radius of the semicircle is equal to $3,$ then find $\frac{AC}{BC}.$
The answer for AC/BC is 30/7.
+1926 
