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 $3$, and $AB = 1$, then find $\frac{AC}{BC}.$
The answer is AC/BC = 15/4.
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