Triangle $ABC$ is inscribed in circle $\omega$. Points $P$ and $Q$ are on side $\overline{AB}$ with $AP<AQ$. Rays $CP$ and $CQ$ meet $\omega$ again at $S$ and $T$ (other than $C$), respectively. If $AP=4,PQ=3,QB=6,BT=5,$ and $AS=7$, then $ST=\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.
Let $\angle ACP=\alpha$, $\angle PCQ=\beta$, and $\angle QCB=\gamma$. Note that since $\triangle ACQ\sim\triangle TBQ$ we have $\tfrac{AC}{CQ}=\tfrac56$, so by the Ratio Lemma \[\dfrac{AP}{PQ}=\dfrac{AC}{CQ}\cdot\dfrac{\sin\alpha}{\sin\beta}\quad\implies\quad \dfrac{\sin\alpha}{\sin\beta}=\dfrac{24}{15}.\]Similarly, we can deduce $\tfrac{PC}{CB}=\tfrac47$ and hence $\tfrac{\sin\beta}{\sin\gamma}=\tfrac{21}{24}$.
Now Law of Sines on $\triangle ACS$, $\triangle SCT$, and $\triangle TCB$ yields \[\dfrac{AS}{\sin\alpha}=\dfrac{ST}{\sin\beta}=\dfrac{TB}{\sin\gamma}.\]Hence \[\dfrac{ST^2}{\sin^2\beta}=\dfrac{TB\cdot AS}{\sin\alpha\sin\gamma},\]so \[TS^2=TB\cdot AS\left(\dfrac{\sin\beta}{\sin\alpha}\dfrac{\sin\beta}{\sin\gamma}\right)=\dfrac{15\cdot 21}{24^2}\cdot 5\cdot 7=\dfrac{35^2}{8^2}.\]Hence $ST=\tfrac{35}8$ and the requested answer is $35+8=\boxed{43}$.
Edit: Note that the finish is much simpler. Once you get $\dfrac{AS}{\sin\alpha}=\dfrac{ST}{\sin\beta}$, you can solve quickly from there getting $ST=\dfrac{AS \sin(\beta)}{\sin(\alpha)}=7\cdot \dfrac{15}{24}=\dfrac{35}{8}$.