13. In $\triangle A B C$, $H$ is the orthocenter, $B C=a$, the circumradius of $\triangle A B C$ is $R$, and $a=2 A H-2 R$, find the value of $\sin A$.


13. From $A H^{2}+a^{2}=4 R^{2}$ and $a=2 A H-2 R$, we have $5 a^{2}+4 R a-12 R^{2}=0$, solving which gives $a=\frac{6}{5} R$. Using the Law of Sines $a=2 R \sin A$, we get $\sin A=\frac{3}{5}$.