jmc

The triangle is shown below:



We have $\sin A=\frac{BC}{AC}$ and $\cos A=\frac{AB}{AC}$, so $2\sin A=3\cos A$ gives us $2\cdot \frac{BC}{AC}=3\cdot \frac{AB}{AC}$. Multiplying both sides by $AC$ gives $2BC=3AB$, so $AB=\frac{2}{3}BC$.

The Pythagorean Theorem gives us $A{B}^2+B{C}^2=A{C}^2$. Substituting $AB=\frac{2}{3}BC$ gives $${\left(\frac{2}{3}BC\right)}^2+B{C}^2=A{C}^2.$$Simplifying the left side gives $\frac{13}{9}B{C}^2=A{C}^2$, so $\frac{B{C}^2}{A{C}^2}=\frac{9}{13}$, which means $$\sin A=\frac{BC}{AC}=\sqrt{\frac{9}{13}}=\frac{\sqrt{9}}{\sqrt{13}}=\frac{3}{\sqrt{13}}=\boxed{\frac{3\sqrt{13}}{13}}.$$We also could have noted that $(\sin A{)}^2+(\cos A{)}^2=1$ for any angle $A$, so $2\sin A=3\cos A$ gives us $\cos A=\frac{2}{3}\sin A$ and $(\sin A{)}^2+{\left(\frac{2}{3}\sin A\right)}^2=1$, which gives $\frac{13}{9}(\sin A{)}^2=1$. Therefore, we have $(\sin A{)}^2=\frac{9}{13}$. Since $A$ is an acute angle, we have $\sin A>0$, so $(\sin A{)}^2=\frac{9}{13}$ gives us $$\sin A=\sqrt{\frac{9}{13}}=\frac{\sqrt{9}}{\sqrt{13}}=\frac{3}{\sqrt{13}}=\boxed{\frac{3\sqrt{13}}{13}}.$$