jmc

By constructing a right triangle with legs 1 and $2\sqrt{2}$ and hypotenuse 3, we see that $\sin \angle BAM$ implies $\tan \angle BAM=\frac{1}{2\sqrt{2}}.$

We can draw right triangle $ABC$ so that $AB=2,$ $AC=2\cos A,$ and $BC=2\sin A.$ Then $BM=CM=\sin A.$



Then $$\begin{array}{rl}\tan \angle BAM& =\tan (\angle BAC-\angle CAM)\\ & =\frac{\tan \angle BAC-\tan \angle CAM}{1+\tan \angle BAC\tan \angle CAM}\\ & =\frac{\tan A-\frac{\tan A}{2}}{1+\tan A\cdot \frac{\tan A}{2}}\\ & =\frac{\tan A}{{\tan }^{2}A+2}.\end{array}$$Thus, $$\frac{\tan A}{{\tan }^{2}A+2}=\frac{1}{2\sqrt{2}}.$$Then $2\sqrt{2}\tan A={\tan }^{2}A+2,$ or $${\tan }^{2}A-2\sqrt{2}\tan A+2=0.$$This factors as $(\tan A-\sqrt{2}{)}^2=0,$ so $\tan A=\sqrt{2}.$

Now, constructing a right triangle where the legs are 1 and $\sqrt{2}$ and the hypotenuse is $\sqrt{3},$ we see that $$\sin A=\frac{\sqrt{2}}{\sqrt{3}}=\boxed{\frac{\sqrt{6}}{3}}.$$