### Instruction: In right triangle $ABC$ with $\angle B = 90^\circ$, we have $$2\sin A = 3\cos A.$$What is $\sin A$?### Response: Let's divide both sides by 2. So $\sin A = \frac{3}{2} \cos A$.
Before we go on, let's remember that for any right triangle, $\sin^2 A + \cos^2 A = 1$.
So if we square both sides of the equation $\sin A = \frac{3}{2} \cos A$, we get $\sin^2 A = \frac{9}{4} \cos^2 A$.
Now, let's substitute $\sin^2 A = 1 - \cos^2 A$ in the equation we just found.
We get $1 - \cos^2 A = \frac{9}{4} \cos^2 A$.
If we add $\cos^2 A$ to both sides, we get $1 = \frac{9}{4} \cos^2 A + \cos^2 A$.
That simplifies to $1 = \frac{13}{4} \cos^2 A$.
So $\cos^2 A = \frac{4}{13}$.
We also know that $\sin^2 A = 1 - \cos^2 A$, so $\sin^2 A = 1 - \frac{4}{13} = \frac{9}{13}$.
So $\sin A = \pm \sqrt{\frac{9}{13}}$. However, since we are looking for the measure of an acute angle, we know that $\sin A$ must be positive.