### Instruction: In right triangle $ABC$ with $\angle B = 90^\circ$, we have $$2\sin A = 3\cos A.$$What is $\sin A$?### Response: I know that $\sin A$ and $\cos A$ are related by the Pythagorean identity: $$\sin^2 A + \cos^2 A = 1.$$
I also know that $\sin A$ and $\cos A$ are both positive, since $\angle A$ is acute in a right triangle.
So, I can use these facts to eliminate one of the trigonometric functions and solve for the other.
I can start by squaring the given equation: $$(2\sin A)^2 = (3\cos A)^2.$$
Expanding and simplifying, I get $$4\sin^2 A = 9\cos^2 A.$$
Now, I can substitute $\cos^2 A = 1 - \sin^2 A$ from the Pythagorean identity and get $$4\sin^2 A = 9(1 - \sin^2 A).$$
Distributing and rearranging, I get $$13\sin^2 A = 9.$$
Dividing by 13 and taking the square root, I get $$\sin A = \pm \sqrt{\frac{9}{13}}.$$
But, I already said that $\sin A$ is positive, so I can discard the negative root.