We are given that $\sin 3y = \cos 2y$ and $0^{\circ} \le y \le 90^{\circ}$. We are asked to find the value of $y$.

1. Problem Description

We are given that $\sin 3y = \cos 2y$ and $0^{\circ} \le y \le 90^{\circ}$. We are asked to find the value of $y$.

2. Solution Steps

We know that $\sin \theta = \cos (90^{\circ} - \theta)$. Therefore, we can write the given equation as:

$\sin 3y = \cos (90^{\circ} - 3y)$

Since $\sin 3y = \cos 2y$, we can write

$\cos (90^{\circ} - 3y) = \cos 2y$

This implies that

$90^{\circ} - 3y = 2y$

Adding $3y$ to both sides, we get

$90^{\circ} = 5y$

Dividing both sides by 5, we have

$y = \frac{90^{\circ}}{5} = 18^{\circ}$

We can check this answer:

$\sin 3y = \sin (3 \cdot 18^{\circ}) = \sin 54^{\circ}$

$\cos 2y = \cos (2 \cdot 18^{\circ}) = \cos 36^{\circ}$

Since $\sin 54^{\circ} = \sin (90^{\circ} - 36^{\circ}) = \cos 36^{\circ}$, our answer is correct.

Also $0^{\circ} \le 18^{\circ} \le 90^{\circ}$

3. Final Answer

$y = 18^{\circ}$

So the answer is A. $18^{\circ}$.