If $\sin (x+y)=1\text{and}\cos (x-y)=\frac{\sqrt{3}}{2},$ find the value of $y$.

The correct answer is 30°.

Let us solve the problem step-by-step.

We are given the following trigonometric equations:

$\sin (x+y)=1$

and

$\cos (x-y)=\frac{\sqrt{3}}{2}$

Step 1: Simplify the first equation.
We know that for acute angles, the sine function equals 1 at 90°:

$\sin (90°)=1$

Therefore, we can write Equation (1) as:

$x+y=90°$

Step 2: Simplify the second equation.
We know that for acute angles, the cosine function equals the square root of 3 divided by 2 at 30°:

$\cos (30°)=\frac{\sqrt{3}}{2}$

Therefore, we can write Equation (2) as:

$x-y=30°$

Step 3: Solve for y by subtracting the two simplified equations.
Subtracting Equation (2) from Equation (1):

$(x+y)-(x-y)=90°-30°$

Simplifying the left-hand side and right-hand side:

$2y=60°$

Dividing by 2:

$y=30°$

Thus, the value of y is 30°.