Solving Trigonometric Equations for Acute Angles

We are given two trigonometric equations:

• $\cos(x - y) = \frac{\sqrt{3}}{2}$
• $\sin(x + y) = 1$

We are also told that $x$ and $y$ are positive acute angles ($0^\circ < x, y < 90^\circ$) and $x \ge y$.

Determining Angle Values

From the first equation, $\cos(x - y) = \frac{\sqrt{3}}{2}$. Since $x$ and $y$ are positive acute angles and $x \ge y$, the angle $(x - y)$ must be in the range $[0^\circ, 90^\circ)$. The angle whose cosine is $\frac{\sqrt{3}}{2}$ is $30^\circ$.

Therefore, $x - y = 30^\circ$. (Equation 1)

From the second equation, $\sin(x + y) = 1$. Since $x$ and $y$ are positive acute angles, their sum $(x + y)$ must be in the range $(0^\circ, 180^\circ)$. The angle whose sine is $1$ is $90^\circ$.

Therefore, $x + y = 90^\circ$. (Equation 2)

Solving the System of Equations

Now we have a system of two linear equations:

1. $x - y = 30^\circ$
2. $x + y = 90^\circ$

Add Equation 1 and Equation 2:

$ (x - y) + (x + y) = 30^\circ + 90^\circ $

$ 2x = 120^\circ $

$ x = \frac{120^\circ}{2} $

$ x = 60^\circ $

Substitute the value of $x$ into Equation 2:

$ 60^\circ + y = 90^\circ $

$ y = 90^\circ - 60^\circ $

$ y = 30^\circ $

Verifying Constraints and Final Answer

The calculated values are $x = 60^\circ$ and $y = 30^\circ$. Let's check the conditions:

• Are $x$ and $y$ positive acute angles? Yes, $60^\circ$ and $30^\circ$ are both between $0^\circ$ and $90^\circ$.
• Is $x \ge y$? Yes, $60^\circ \ge 30^\circ$.

The values $x = 60^\circ$ and $y = 30^\circ$ satisfy all given conditions.

This corresponds to Option A.