If cos(x + y)=$$\frac{1}{2}$$​ and sin(x-y) = 0, where x and y are positive acute angles and x ≥y, then x and y are:

Correct option is C

Given:
$$• \cos(x + y) = \frac{1}{2} \\\ \\• \sin(x - y) = 0$$​​
• x and y are positive acute angles, and  .$$x \geq y$$​​

Formula Used:

$$\cos(\theta) = \frac{1}{2} \ when \ \theta = 60^\circ or \theta = 300^\circ .$$​ Since x and y are acute angles,
we only consider $$\theta = 60^\circ$$​.
$$\sin(\theta) = 0 \ when\ \theta = 0^\circ \ or \ \theta = 180^\circ$$​ . Since x and y are acute angles,
we only consider $$\theta = 0^\circ$$​​
Solution:

Using $$\cos(x + y) = \frac{1}{2}$$​:

Since $$\cos(x + y) = \frac{1}{2} ,$$​ we deduce that:

​$$x + y = 60^\circ$$​​

Using $$\sin(x - y) = 0 :$$​​

​$$sin(x - y) = 0 ,$$​​

​$$x - y = 0^\circ$$​​
 $$\sin(0^\circ) = 0 .$$​​
x = y

We now have two equations:

Adding these two equations:

​$$(x + y) + (x - y) = 60^\circ \\\ \\2x = 60^\circ\\\ \\x = 30^\circ$$​​

Substitute $$x = 30^\circ into \ x - y = 0^\circ :$$​​

​$$30^\circ - y = 0 \Rightarrow y = 30^\circ$$​​

The values of x and y are both $$30^\circ$$​.