If sin (x-y)= $$\frac{\sqrt{3}}{2}$$​ and cos (x+y)=$$\frac12$$​ where x and y are positive acute angles and x ≥ y, then the value of x is:

Correct option is B

Given:
$$\text {sin (x-y)}= \frac{\sqrt{3}}{2}$$​​
$$\cos (x+y) =\frac 12$$​​
Solution:
We know that:
$$\text {sin (x-y)}= \frac{\sqrt{3}}{2}$$​ 

Since, sin60∘ = $$\frac{\sqrt{3}}{2}$$​
x-y= $$60\degree$$..............(1)​
Similarly, we know:

​$$\cos (x+y) =\frac 12$$​​

Since, cos 60∘= $$\frac 12$$​​
x+y=$$60\degree.......(2)$$​​
Add (1) and (2)
x-y=$$60\degree$$​​
x+y=$$60\degree$$​
Adding these equations
(x-y)+(x+y)= $$60\degree +60\degree$$​​
2x=$$120\degree$$​
x=$$60\degree$$​ ​​
The value of x = $$60\degree$$​​