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

Correct option is D

Given:

$$cos(x + y) = \frac{1}{2}$$ 

​$$sin(x - y) = 0$$ 

where x and y are positive acute angle and x ≥ y,  

Formula Used: 

​$$\cos 60^0 = \frac{1}{2} \\ \ \\ \sin 0^0 = 0$$ 

Solution:

​$$cos(x + y) = \frac{1}{2}$$ 

cos( x + y ) = cos 60°

x + y = 60° ......(1)

Or sin(x - y ) = 0 

sin ( x - y) = sin 0° 

x  - y = 0°.......(2) 

Solving eq. 1 and 2 we get; 

x = y = 30°​