If $\cos (x-y)=\frac{\sqrt{3}}{2}$ and $\sin (x+y)=1$ where $x > y$, then the value of $y$ is:
Answer & explanation
Correct answer: option 2
We have,
$\cos (x-y)=\frac{\sqrt{3}}{2}$ and $\sin (x+y)=1$
(x-y)= $\frac{\sqrt{3}}{2}$ = 30o -------(1)
(x+y) = 1 = 90o -------(2)
From 1 and 2,
x = 60o and y = 30o