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If $\cos (x-y)=\frac{\sqrt{3}}{2}$ and $\sin (x+y)=1$ where $x > y$, then the value of $y$ is: |
0° 30° 60° 90° |
30° |
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 |
