Why does x have to be acute?

Given that $\sin{x} = \frac{1}{\sqrt{5}}$ where $x$ is acute, and that $\cos(x-y) = \sin{y}$, show that $\tan{y} = \frac{\sqrt{5} + 1}{2}$

I can derive the required equation but why does $x$ have to be acute?

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$sin x$ takes on the same value at $x+\pi$, however at that value $cos(x)$ takes on the negative of the value in question. –  Foo Barrigno Mar 11 '14 at 13:30

As $\sin x=\sin(\pi-x)$
and $\sin x=\frac1{\sqrt5}>0,$ one value of $x$ lies in the First Quadrant ( where $\cos x>0$ ), and the other in the Second (where $\cos x<0$)
Now as $x$ is acute, we can safely discard the value in the Second Quadrant