In higher level mathematics, if we know two angles of a triangle and one of the sides, we can calculate the two missing sides. The good news is we don't have to worry about that on the GRE. However, we do have to memorize two specific cases. The first is below:

$30$-$60$-$90$ Triangles

For example, if the side opposite a $30$° angle in a right triangle is $6$, we know the other side is $6\sqrt{3}$ and the hypotenuse is $12$.

If we're given the side length opposite the $60$° side, no problem. We can still calculate the other two sides.

For example, if the side length opposite the $60$° side is $5$, the shortest side is $\frac{5}{\sqrt{3}}$ (notice we divided by $\sqrt{3}$ here) and the hypotenuse is $2 \times \frac{5}{\sqrt{3}} = \frac{10}{\sqrt{3}}$

Trignomentry?

If you're more comfortable with trignomentry, notice that what we've said above is just a "dressed" version of trignomentry. So in the above example, notice that

$$\frac{\textrm{opposite side}}{\textrm{adjacent side}} = \frac{x}{\sqrt{3}x} = \frac{1}{\sqrt{3}}$$

which is just ...

$$\tan{30^{\circ}} = \frac{1}{\sqrt{3}}$$

and you can use the other trignomentric ratios to find the other sides, or just use Pythagoras' theorem at that point.

If you are comfortable with trignomentry, there's no need to memorize the ratios above (because you already know it). If you don't know it, no big deal, just memorize the triangle ratios above. You don't need trignomentry for the GRE.