Here's a link to the figure:
http://oi56.tinypic.com/15civs0.jpg

I can't solve for X, IDK how.

The answer is ((16 square root of 6)/3)

Help!

The first step is to find the length of the side shared by the two triangles. It forms the hypotenuse of the triangle on the right, so let's call it H. That triangle has a leg of length 16, adjacent to a 45 degree angle. Since we know that cosine=adjacent/hypotenuse, we can say

\cos 45 = \frac{16}{H}

What does that give you for a value for H?

Notice how we were able to find the length of one side of the triangle by simply knowing one angle and the length of one other side? Now that you know H, you just have to repeat that process for the triangle on the left (since H also comprises a leg of THAT triangle). You'll have to use a different trig identity than cos=adj./hyp. but it's the same idea.

Let me know if you still can't get it after finding H. I'll help you through the next step.