Simple trigonometry question (angles) I am starting again with trigonometry just for fun and remember the old days. I was not bad at maths, but however I remember nothing about trigonometry...
And I'm missing something in this simple question, and I hope you can tell me what.

One corner of a triangle has a 60º angle, and the length of the two
  adjacent sides are in ratio 1:3. Calculate the angles of the other
  triangle corners.

So what we have is the main angle, $60^\circ$, and the adjacent sides, which are $20$ meters (meters for instance). We can calculate the hypotaneous just using $a^2 + b^2 = h^2$. But how to calculate the other angles?
Thank you very much and sorry for this very basic question...

There is no "hyp", as no one has said this is a right-angle triangle. There is also no reason to think the adjacent sides are the same length - quite the opposite, you're told their lengths are in the ratio 1:3. So make believe those sides are 1 and 3, and then use the law of cosines and the law of sines to work out the other sides and angles. 
