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...
 A: Since we are only interested in the angles, the actual lengths of the two sides do not matter, as long as we get their ratio right. So we can take the lengths of the adjacent sides to be $1$ and $3$, in whatever units you prefer.  If you want the shorter of the two adjacent sides to be $20$ metres, then the other adjacent side will need to be $60$ metres. But we might as well work with the simpler numbers $1$ and $3$.
To compute the length of the third side, we use a generalization of the Pythagorean Theorem called the Cosine Law.  Let the vertices of a triangle be $A$, $B$, and $C$, and let the sides opposite to these vertices be $a$, $b$, and $c$. For brevity, let the angle at $A$ be called $A$, the angle at $B$ be called $B$, and the angle at $C$ be called $C$. The Cosine Law says that
$$c^2=a^2+b^2-2ab\cos C.$$
Take $C=60^\circ$, and $a=1$, $b=3$.
Since $\cos(60^\circ)=1/2$, we get
$$c^2=1^2+3^2-2(1)(3)(1/2),$$
so $c^2=7$ and therefore $c=\sqrt{7}$.  We now know all the sides.
To find angles $A$ and $B$, we could use the Cosine Law again. We ilustrate the procedure by finding  $\cos A$.
By the Cosine Law,
$$a^2=b^2+c^2-2bc\cos A.$$
But $a=1$, $b=3$, and by our previous work $c=\sqrt{7}$. It follows that
$$1=9+7-2(3)(\sqrt{7})\cos A,$$
and therefore 
$$\cos A= \frac{5}{2\sqrt{7}}.$$
The angle in the interval from $0$ to $180^\circ$ whose cosine is $5/(2\sqrt{7})$ is not a "nice" angle. The calculator (we press the $\cos^{-1}$ button) says that this angle is about $19.1066$ degrees.
Another way to proceed, once we have found $c$, is to use the Sine Law
$$\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}.$$
From this we obtain that
$$\frac{\sin A}{1}=\frac{\sqrt{3}/2}{\sqrt{7}}.$$
The calculator now says that $\sin A$ is approximately $0.3273268$, and then the calculator gives that $A$ is approximately $19.1066$ degrees.   In the old days, the Cosine Law was not liked very much, and the Sine Law was preferred, because the Sine Law involves only multiplication and division, which can be done easily using tables or a slide rule.  A Cosine Law calculation with ugly numbers is usually more tedious. 
The third angle of the triangle (angle $B$) can be found in the same way. But it is easier to use the fact that the angles of a triangle add up to $180^\circ$.  So angle $B$ is about $100.8934$ degrees.
A: This is indeed a very basic question. The information you have is "Side-Angle-Side (SAS)", which is enough to classify your triangle. (You may assume that one side has length 1 and the other has length 3 or just keep it as a variable x and 3x). Then you essentially have to apply the law of cosines, see this page for a detailed solution, or just google for SAS.
A: 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. 
A: 
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 basically this is the information we have so far:

The first thing to do here is realize that the two adjacent sides are in a ratio 1:3. For the purpose of solving the problem let us give the two adjacent side lengths 1 and 3. Using the cosine rule $a^2=b^2+c^2-2bc\cos A$ we can determine the last length: $$a^2=3^3+1^2-2\cdot 3\cdot 1\cdot \cos 60$$
$$a^2=10-6\cdot \cos 60$$
$$a^2=10-6\cdot {1\over 2}$$
$$a=\sqrt7$$
Now, using the sine rule we can find the remaining angles.
$${\sin 60\over \sqrt7}={\sin \phi \over 1}$$
$$\sin \phi = {1\cdot \sin 60 \over \sqrt7}$$
$$\phi \approx 19.1$$
$$\theta =180-60-19.1=100.9$$
