Is it possible to have two triangles with equal sides but with different angles? Can there be two triangles, for example $ \triangle ABC $ and $ \triangle DEF$ so that the measure of their sides is the same but their angles are different? Thanks.
 A: Yes, if the triangles aren't necessarily both Euclidean. This may not be what you meant, and the other answers may answer your question more appropriately, but this seems a good opportunity to point out some more exotic scenarios that can exist.
One thing we could do is take triangles from different geometries. For instance, the sum of the angles of a triangle drawn on a sphere will always be greater than $180$ degrees. So you could draw one triangle on a flat plane, another on a sphere, make the side lengths match, and you will get different sized angles.
Be aware though that it isn't sufficient to just scale the metric on a Euclidean space. That is, if I define two different planes where distance is greater in one but they are both flat, and I draw two triangles with matching side length, the angles will still be the same. The important thing about the example of the previous paragraph is that we changed the curvature of the space.
Maybe it's cheating to allow the two triangles to come from different geometries. Well it is in fact possible to draw two triangles with matching side lengths and different angles, in the same geometry. You can do this in the hyperbolic plane by allowing vertices on the boundary of the plane.
The boundary of the hyperbolic plane is like a circle that is infinitely far away from any point in the space. If you put two vertices on this boundary, then pick whatever point you want within the space for your third vertex, each side of the triangle will have infinite length. So the lengths of the sides would be unaffected by where you choose to put your third vertex. Also, the two angles at the vertices on the boundary would actually be $0$ degrees, and the third one could be any number up to and including $180$ degrees, depending on where you put it! This wouldn't work in Euclidean space. You can also talk about a circle boundary of $\mathbb{R}^2$ infinitely far away, but there, you can't draw a straight line connecting arbitrary boundary points and passing through the plane.
There's lots of Google-able info out there (and in Wikipedia too) about the hyperbolic plane, if this concept interests you and you want to know more!
A: No: given three sides of a triangle, the cosine rule determines the angles unambiguously:
$$ c^2 = a^2+b^2 - 2ab\cos{\gamma}, $$
so
$$ \cos{\gamma} = \frac{a^2+b^2-c^2}{2ab}, $$
where $\gamma$ is one of the angles, $c$ is the side opposite $\gamma$, and $a,b$ the other two sides. Cosine is one-to-one from the angles between $0$ and $\pi$ to the real interval $(-1,1)$, and one can check that the right-hand side lies strictly between $-1$ and $1$ when $a,b,c$ form a triangle, so one has
$$ \gamma = \arccos{\left( \frac{a^2+b^2-c^2}{2ab} \right)}. $$
A: Take 3 sticks, each must be shorter than the sum of the other 2 sticks. Try to construct more than one triangle, can you? No. If you take 2 sticks and joins their ends, then the third stick will be the constraint for the other ends of the 2 sticks.
A: No, this is not possible in normal Euclidean geometry. Two triangles are congruent if all the corresponding sides are pairwise equal, so their angles will be equal too.
