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!
Note that $$\sin (\angle ABC)=\frac{\sqrt3}2\quad\Rightarrow \angle ABC= 60^{\circ}\quad\text{or}\quad120^{\circ}$$Hence there is another triangle with angles $30^{\circ},30^{\circ},120^{\circ} $.
Best Answer
From the midpoint $M$ of $AC$ construct the perpendicular bisector of $AC$, and take on it (outside triangle $ABC$) a point $P$ such that $\angle CPM=\angle COA$.
From the midpoint $N$ of $BC$ construct the perpendicular bisector of $BC$, and take on it (outside triangle $ABC$) a point $Q$ such that $\angle CQN=\angle COB$.
Point $O$ is the second intersection of the circle centred at $P$ through $C$, with the circle centred at $Q$ through $C$.