Calculate position based on angles between three known points

Yes, you can. First, make sure your $\theta$ angles are determined by counterclockwise orientation and are between $0$ and $\pi$. Also make sure $$\theta_{AB} + \theta_{BC} + \theta_{CA} = 2 \pi$$ The three circles on the picture intersect at the point you want. These circles are determined by the position of two vertices of the triangle and the corresponding angle $\theta$.

To determine the point $O$ it is enough to construct only two of the three circles. The third one is guaranteed to pass through $O$ as well if and only if $$\theta_{AB} + \theta_{BC} + \theta_{CA} = 2 \pi$$ So basically, given say $\theta_{CA}$ and $\theta_{AB}$, as well as the positions of the points $A, B, C$, you can determine the coordinates of $O$ as the solution to a system of quadratic equations for the two circles, circle $k_{CA}$ with center $O_{CA}$ and radius $AO_{CA}$ and circle $k_{AB}$ with center $O_{AB}$ and radius $AO_{AB}$.

To determine circle $k_{AB}$ for instance, you need to find the position of point $O_{AB}$ and find the length of $AO_{AB}$. Notice that triangle $ABO_{AB}$ as well as triangle $CAO_{CA}$ are isosceles, the points $M_{AB}$ and $M_{CA}$ are the orthogonal projections of the centers $O_{AB}$ and $O_{CA}$ on the edges $AB$ and $CA$ respectively and as such $M_{AB}$ and $M_{CA}$ are the midpoints of the edges $AB$ and $CA$ respectively. Angles $\angle \, O_{AB}AB = \theta_{AB} - \frac{\pi}{2}$ and $\angle \, O_{CA}AC = \theta_{CA} - \frac{\pi}{2}$. Hence in the right angled triangle $M_{AB}AO_{AB}$ you can find $$M_{AB}O_{AB} = \frac{1}{2}AB \tan\left(\theta_{AB} - \frac{\pi}{2}\right)$$ $$AO_{AB} =\frac{AB}{2 \,\cos\left(\theta_{AB} - \frac{\pi}{2}\right)}$$

Analogously

$$M_{CA}O_{CA} = \frac{1}{2}CA \tan\left(\theta_{CA} - \frac{\pi}{2}\right)$$ $$AO_{CA} =\frac{CA}{2 \,\cos\left(\theta_{CA} - \frac{\pi}{2}\right)}$$ This determines the locations of the centers $O_{AB}$ and $O_{CA}$ and $AO_{CA}$ and $AO_{AB}$ are the radii of the corresponding circles $k_{AB}$ and $k_{CA}$. The point $O$ is the first intersection point of circles $k_{AB}$ and $k_{CA}$. Have in mind the second one is $A$ so you know it.

It's possible. Assuming all measured angles are greater than 0, the following should work for any position of the observer.

Given a line segment AB, the loci of points P such that the angle APB has a constant value are circular arcs that pass through the points A and B. In other words, if you have measured an angle $\theta_{AB}$ between the points $A$ and $B$, your position must be on a particular circular arc passing through $A$ and $B$. This arc can be determined when you know the angle and the position of $A$ and $B$. And when you have $3$ points, you can make $3$ circular arcs and find their common intersection point. That point will be the position of the observer.

Step 1 (AB)

To find the circular arc passing through $A$ and $B$, first find the bisector of $AB$. Now find the point on the bisector which creates an angle of $\theta_{AB}$ to the cord $AB$. This can be determined using the formula $$h_{AB} = x_{AB}*cot(\frac{\theta_{AB}}{2})$$

Draw a circle through this point and points $A$ and $B$. The observer's position will be on this circle.

Step 2 (BC)

In the same way as was done above, we find the circular arc passing through $B$ and $C$ (the red circle is the one found above).

We see that the new circle intersects the first circle at point $B$ (not surprising) and at the observer's position. We have now determined the observer's position.

To make absolutely sure, we can do the last circular arc:

And we have a match.