HI
Did u mean these three angles , $\displaystyle \angle ACB$ , $\displaystyle \angle CAB $and $\displaystyle \angle ABC$ .
If so , you are correct by saying $\displaystyle \angle ABC$ is $\displaystyle 80^o $.
Let D be another point intersecting the circle ,
since ABCD is a cyclic quadrilateral , it should be obvious that
$\displaystyle \angle ACB=180-120 $
Then you can finish off the last one .
Hello kevin3000You say that you can see how $\displaystyle \angle ABC = 80^o$. You are right, and the reason that it's equal to the marked angle of $\displaystyle 80^o$ is that they are both subtended by the arc $\displaystyle AC$.
We say that angles like this are Angles in the Same Segment - and they're equal.
(When you divide a circle into two by a chord you get two segments - the bigger one is the major segment, and the smaller one is the minor segment. Here the chord is $\displaystyle AC$, and the $\displaystyle 80^o$ angle and $\displaystyle \angle ABC$ are both in its major segment.)
Well now, can you spot the arc that is subtending the marked $\displaystyle 40^o$ angle? It is the arc $\displaystyle BC$. And this arc also subtends $\displaystyle \angle BAC$. So these angles are also equal - they're in the major segment of the chord $\displaystyle BC$. So $\displaystyle \angle BAC = 40^o$.
I'm sure you know how to find the third angle of the triangle, $\displaystyle \angle ACB$, so I'll leave it up to you.
Grandad