1. ## Calculate the angles

Ok I can see how angle B is 80 degrees. Thats all I can solve on my own hand right now. I wonder If you can explain and help me notice other obvious things with the problem.

2. Originally Posted by kevin3000
Ok I can see how angle B is 80 degrees. Thats all I can solve on my own hand right now. I wonder If you can explain and help me notice other obvious things with the problem.

HI

Did u mean these three angles , $\angle ACB$ , $\angle CAB$and $\angle ABC$ .

If so , you are correct by saying $\angle ABC$ is $80^o$.

Let D be another point intersecting the circle ,

since ABCD is a cyclic quadrilateral , it should be obvious that
$\angle ACB=180-120$

Then you can finish off the last one .

HI

Did u mean these three angles , $\angle ACB$ , $\angle CAB$and $\angle ABC$ .

If so , you are correct by saying $\angle ABC$ is $80^o$.

Let D be another point intersecting the circle ,

since ABCD is a cyclic quadrilateral , it should be obvious that
$\angle ACB=180-120$

Then you can finish off the last one .
But where is the point D ?

4. ## Angles in the Same Segment

Hello kevin3000
Originally Posted by kevin3000
Ok I can see how angle B is 80 degrees. Thats all I can solve on my own hand right now. I wonder If you can explain and help me notice other obvious things with the problem.
You say that you can see how $\angle ABC = 80^o$. You are right, and the reason that it's equal to the marked angle of $80^o$ is that they are both subtended by the arc $AC$.

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 $AC$, and the $80^o$ angle and $\angle ABC$ are both in its major segment.)

Well now, can you spot the arc that is subtending the marked $40^o$ angle? It is the arc $BC$. And this arc also subtends $\angle BAC$. So these angles are also equal - they're in the major segment of the chord $BC$. So $\angle BAC = 40^o$.

I'm sure you know how to find the third angle of the triangle, $\angle ACB$, so I'll leave it up to you.