Problem with degrees of a circle

Question

$A,B,C$ are points on the $O$ center circle, it is given that $BAC \sphericalangle =71^o$. How to prove that $CBO \sphericalangle =19^o$. Obviously, $OB=OC=OA=r$ and we have three isoscels triangles $ABO,AOC$ and $BOC$. How to go form here?

Here is a geogebra image that I made:

Answer 1

Given your triangle with one (inscribed) angle $A = 71^o$ , the central angle $BOC$ is twice angle $A$. This is the inscribed angle theorem:

https://en.wikipedia.org/wiki/Inscribed_angle#:~:text=The%20inscribed%20angle%20theorem%20states,different%20positions%20on%20the%20circle.

So you have an isosceles triangle $BOC$ with an angle $142^o$ and the two remaining angles are $(180^o-142^o)/2$ each.

Answer 2

Angle BOC = $71^0 \times 2 = 142^0$ (I'm guessing you already know this fact about circle geometry). As you deduced, BOC is an isoscles triangle and thus, $\angle OBC = \angle OCB$. Therefore, if the desired angle is x, $142^0 + 2x = 180^0$. It follows that x is $19^0$