Finding the measure of an arc on a circle If someone could work me through how to solve this, that would be great because I am stumped on this one. 


I know it looks like there is a lot of useless information in the picture, but there are many problems pertaining to this picture. I already know that AO = AG because they are both tangents. Although I don't see how this helps me to find the angles anyways. I also know that angle OFH is 100 degrees, because it is an inscribed angle. Besides that, I am absolutely stumped. I am not fishing for an answer, but if anyone could put me on the right path, that would be great. Thanks!  
 A: Hints:
Note that
$$m(\angle OAF)= m(\angle BAQ),$$
$$m(\angle AOF)=\frac{1}{2} (110 ^\circ),$$
$$m(\angle OFH)=\frac{1}{2} (200 ^\circ) $$
and that $\angle OFH$ is an exterior angle of $\triangle AOF$.
A: Here's an alternative applying other geometric theorems.  A couple of segments need to be added to the diagram, so I've removed everything unrelated to the question.

From the known measures of the arcs of circle $ \ N \ $ , we then have that the measure of arc $ \ FH \ $ is 50º .  Segments $ \ \overline{ON} \ , \ \overline{NF} \ , \ \text{and} \ \ \overline{NH} \ $ are radii of this circle, and $ \ \overline{OF} \ \  \text{and}  \ \ \overline{FH} \ $ are chords, so $ \ \Delta ONF \ \ \text{and} \ \ \Delta NFH \ $ are isosceles triangles.  Thus, 
$$ \ m(\angle ONF) \ = \ 110º \ \ \Rightarrow \ \ m(\angle NOF) \ = \ m(\angle OFN) \ = \ 35º \ \ \ \text{and} $$
$$ \ m(\angle FNH) \ = \ 50º \ \ \Rightarrow \ \ m(\angle NFH) \ = \ m(\angle NHF) \ = \ 65º \ \ . $$
Angle $ \ OFA \ $ is supplementary to angles $ \ OFN \ \ \text{and} \ \ NFH \ $ , so $ \ m(\angle OFA) \ = \ 80º \ . $
Line $ \ \overline{AB} \ $ is tangent to circle $ \ N \ $ at point $ \ O \ $ , so radius $ \ \overline{NO} \ $ is perpendicular to $ \ \overline{AB} \ $ there.  Angle $ \ AOF \ $ is complementary to angle $ \ NOF \ $ , hence $ \ m(\angle AOF) \ = \ 55º \ $ .
Therefore, in $ \ \Delta AOF \ $ , 
$$ \ m(\angle OAF) \ = \ m(\angle BAQ) \ = \ 180º \ - \ 80º \ - \ 55º \ = \ 45º \ \ . $$
Quite a few of the angle-arc "circle theorems" are tied together, so there are a number of approaches that can be taken.
