How can absolute value of azimuth exceed distance from pole? I'm sure I'm just doing something dumb, but: 


Yildun has a declination of approximately 86.5. This puts it 3.5 
degrees from the celestial north pole. 
Per the snapshot below, Yildun's azimuth can be as high as 4.17 
degrees. 
Since the celestial north pole always has an azimuth of 0, 
wouldn't this put Yildun 4.17 degrees from the pole, contradicting 
the distance calculated from declination? 


What am I missing? 



EDIT: I thought I'd figured this out:


On Earth, one degree of latitude is not the same as one degree of longitude (except at the equator). 
Analogously, in the sky, one declination degree is not the same as one right ascension degree. 


I thought the situation for right ascension applied to azimuth, but it doesn't.

Both azimuth and declination are measured on "great circles" (declination is measured on a half circle, but same general idea), so degrees in azimuth should equal degrees in declination.
It turns out I was right when I thought I was wrong.

Consider the azimuth circle for 85 degrees elevation. It's very small. For example, 85 degrees elevation and 0 azimuth is only 10 degrees away from 85 degrees elevation and 180 azimuth.

Therefore, the azimuth circle for a given elevation is smaller than a great circle. It's only cosine(elevation) the size of a great circle.

At 35 degrees, the azimuth circle is cosine(35 degrees) = 0.819152 the size of a great circle, so 4.17 degrees in azimuth = 4.17 degrees * 0.819152 = right around 3.5 degrees on a great circle.