# Relation of length of small circle with that of a great circle

This is not the best place to ask such doubts, so please recommend where I can. I currently don't have the facility to ask a teacher specialized for astronomy.

So I just got reading- "Astronomy: Principles and Practice, Fourth Edition by A. Roy and D. Clarke" and I couldn't understand some lines (a possible derivation for some geometrical formula, which I have no clue about)

It stated the following:-

""The length of a small circle arc, such as $$FG$$ is related simply to the length of an arc of the great circle whose plane is parallel to that of the small circle In figure 7.1, let $$r$$ be the radius of the small circle $$EFGHE$$. Then $$FG = r × ∠FKG$$.

Also $$BC= R×∠BOC$$.

Both $$OB$$ and $$KF$$ lie on plane $$PFBQ$$; $$KF$$ also lies on plane $$EFGH$$ while $$OB$$ lies on plane $$ABCD$$. Therefore, $$KF$$ must be parallel to $$OB$$, since plane $$EFGH$$ is parallel to plane $$ABCD$$. Similarly, $$KG$$ is parallel to $$OC$$ Then $$∠FKG = ∠BOC$$. Hence, $$FG = BC × r/R$$.

In the plane triangle $$KOF$$, right-angled at $$K$$, $$KF=r$$; $$OF = R$$. Hence, $$FG = BC × \sin KOF$$. But $$∠POB = 90^\circ$$ so that we may write alternatively $$FG = BC\cos FB$$. If the radius of the sphere is unity, $$PF = ∠POF = ∠KOF$$ and $$FB = ∠FOB$$, so that we have $$FG = BC\sin PF$$ and $$FG = BC\cos FB$$.""

Here is the figure 7.1 it was talking about:-

I couldn't understand what it wanted to derive. Also how can $$\sin PF$$ and $$\cos FB$$ be possible? aren't $$PF$$ and $$FB$$ arcs? Could anyone please help, I will be grateful!

• The author says FB=∠FOB, because for a unit sphere or circle (and angles measured in radians) the length of an arc is identical to the measure of angle subtending it. So cos BF is cos∠FOB. As for what the author is trying to derive, he says it right at the top: it's to relate the length FG to BC. Which he does in the last two lines. Commented Mar 10 at 14:28
• @ScienceSnake this addresses the issue, thanks, you can answer this question if you want to, instead Commented Mar 10 at 14:39
• I finally got one of the possible reasons why- the author wanted to derive the formula for departure Commented Mar 11 at 3:19