Length of Arc of Small Circle

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $S$ be a sphere.

Let $\bigcirc FCD$ be a small circle on $S$.

Let $C$ and $D$ be the points on $\bigcirc FCD$ such that $CD$ is the arc of $\bigcirc FCD$ whose length is to be determined.


Construction

Let $P$ and $Q$ be the poles of $\bigcirc FCD$.

Let $\bigcirc PCQ$ and $\bigcirc PDQ$ be great circles on $S$.

Let $\bigcirc EAB$ be the great circle whose poles are $P$ and $Q$.

Let $A$ and $B$ be the points on $\bigcirc EAB$ which intersect $\bigcirc PCQ$ and $\bigcirc PDQ$.


The length of the arc $CD$ of $\bigcirc FCD$ is given by:

$CD = AB \cos AC$

or:

$CD = AB \sin PC$


Proof

Size-of-small-circle.png

Let $R$ denote the center of $\bigcirc FCD$.

Let $O$ denote the center of $S$, which is also the center of $\bigcirc EAB$.

We have:

$CD = RC \times \angle CRD$

Similarly:

$AB = OA \times \angle AOB$

By Circles with Same Poles are Parallel:

$\bigcirc FCD \parallel \bigcirc EAB$


Hence $RC$ and $RD$ are parallel to $OA$ and $OB$ respectively.

Thus:

\(\ds \angle CRD\) \(=\) \(\ds \angle AOB\)
\(\ds \leadsto \ \ \) \(\ds CD\) \(=\) \(\ds \dfrac {RC} {OA} AB\)
\(\ds \) \(=\) \(\ds \dfrac {RC} {OC} AB\) as $OA = OC$ are both radii of $S$


We also have that:

\(\ds RC\) \(\perp\) \(\ds OR\)
\(\ds \leadsto \ \ \) \(\ds RC\) \(=\) \(\ds OC \cos \angle RCO\)


and that:

\(\ds RC\) \(\parallel\) \(\ds OA\)
\(\ds \leadsto \ \ \) \(\ds \angle RCO\) \(=\) \(\ds \angle AOC\)

We have that $\angle AOC$ is the (plane) angle subtended at $O$ by the arc $AC$ of $\bigcirc EAB$.


Thus:

\(\ds CD\) \(=\) \(\ds AB \cos AC\)
\(\ds \) \(=\) \(\ds AB \, \map \cos {PA - PC}\)
\(\ds \) \(=\) \(\ds AB \sin PC\) as $PA$ is a right angle, and Cosine of Complement equals Sine‎


Hence the result.

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Length_of_Arc_of_Small_Circle&oldid=428611"