Spherical Law of Cosines/Angles

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $\triangle ABC$ be a spherical triangle on the surface of a sphere whose center is $O$.

Let the sides $a, b, c$ of $\triangle ABC$ be measured by the angles subtended at $O$, where $a, b, c$ are opposite $A, B, C$ respectively.


Then:

$\cos A = -\cos B \cos C + \sin B \sin C \cos a$


Proof

Let $\triangle A'B'C'$ be the polar triangle of $\triangle ABC$.

Let the sides $a', b', c'$ of $\triangle A'B'C'$ be opposite $A', B', C'$ respectively.


From Spherical Triangle is Polar Triangle of its Polar Triangle we have that:

not only is $\triangle A'B'C'$ be the polar triangle of $\triangle ABC$
but also $\triangle ABC$ is the polar triangle of $\triangle A'B'C'$.


We have:

\(\ds \cos a'\) \(=\) \(\ds \cos b' \cos c' + \sin b' \sin c' \cos A'\) Spherical Law of Cosines
\(\ds \leadsto \ \ \) \(\ds \map \cos {\pi - A}\) \(=\) \(\ds \map \cos {\pi - B} \, \map \cos {\pi - C} + \map \sin {\pi - B} \, \map \sin {\pi - C} \, \map \cos {\pi - a}\) Side of Spherical Triangle is Supplement of Angle of Polar Triangle
\(\ds \leadsto \ \ \) \(\ds -\cos A\) \(=\) \(\ds \paren {-\cos B} \paren {-\cos C} + \map \sin {\pi - B} \, \map \sin {\pi - C} \, \paren {-\cos a}\) Cosine of Supplementary Angle
\(\ds \leadsto \ \ \) \(\ds -\cos A\) \(=\) \(\ds \paren {-\cos B} \paren {-\cos C} + \sin B \sin C \paren {-\cos a}\) Sine of Supplementary Angle
\(\ds \leadsto \ \ \) \(\ds \cos A\) \(=\) \(\ds -\cos B \cos C + \sin B \sin C \cos a\) simplifying and rearranging

$\blacksquare$


Historical Note

The Spherical Law of Cosines was first stated by Regiomontanus in his De Triangulis Omnimodus of $1464$.


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Spherical_Law_of_Cosines/Angles&oldid=720276"