Four-Parts Formula

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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.


We have:

$\cos a \cos C = \sin a \cot b - \sin C \cot B$


That is:

$\map \cos {\text {inner side} } \cdot \map \cos {\text {inner angle} } = \map \sin {\text {inner side} } \cdot \map \cot {\text {other side} } - \map \sin {\text {inner angle} } \cdot \map \cot {\text {other angle} }$


This is known as the four-parts formula, as it defines the relationship between each of four consecutive parts of $\triangle ABC$.


Corollary

$\cos A \cos c = \sin A \cot B - \sin c \cot b$


Proof

SphericalTriangle-FourParts.png
\(\ds \cos b\) \(=\) \(\ds \cos a \cos c + \sin a \sin c \cos B\) Spherical Law of Cosines
\(\ds \cos c\) \(=\) \(\ds \cos b \cos a + \sin b \sin a \cos C\) Spherical Law of Cosines
\(\ds \leadsto \ \ \) \(\ds \cos b\) \(=\) \(\ds \cos a \paren {\cos b \cos a + \sin b \sin a \cos C} + \sin a \sin c \cos B\) substituting for $\cos c$
\(\ds \leadsto \ \ \) \(\ds \cos b\) \(=\) \(\ds \cos^2 a \cos b + \cos a \sin b \sin a \cos C + \sin a \sin c \cos B\)
\(\ds \leadsto \ \ \) \(\ds \cos b \paren {1 - \cos^2 a}\) \(=\) \(\ds \cos a \sin b \sin a \cos C + \sin a \sin c \cos B\)
\(\ds \leadsto \ \ \) \(\ds \cos b \sin^2 a\) \(=\) \(\ds \cos a \sin b \sin a \cos C + \sin a \sin c \cos B\) Sum of Squares of Sine and Cosine
\(\ds \leadsto \ \ \) \(\ds \cot b \sin a\) \(=\) \(\ds \cos a \cos C + \dfrac {\sin c} {\sin b} \cos B\) dividing both sides by $\sin a \sin b$
\(\ds \leadsto \ \ \) \(\ds \cot b \sin a\) \(=\) \(\ds \cos a \cos C + \dfrac {\sin C} {\sin B} \cos B\) Spherical Law of Sines
\(\ds \leadsto \ \ \) \(\ds \cos a \cos C\) \(=\) \(\ds \sin a \cot b - \sin C \cot B\) simplification

$\blacksquare$


Also see


Sources