This article will derive fundamental equations for a spherical triangle. These equations express the ratio between the angles and curved sides of the triangle drawn on a celestial sphere. We will need these formulas for calculating the primary directions.
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Consider a right-angled triangle on a sphere with sides $\alpha$, $\beta$, and $\gamma$, as well as interior angles $A$, $B$, and $C$.
How will its sides and angles be related to each other?
Rotated Coordinate Systems
First, we introduce a cartesian vector $\vec{v}$ in $XYZ$ coordinate system.
From the conversion equation (1) it follows that $\vec{v}$ has coordinates
$$ \vec{v} = \left( \begin{aligned} & x = R\cos\beta \\ & y = 0 \\ & z = R\sin\beta \end{aligned} \right) $$
Now, let's use the rotation matrix to rotate $XYZ$ by angle $\alpha$ and express this vector in $\grave{X}\grave{Y}\grave{Z}$ axis. As a result, we have the following:
$$ \begin{aligned} \vec{v} & = R \cos\alpha \cos\beta~ \grave{\vec{X}} \\ & - R \sin\alpha \cos\beta~ \grave{\vec{Y}} \\ & + R\sin\beta~ \grave{\vec{Z}} \\ \end{aligned} $$
But on the other hand, the same $\grave{x}$ component of the vector $\vec{v}$ is just a projection of that vector to the $\grave{X}$-axis in the $\gamma$-angle plane.
$$ v_{\grave{x}} = R\cos{\gamma} $$
It gives us the first equation:
$$ \cos\gamma = \cos\alpha\cos\beta\tag{1} $$
Now we move to the next step and introduce a new coordinate system $\tilde{X}\tilde{Y}\tilde{Z}$, which is $\grave{X}\grave{Y}\grave{Z}$, rotated by angle $-B$ (minus B) in $\grave{Y}\grave{Z}$ plane.
If we apply rotation matrix to $\grave{X}\grave{Y}\grave{Z}$ coordinate system, we will have a $\tilde{z}$-component of the vector $\vec{v}$ to be equal to
$$ \begin{aligned} \tilde{z} & = \sin B~ v_{\grave{y}} + \cos B~ v_{\grave{z}} = \\ & - R \sin B \sin\alpha \cos\beta \\ & + R \cos B \sin \beta \end{aligned} $$
On the other hand, $\tilde{z}$-component of vector $\vec{v}$ is equal to zero. It means that
$$ \sin\alpha = \frac{\tan\beta}{\tan B} \tag{2} $$
The exact ratio applies to $\beta$ angle:
$$ \sin \beta = \frac{\tan \alpha} {\tan A }\tag{3} $$
Other Equations
We have set the main equations for spherical triangles:
$$ \begin{cases} \cos \gamma = \cos \alpha \cos\beta \\ \sin \alpha = \tan\beta/ \tan B \\ \sin \beta = \tan\alpha/ \tan A \end{cases} $$
All the rest is just a consequence of these three equations.
First, let's multiply $(1)$ by $(2)$, and we will get
$$ \cos \gamma = \frac{1} {\tan A\tan B} \tag{4} $$
Now, let's consider $\sin^2 \gamma$:
Similarly, we can write
$$ \sin^2\gamma = \sin^2\beta + \cos^2\alpha \sin^2\beta\tag{5} $$
Now, let's divide both sides of the equation by $\cos^2(\gamma)$:
$$ \begin{aligned} \tan^2\gamma & = \tan^2\alpha \frac {1} {\cos^2\beta} + \tan^2\beta \\ & = \tan^2\beta \tan^2A + \sin^2\alpha\tan^2B \\ & = \sin^2\alpha\tan^2B \left[ \tan^2A + 1\right] \\ & = \frac{\sin\alpha \tan B } {\cos A} \end{aligned} $$
which gives
$$ \sin\gamma = \frac{\sin\alpha} {\sin A}\tag{6} $$
With the same approach, we get from the equation $(5)$
$$ \sin\gamma = \frac{\sin\beta} {\sin B}\tag{7} $$
This equality
$$ \frac{\sin\alpha} {\sin A} = \frac{\sin\beta} {\sin B} $$
is also called the sine theorem.
Now from $(3)$ it follows
$$ \sin A = \frac{\tan\alpha} {\sin\beta}\cos A $$
If we sunstitude $\sin A$ from $(6)$ we will get
$$ \cos A = \frac{\cos\alpha\sin\beta}{\sin\gamma}\tag{8} $$
or
$$ \cos A = \frac{\tan\beta} {\tan\gamma}\tag{9} $$
In a similar manner from (2) follows that
$$ \cos B = \frac{\tan\alpha} {\tan\gamma}\tag{10} $$
If we substitute (8) with (7), we will get
$$ \cos A = \sin B \cos\alpha\tag{11} $$
Similarly,
$$ \cos B = \sin A \cos\beta\tag{12} $$
Bottom Line
We have derived a set of handy equations necessary for calculating the primary directions. Here is the recap of what we got
