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.

Consider a right-angled triangle on a sphere with sides α\alpha, β\beta, and γ\gamma, as well as interior angles AA, BB, and CC.

How will its sides and angles be related to each other?

A B α γ β

Rotated Coordinate Systems

First, we introduce a cartesian vector v⃗\vec{v} in XYZ\textbf{XYZ} coordinate system.

B α γ β X Z v X' Y'

From the conversion equation (1) it follows that v⃗\vec{v} has coordinates

v⃗=(x=Rcos⁡β y=0 z=Rsin⁡β)\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\textbf{XYZ} by angle α\alpha along XY\textbf{XY} and express this vector in X′Y′Z′\textbf X' \textbf Y' \textbf Z' axis. As a result, we have the following:

v⃗=Rcos⁡αcos⁡β X′ −Rsin⁡αcos⁡β Y′ +Rsin⁡β Z′ \begin{aligned} \vec{v} & = R \cos\alpha \cos\beta~ \textbf X' \\\ & - R \sin\alpha \cos\beta~ \textbf Y' \\\ & + R\sin\beta~ \textbf Z' \\\ \end{aligned}

But on the other hand, the same x′x' component of the vector v⃗\vec{v} is just a projection of that vector to the X′\textbf X'-axis in the γ\gamma-angle plane.

b α γ β X Z v X' Y'
x′=Rcos⁡γx' = R\cos{\gamma}

It gives us the first equation:

cos⁡γ=cos⁡αcos⁡β(1)\cos\gamma = \cos\alpha\cos\beta\tag{1}

Now we move to the next step and introduce a new coordinate system X′′Y′′Z′′\textbf X'' \textbf Y'' \textbf Z'', which is X′Y′Z′\textbf X' \textbf Y' \textbf Z', rotated by angle −B-B (minus B) in Y′Z′\textbf Y' \textbf Z' plane.

B α β Z v X'' Y' Z'' Y''

If we apply rotation matrix to X′Y′Z′\textbf X' \textbf Y' \textbf Z' coordinate system, we will have a z′′z''-component of the vector v⃗\vec{v} to be equal to

z′′=sin⁡B y′+cos⁡B z′= −Rsin⁡Bsin⁡αcos⁡β +Rcos⁡Bsin⁡β\begin{aligned} z'' & = \sin B~ y' + \cos B~ z' = \\\ & - R \sin B \sin\alpha \cos\beta \\\ & + R \cos B \sin \beta \end{aligned}

On the other hand, z′′z''-component of vector v⃗\vec{v} is equal to zero. It means that

sin⁡α=tan⁡βtan⁡B(2)\sin\alpha = \frac{\tan\beta}{\tan B} \tag{2}

The exact ratio applies to β\beta angle:

B α β A γ B α β A γ
sin⁡β=tan⁡αtan⁡A(3)\sin \beta = \frac{\tan \alpha} {\tan A }\tag{3}

Other Equations

We have set the main equations for spherical triangles:

{cos⁡γ=cos⁡αcos⁡β sin⁡α=tan⁡β/tan⁡B sin⁡β=tan⁡α/tan⁡A\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)(1) by (2)(2), and we will get

cos⁡γ=1tan⁡Atan⁡B(4)\cos \gamma = \frac{1} {\tan A\tan B} \tag{4}

Now, let’s consider sin⁡2γ\sin^2 \gamma:

sin⁡2γ=1−cos⁡2γ =cos⁡2α+sin⁡2α−cos⁡2αcos⁡2β =sin⁡2α+cos⁡2αsin⁡2β\begin{aligned} \sin^2 \gamma & = 1 - \cos^2 \gamma \\\ & = \cos^2\alpha + \sin^2\alpha - \cos^2\alpha \cos^2\beta \\\ & = \sin^2\alpha + \cos^2\alpha \sin^2\beta \end{aligned}

Similarly, we can write

sin⁡2γ=sin⁡2β+cos⁡2αsin⁡2β(5)\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(γ)\cos^2(\gamma):

tan⁡2γ=tan⁡2α1cos⁡2β+tan⁡2β =tan⁡2βtan⁡2A+sin⁡2αtan⁡2B =sin⁡2αtan⁡2B[tan⁡2A+1] =sin⁡αtan⁡Bcos⁡A\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⁡γ=sin⁡αsin⁡A(6)\sin\gamma = \frac{\sin\alpha} {\sin A}\tag{6}

With the same approach, we get from the equation (5)(5)

sin⁡γ=sin⁡βsin⁡B(7)\sin\gamma = \frac{\sin\beta} {\sin B}\tag{7}

This equality

sin⁡αsin⁡A=sin⁡βsin⁡B\frac{\sin\alpha} {\sin A} = \frac{\sin\beta} {\sin B}

is also called the sine theorem.

Now from (3)(3) it follows

sin⁡A=tan⁡αsin⁡βcos⁡A\sin A = \frac{\tan\alpha} {\sin\beta}\cos A

If we sunstitude sin⁡A\sin A from (6)(6) we will get

cos⁡A=cos⁡αsin⁡βsin⁡γ(8)\cos A = \frac{\cos\alpha\sin\beta}{\sin\gamma}\tag{8}

or

cos⁡A=tan⁡βtan⁡γ(9)\cos A = \frac{\tan\beta} {\tan\gamma}\tag{9}

In a similar manner from (2) follows that

cos⁡B=tan⁡αtan⁡γ(10)\cos B = \frac{\tan\alpha} {\tan\gamma}\tag{10}

If we substitute (8) with (7), we will get

cos⁡A=sin⁡Bcos⁡α(11)\cos A = \sin B \cos\alpha\tag{11}

Similarly,

cos⁡B=sin⁡Acos⁡β(12)\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

B α β A γ B α β A γ

Spherical triangle.

γ,α,β:cos⁡γ=cos⁡αcos⁡β α,β,B:tan⁡β=sin⁡αtan⁡B β,α,A:tan⁡α=sin⁡βtan⁡A γ,A,B:cos⁡γ=1tan⁡Atan⁡B γ,α,A:sin⁡α=sin⁡γsin⁡A γ,β,B:sin⁡β=sin⁡γsin⁡B A,β,γ:tan⁡β=cos⁡Atan⁡γ B,α,γ:tan⁡α=cos⁡Btan⁡γ A,B,α:cos⁡A=sin⁡Bcos⁡α B,A,β:cos⁡B=sin⁡Acos⁡β\begin{gather*} \gamma, \alpha, \beta: & \cos\gamma = \cos\alpha\cos\beta\tag{1}\\\ \alpha, \beta, B: & \tan\beta = \sin\alpha \tan B\tag{2}\\\ \beta, \alpha, A: & \tan \alpha = \sin \beta \tan A \tag{3}\\\ \gamma, A, B: & \cos \gamma = \frac{1} {\tan A\tan B} \tag{4}\\\ \gamma, \alpha, A: & \sin\alpha = \sin\gamma \sin A\tag{6}\\\ \gamma, \beta, B: & \sin\beta = \sin\gamma \sin B\tag{7}\\\ A, \beta, \gamma: & \tan\beta = \cos A \tan\gamma\tag{9}\\\ B, \alpha, \gamma: & \tan\alpha = \cos B \tan\gamma\tag{10}\\\ A, B, \alpha: & \cos A = \sin B \cos\alpha\tag{11}\\\ B, A, \beta: & \cos B = \sin A \cos\beta\tag{12} \end{gather*}