Spherical Triangles Equations

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 $\alpha$, $\beta$, 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 $\text{XYZ}$ coordinate system.

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

$$\vec{v}=\left(\begin{array}{rl}& x=R\cos \beta \\ & y=0\\ & z=R\sin \beta \end{array}\right)$$

Now, let’s use the rotation matrix to rotate $\text{XYZ}$ by angle $\alpha$ along $\text{XY}$ and express this vector in ${\text{X}}^{\prime }{\text{Y}}^{\prime }{\text{Z}}^{\prime }$ axis. As a result, we have the following:

$$\begin{array}{rl}\vec{v}& =R\cos \alpha \cos \beta {\text{X}}^{\prime }\\ & -R\sin \alpha \cos \beta {\text{Y}}^{\prime }\\ & +R\sin \beta {\text{Z}}^{\prime }\\ & \end{array}$$

But on the other hand, the same $x^{\prime }$ component of the vector $\vec{v}$ is just a projection of that vector to the ${\text{X}}^{\prime }$-axis in the $\gamma$-angle plane.

$$x^{\prime }=R\cos \gamma$$

It gives us the first equation:

$$\cos \gamma =\cos \alpha \cos \beta \text{\hspace{1em}(1)}$$

Now we move to the next step and introduce a new coordinate system ${\text{X}}^{\prime \prime }{\text{Y}}^{\prime \prime }{\text{Z}}^{\prime \prime }$, which is ${\text{X}}^{\prime }{\text{Y}}^{\prime }{\text{Z}}^{\prime }$, rotated by angle $-B$ (minus B) in ${\text{Y}}^{\prime }{\text{Z}}^{\prime }$ plane.

If we apply rotation matrix to ${\text{X}}^{\prime }{\text{Y}}^{\prime }{\text{Z}}^{\prime }$ coordinate system, we will have a $z^{\prime \prime }$-component of the vector $\vec{v}$ to be equal to

$$\begin{array}{rl}{z}^{\prime \prime }& =\sin B{y}^{\prime }+\cos B{z}^{\prime }=\\ & -R\sin B\sin \alpha \cos \beta \\ & +R\cos B\sin \beta \end{array}$$

On the other hand, $z^{\prime \prime }$-component of vector $\vec{v}$ is equal to zero. It means that

$$\sin \alpha =\frac{\tan \beta }{\tan B}\text{\hspace{1em}(2)}$$

The exact ratio applies to $\beta$ angle:

$$\sin \beta =\frac{\tan \alpha }{\tan A}\text{\hspace{1em}(3)}$$

Other Equations

We have set the main equations for spherical triangles:

$$\{\begin{array}{l}\cos \gamma =\cos \alpha \cos \beta \\ \sin \alpha =\tan \beta /\tan B\\ \sin \beta =\tan \alpha /\tan A\end{array}$$

All the rest is just a consequence of these three equations.

First, let’s multiply $(1)$ by $(2)$, and we will get

$$\cos \gamma =\frac{1}{\tan A\tan B}\text{\hspace{1em}(4)}$$

Now, let’s consider ${\sin }^2\gamma$:

$$\begin{array}{rl}{\sin }^2\gamma & =1-{\cos }^2\gamma \\ & ={\cos }^2\alpha +{\sin }^2\alpha -{\cos }^2\alpha {\cos }^2\beta \\ & ={\sin }^2\alpha +{\cos }^2\alpha {\sin }^2\beta \end{array}$$

Similarly, we can write

$${\sin }^2\gamma ={\sin }^2\beta +{\cos }^2\alpha {\sin }^2\beta \text{\hspace{1em}(5)}$$

Now, let’s divide both sides of the equation by ${\cos }^2(\gamma )$:

$$\begin{array}{rl}{\tan }^2\gamma & ={\tan }^2\alpha \frac{1}{{\cos }^2\beta }+{\tan }^2\beta \\ & ={\tan }^2\beta {\tan }^{2}A+{\sin }^2\alpha {\tan }^{2}B\\ & ={\sin }^2\alpha {\tan }^{2}B\left[{\tan }^{2}A+1\right]\\ & =\frac{\sin \alpha \tan B}{\cos A}\end{array}$$

which gives

$$\sin \gamma =\frac{\sin \alpha }{\sin A}\text{\hspace{1em}(6)}$$

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

$$\sin \gamma =\frac{\sin \beta }{\sin B}\text{\hspace{1em}(7)}$$

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

$$\cos A=\frac{\cos \alpha \sin \beta }{\sin \gamma }\text{\hspace{1em}(8)}$$

or

$$\cos A=\frac{\tan \beta }{\tan \gamma }\text{\hspace{1em}(9)}$$

In a similar manner from (2) follows that

$$\cos B=\frac{\tan \alpha }{\tan \gamma }\text{\hspace{1em}(10)}$$

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

$$\cos A=\sin B\cos \alpha \text{\hspace{1em}(11)}$$

Similarly,

$$\cos B=\sin A\cos \beta \text{\hspace{1em}(12)}$$

Bottom Line

We have derived a set of handy equations necessary for calculating the primary directions. Here is the recap of what we got

$$\begin{array}{cc}\gamma ,\alpha ,\beta :& \cos \gamma =\cos \alpha \cos \beta \\ \alpha ,\beta ,B:& \tan \beta =\sin \alpha \tan B\\ \beta ,\alpha ,A:& \tan \alpha =\sin \beta \tan A\\ \gamma ,A,B:& \cos \gamma =\frac{1}{\tan A\tan B}\\ \gamma ,\alpha ,A:& \sin \alpha =\sin \gamma \sin A\\ \gamma ,\beta ,B:& \sin \beta =\sin \gamma \sin B\\ A,\beta ,\gamma :& \tan \beta =\cos A\tan \gamma \\ B,\alpha ,\gamma :& \tan \alpha =\cos B\tan \gamma \\ A,B,\alpha :& \cos A=\sin B\cos \alpha \\ B,A,\beta :& \cos B=\sin A\cos \beta \end{array}\text{\hspace{1em}(1)(2)(3)(4)(6)(7)(9)(10)(11)(12)}$$