Page:General Investigations of Curved Surfaces, by Carl Friedrich Gauss, translated into English by Adam Miller Hiltebeitel and James Caddall Morehead.djvu/19

 But as there are for each great circle two branches going out from the point $A,$ these two branches form at this point two angles whose sum is $180^{\circ}.$  But our analysis shows that those branches are to be taken whose directions are in the sense from the point $L$  to $L',$  and from the point $L$  to $L';$  and since great circles intersect in two points, it is clear that either of the two points can be chosen arbitrarily. Also, instead of the angle $A,$ we can take the arc between the poles of the great circles of which the arcs $LL',$  $LL'$  are parts. But it is evident that those poles are to be chosen which are similarly placed with respect to these arcs; that is to say, when we go from $L$  to $L'$ and from $L$  to $L',$  both of the two poles are to be on the right, or both on the left.

VII. Let $L,$  $L',$  $L''$ be the three points on the sphere and set, for brevity,

$$\begin{alignedat}{3} &\cos (1)L  &&= x,\quad&& \cos (2)L   &&= y,\quad&& \cos (3)L   &&= z \\ &\cos (1)L' &&= x',    && \cos (2)L'  &&= y',    && \cos (3)L'  &&= z' \\ &\cos (1)L &&= x,  && \cos (2)L &&= y,   && \cos (3)L &&= z \\ \end{alignedat}$$ and also

$$x y' z + x' y z + x y z' - x y z' - x' y z - x y' z = \Delta$$

Let $\lambda$  denote the pole of the great circle of which $LL'$  is a part, this pole being the one that is placed in the same position with respect to this arc as the point $(1)$ is with respect to the arc $(2)(3).$  Then we shall have, by the preceding theorem,

$$y z' - y' z = \cos (1)\lambda. \sin (2)(3). \sin LL',$$

or, because $(2)(3) = 90^{\circ},$

$$\begin{aligned} y z' - y' z &= \cos (1)\lambda. \sin LL', \end{aligned} $$

and similarly,

$$ \begin{aligned} z x' - z' x &= \cos (2)\lambda. \sin LL' \\ x y' - x' y &= \cos (3)\lambda. \sin LL' \end{aligned} $$

Multiplying these equations by $x,$  $y,$  $z''$ respectively, and adding, we obtain, by means of the second of the theorems deduced in V,

$$\Delta = \cos \lambda L''. \sin LL'$$

Now there are three cases to be distinguished. First, when $L$  lies on the great circle of which the arc $LL'$ is a part, we shall have $\lambda L = 90^{\circ},$  and consequently, $\Delta = 0.$  If $L$  does not lie on that great circle, the second case will be when $L$  is on the same side as $\lambda;$  the third case when they are on opposite sides. In the last two cases the points $L,$  $L',$  $L''$ will form a spherical triangle, and in the second case these points will lie in the same order as the points $(1),$  $(2),$  $(3),$  and in the opposite order in the third case.