Page:General Investigations of Curved Surfaces, by Carl Friedrich Gauss, translated into English by Adam Miller Hiltebeitel and James Caddall Morehead.djvu/18

III. The angle between two planes is equal to the spherical angle between the great circles representing them, and, consequently, is also measured by the arc intercepted between the poles of these great circles. And, in like manner, the angle of inclination of a straight line to a plane is measured by the arc drawn from the point which corresponds to the direction of the line, perpendicular to the great circle which represents the orientation of the plane.

IV. Letting ${\textstyle x,}$ ${\textstyle y,}$ ${\textstyle z;}$ ${\textstyle x',}$ ${\textstyle y',}$ ${\textstyle z'}$ denote the coordinates of two points, ${\textstyle r}$ the distance between them, and ${\textstyle L}$ the point on the sphere which represents the direction of the line drawn from the first point to the second, we shall have

$${\begin{aligned}x'&=x+r\cos(1)L\\y'&=y+r\cos(2)L\\z'&=z+r\cos(3)L\end{aligned}}$$

V. From this it follows at once that, generally,

$$\cos ^{2}(1)L+\cos ^{2}(2)L+\cos ^{2}(3)L=1$$

and also, if ${\textstyle L'}$ denote any other point on the sphere,

$$\cos(1)L.\cos(1)L'+\cos(2)L.\cos(2)L'+\cos(3)L.\cos(3)L'=\cos LL'.$$

VI. Theorem. If ${\textstyle L,}$ ${\textstyle L',}$ ${\textstyle L'',}$ ${\textstyle L'''}$ denote four points on the sphere, and ${\textstyle A}$ the angle which the arcs ${\textstyle LL',}$ ${\textstyle L''L'''}$ make at their point of intersection, then we shall have

$$\cos LL''.\cos L'L'''-\cos LL'''.\cos L'L''=\sin LL'.\sin L''L'''.\cos A$$

Demonstration. Let ${\textstyle A}$ denote also the point of intersection itself, and set

$$AL=t,\quad AL'=t',\quad AL''=t'',\quad AL'''=t'''.$$

Then we shall have

$${\begin{alignedat}{3}&\cos LL''&&=\cos t.&&\cos t''&&+\sin t&&\sin t''&&\cos A\\&\cos L'L'''&&=\cos t'&&\cos t'''&&+\sin t'&&\sin t'''&&\cos A\\&\cos LL'''&&=\cos t&&\cos t'''&&+\sin t&&\sin t'''&&\cos A\\&\cos L'L''&&=\cos t'&&\cos t''&&+\sin t'&&\sin t''&&\cos A\end{alignedat}}$$

and consequently,

$${\begin{aligned}&\cos LL''.\cos L'L'''-\cos LL'''.\cos L'L''\\&\quad {\begin{aligned}&=\cos A(\cos t\cos t''\sin t'\sin t'''+\cos t'\cos t'''\sin t\sin t''\\&\qquad -\cos t\cos t'''\sin t'\sin t''-\cos t'\cos t''\sin t\sin t''')\\&=\cos A(\cos t\sin t'-\sin t\cos t')(\cos t''\sin t'''-\sin t''\cos t''')\\&=\cos A.\sin(t'-t).\sin(t'''-t'')\\&=\cos A.\sin LL'.\sin L''L'''\end{aligned}}\end{aligned}}$$