Page:General Investigations of Curved Surfaces, by Carl Friedrich Gauss, translated into English by Adam Miller Hiltebeitel and James Caddall Morehead.djvu/57

Thus to the angles ${\textstyle A,}$ ${\textstyle B,}$ ${\textstyle C}$ on a non-spherical surface, unequal reductions must be applied, so that the sines of the changed angles become proportional to the sides opposite. The inequality, generally speaking, will be of the third order; but if the surface differs little from a sphere, the inequality will be of a higher order. Even in the greatest triangles on the earth’s surface, whose angles it is possible to measure, the difference can always be regarded as insensible. Thus, e.g., in the greatest of the triangles which we have measured in recent years, namely, that between the points Hohehagen, Brocken, Inselberg, where the excess of the sum of the angles was ${\textstyle 14''.85348,}$ the calculation gave the following reductions to be applied to the angles:

Hohehagen ${\textstyle \quad .\quad .\quad .\quad .\quad .\quad .-4''{.}95113}$
Brocken ${\textstyle \quad .\quad .\quad .\quad .\quad .\quad .-4''{.}95104}$
Inselberg ${\textstyle \quad .\quad .\quad .\quad .\quad .\quad .-4''{.}95131}$

29.

We shall conclude this study by comparing the area of a triangle on a curved surface with the area of the rectilinear triangle whose sides are ${\textstyle a,}$ ${\textstyle b,}$ ${\textstyle c.}$ We shall denote the area of the latter by ${\textstyle \sigma ^{*};}$ hence

$$\sigma ^{*}={\tfrac {1}{2}}bc\sin A^{*}={\tfrac {1}{2}}ac\sin B^{*}={\tfrac {1}{2}}ab\sin C^{*}$$

We have, to quantities of the fourth order,

$$\sin A^{*}=\sin A-{\tfrac {1}{12}}\sigma \cos A.(2\alpha +\beta +\gamma )$$

or, with equal exactness,

$$\sin A=\sin A^{*}.{\bigl (}1+{\tfrac {1}{24}}bc\cos A.(2\alpha +\beta +\gamma ){\bigr )}$$

Substituting this value in formula [9], we shall have, to quantities of the sixth order,

$${\begin{aligned}\sigma ={\tfrac {1}{2}}bc\sin A^{*}.{\bigl (}1&+{\tfrac {1}{120}}\alpha (3b^{2}+3c^{2}-2bc\cos A)\\&+{\tfrac {1}{120}}\beta (3b^{2}+4c^{2}-4bc\cos A)\\&+{\tfrac {1}{120}}\gamma (4b^{2}+3c^{2}-4bc\cos A){\bigr )},\end{aligned}}$$

or, with equal exactness,

$$\sigma =\sigma ^{*}{\bigl (}1+{\tfrac {1}{120}}\alpha (a^{2}+2b^{2}+2c^{2})+{\tfrac {1}{120}}\beta (2a^{2}+b^{2}+2c^{2})+{\tfrac {1}{120}}\gamma (2a^{2}+2b^{2}+c^{2})$$

For the sphere this formula goes over into the following form:

$$\sigma =\sigma ^{*}{\bigl (}1+{\tfrac {1}{24}}\alpha (a^{2}+b^{2}+c^{2}){\bigr )}.$$