Let [alpha], [alpha]' be the mutual azimuths of two points A, B on a spheroid, k the chord line joining them, [mu], [mu]' the angles made by the chord with the normals at A and B, [phi], [phi]', [omega] their latitudes and difference of longitude, and (x^2 + y^2)/a^2 + z^2 b^2 = 1 the equation of the surface; then if the plane xz passes through A the co-ordinates of A and B will be x = (a/[Delta]) cos [phi], x' = (a/[Delta]') cos [phi]' cos [omega], y = 0 y' = (a/[Delta]') cos [phi]' sin [omega], z = (a/[Delta]) (1 - e^2) sin [phi], z' = (a/[Delta]') (1 - e^2) sin [phi]', where [Delta] = (1 - e^2 sin^2 [phi])^1/2, [Delta]' = (1 - e^2 sin^2 [phi]')^1/2, and e is the eccentricity. Let f, g, h be the direction cosines of the normal to that plane which contains the normal at A and the point B, and whose inclinations to the meridian plane of A is = [alpha]; let also l, m, n and l', m', n' be the direction cosines of the normal at A, and of the tangent to the surface at A which lies in the plane passing through B, then since the first line is perpendicular to each of the other two and to the chord k, whose direction cosines are proportional to x' - x, y' - y, z' - z, we have these three equations f(x' - x) + gy' + h(z' - z) = 0 fl + gm + hn = 0 fl' + gm' + hn' = 0. Eliminate f, g, h from these equations, and substitute l = cos [phi] l' = - sin [phi] cos [alpha] m = 0 m' = sin [alpha] n = sin [phi] n' = cos [phi] cos [alpha], and we get (x' - x) sin [phi] + y' cot [alpha] - (z' - z) cos [phi] = 0. The substitution of the values of x, z, x', y', z' in this equation will give immediately the value of cot [alpha]; and if we put [zeta], [zeta]' for the corresponding azimuths on a sphere, or on the supposition e = 0, the following relations exist cos [phi] Q cot [alpha] - cot [zeta] = e^2 ------------------ cos [phi]' [Delta] cos [phi]' Q cot [alpha]' - cot [zeta]' = e^2 ------------------ cos [phi] [Delta]' [Delta]' sin [phi] - [Delta] sin [phi]' = Q sin [omega]. If from B we let fall a perpendicular on the meridian plane of A, and from A let fall a perpendicular on the meridian plane of B, then the following equations become geometrically evident: k sin [mu] sin [alpha] = (a/[Delta]') cos [phi]' sin [omega] k sin [mu]' sin [alpha]' = (a/[Delta]) cos [phi] sin [omega]. Now in any surface u = 0 we have