82 On finding the Latitude. 
which adds to the length and perplexity of the operations, before 
too complicated. In the Quarterly Journal of Science, No. 22, 
Dr. Brinkley has computed an example making allowance for the 
change of declination ; and if a fair comparison be instituted be- 
tween his calculation and that by the direct method, as applied 
to the same example in No. 21 of the same Journal, it appears 
hardly possible to avoid giving the preference in every respect to 
the latter. 
I have now to propose a direct solution of the problem, sim- 
pler and easier in the calculation than that recommended by 
Delambre, and which, I conceive, will be found more convenient 
in practice than the indirect process commonly used. In ex- 
plaining this solution I shall first assume that the sun’s declina- 
tion is the same at both the observations and equal to the mean 
quantity between the two times; and I shall afterwards point 
out an easy way of correcting the error which this assumption 
introduces in the result. 
The principles of the method are contained in this preliminary 
proposition. | 
Lemma. (See figure 3, Plate II.) Let the base AB of a 
spherical triangle A ZB, be bisected in O, and through O draw 
a great circle perpendicular to AB; then having let fall upon 
this circle the perpendicular Z D from the vertex of the triangle, 
we shall have these two formula, viz. 
4 cos ZB—cos ZA 
SinZD = — =a? 
Cos ZB + cosZA 
Cos DO = Fn AO ce ED 
Conceive a great circle to pass through the points Z and O: 
_ then cos ZOB = — cos ZOA= sin ZOD. Now, from the two 
spherical triangles ZOB and ZOA, we get, 
Cos ZB = cos BO cosZO + sin BO sin ZO sin ZOD, 
Cos ZA = cos AO cos ZO —sin AO sin ZO sin ZOD; 
wherefore, by subtracting, r 
Cos ZB — cos ZA= 2 sin AO sin ZO sin ZOD; 
but, in the right-angled triangle ZOD, sin ZD = sin ZO sin 
ZOD; consequently, 
_ CosZB—cos ZA = 2sinAO sinZD, 
from which the first of the two formule is derived. 
Agaiu, by adding the same two equations, we obtain, 
Cos ZB + cos ZA = 2cos AO cos ZO; 
but, in the triangle ZOD, cos ZO = cos ZD x cos OD; 
wherefore, 
Cos ZB + cosZA = 2 cos AO cos ZD cosDO, 
from which the second formula is deduced. - Now 
