190 Bond, Graham, and Peirce, on the Latitude of Cambridge. 
The triangles 4P H and BPH give 
sin. dt = sin. D cos. L — cos. D sin. L cos. h, 
= — sin. D cos. L + cos. Dain. L cos. h, 5 
whence 
tan. D cot. L = i (cos, h, + cos. h,). 
But the right triangle P D Z gives, by putting 
is ID IP YZ, 
eos: h = tan. D cot. L. 
Hence 
cos. h = 4 (cos. h,, + cos. h,) 
= cos. 3 (h, +h,) cos. 3 (h, — h,}- 
cos. hk 
cos..3 (h, + h,) 
1—cos.4(h,—h,) cos. 3 (h, + h,) — cos. h 
1+ cos. 4 (h,—h,) ~ cos. 4 (h, + h,) + cos. h 
tan.? } (h, —h,) = tan. } [h — 3 (h, +h,)] tan. 4 [h + 3 (h, + h,)]. 
But h, —h, is very small, and A differs very little from 2 (h, +h.) 
whence 
cos. } (h, —h,) = 
h—4(h, +h.) = # (hk, —f,)? tan. 1’ cot. hk 5 
and the second member of this equation, which may be denoted 
by oh, is a correction which must be added to } (h, +h,) to ob- 
tain h. And in the same way, the corrections for reducing h,, and 
h, to h are obtained by the following formule : — 
h, —h= 4 (hk, —h,) — oh, 
hk —h, =} (h,—Ah,) + 5h. 
These corrections need be computed but once; and the compu- 
tation of them for « Lyra, from the mean of all the observations on 
those days on which the east and west transits were both observed, 
is contained in Table VII. The corrections of A for changes in 
the declination of the point Z, that is, for changes in the level, are 
obtained from the differential of the above value of cos. h, which is, 
if Dh and DL denote these corresponding changes, 
tan. D 
= DL. 
sin.” L 
sin. hh. Dh = 
