Trigonometrical Survey. 523 
also the difference of longitude being the same on both fi- 
gures. We therefore shall have recourse to that determina- 
tion, and apply it to the present question. 
The co-latitudes of D and B, or the arches D P and B P, 
are 39 0 22' 52 ",69, and 3 9 0 15' 36", 29, therefore half their sum 
is 39 0 19' 14//, 49, and half their difference 3' 38", 2. 
Plalf the sum of the angles PDB and PBD is 89° 2 6' 25", 5 ; 
therefore, as tang. 39 0 19' 14", 49 : tang. 3' 38",2 : : tang. 
8 9 0 2 6' 2 5", 5 : tang. 7 0 31' or half the difference of 
the angles : hence the angles for computation are 8i° 54' 27",79, 
and 96° 58' 23",2i, which, with the co-latitudes of D and B, 
give the difference of longitude between Beachy Head and 
Dunnose, or the angle DPB = i° 26' 47", 93. 
We have now two right angled triangles, which may be 
considered spherical, namely, PBW, and PDR, in which the 
angle at the pole P is given, and likewise the sides P B and 
PD ; therefore, using these data , we find the arc BW = 
54' 56", 31, and the arc DR = 55' -4", 7 4. 
The chords of the two perpendicular arcs are about 3^ feet 
less than the arcs themselves; therefore BW = 336*119,1 
feet, and DR = 33698355 feet ; and by proportioning these 
arcs to their respective values in fathoms, we get the length 
of the degree of the great circle perpendicular to the meridian 
in the middle point between W and B == 61182,8 fathoms, and 
in the middle point between R and D = 61181,8 fathoms. 
Therefore 61182,3 fathoms is the length of a degree of the 
great circle perpendicular to the meridian, in latitude 50° 41', 
which is nearly that of the middle point between Beachy Head 
and Dunnose. 
If the horizontal angles, or the directions of the meridians. 
