so
CALCULATIONS TO ASCERTAIN THE
TRACT 26.
16. Depressions
below p
in the
S.E.
quarter.
Rings
1
2
3
4
5
6
7
8
9
10
11
12
.Radii.
7
80
4333
8
20
30
30
210
5000
9
26'0
290
290
280
240
150
30
•
.
.
270
5667
10
420
440
450
440
420
3/0
270
140
•
330
6333
11 :
530
540
560
560
550
480
430
330
150
40
40
430
7000
12 ;
500
510
520
550
630
600
500
430
29O
230
200
630
7667
13
450
430
420
410
430
570
630
530
430
480
340
710
8333
14
360
330
310
290
280
330
510
670
590
630
570
830
9000
15
240
230
220
200
180
200
S30
530
770
760
710
S70
9667
16
180
160
150
130
110
140
230
330
630
830
790
880
10333
1/
110
80
50
40
30
90
190
280
500
860
830
860
11000
18
10
•
'•
10
150
260
400
760
830
760
11667
19
.
•
•
.
t
70
230
330
600
770
630
12333
20
•
•
•
•
•
•
10
ISO
290
530
69O
530
13000
It remains next to find the sines of the vertical angles, sub-tended by all the foregoing altitudes and depressions ; sincethe sum of these sines is the thing we are in quest of. Now,each altitude, or depression, is the perpendicular of a right-angled triangle, of which the given radius, standing on thesame line with it, in the right-hand margin, is the base, orthe other side about the right angle; and by the resolutionof the right-angled triangle, for each perpendicular, the samenumber of corresponding sines will be found. But with suchdata, the tangent of the angle is much easier to be found,than the sine, and the analogy for that purpose is this, as thebase : to the perpendicular : : radius 1 : the tangent, whichwill therefore be found, by barely dividing the given perpen-dicular by the base ; and if we find this number in its propercolumn, in a table of sines and tangents, then on the sameline with it, in the column of sines, will be found the sine ofthe angle required. This seems to be the easiest way of re-solving all the triangles, when computed separately. Butasthe labour would be very great, in performing so many hun-dreds of arithmetical divisions, &c, either by logarithms, or
■c* .