Having added the figures in four of the columns, we find the sum of the northings 268.21, the sum
of the southings 141.32, and their difference=126.89-2° 6'.89 the difference of latitude.
The sum
of the eastings=289.9, the sum of the westings=1634.6, and their differences=1344.7=22° 24'.8=
the difference of longitude. By Mercator's sailing, (Art. 72.) Merid. Dif. Lat. : Dif. Lon.::R: tan
Course; that is, 686.5: 1344.8::1: tan Course; and, by the tables, 2.8366405: 3.1286577::10.000
0000: 10.2920172-62° 19'+the direct course. Again, by example 3d, Art. 74, R: Prop. Dif.
Lat.::sec Course: Dist.; that is, 1 : 126.89::sec 62° 59′+ : Dist. ; and, by the tables, 10.0000000 :
2.1034274::10.3427050 : 2.1044632=279.34 the distance. Therefore,

The difference of latitude-2° 6' .89
The difference of longitude=22° 24′ .8

The direct course 62° 19′+
The distance=279.34

OBLIQUE SAILING.

Example 1. See Figure 27. By the conditions given, the angle ABC=56° 15', CAB=45°, and ACB=78° 45'. By Trig. 143, sin C: AB: :sin B: AC; that is, sin 78° 45′ : 9::sin 50° 15'. AC; and, by the tables, 9.991574 : 0.954 243: 9.919846 0.882515=7.63 miles.

Example 2. See Figure 27.

CAN+BAS=76° 16′ + 35° 10′=111° 26′, which subtracted from 180°, leaves 68° 34'-BAC. The angle CBN'=17° 13, and this has the same ratio BAS, that it has to ABN'=35° 10′, and this added to 17° 13′ 52° 23′=CBA. Therefore, (180° – CAB +CBA)=180°-(69° 34′+52° 23′)=59° 03'=ACB. By Trig. 143, sin ACB: AB::sin CBÁ: AC; that is, 59° 03′ : 8::sin 52° 23′: AC; and, by the tables, 9.933293 : 0.9 03090: 9.898787: 0.685836 = 7.3889 miles. Again, by Plane Sailing, R: Dist.::cos Course: Dif. Lat.; that is, 1 : 7.3889: cos 76° 16': Dif. Lat.; and, by the tables, 10.00 0000 0.868584: 9.375487: 0.244070 1.754, which subtracted from 40° 28', leaves 40° 26′.246=the latitude of the first observation. By Mid. Lat. Sailing, cos Mid. Lat. Dif. Lat.::tan. Course: Dif. Lon. By Art. 57, Mid. Lat. 40° 27'.1; Therefore, 40° 27'.1 : 1.754::tan 76° 16′ : Dif. Lon.; and, by the tables, 9.881369: 0.240704::10. 611916 0.974920-9'.4589 which subtracted from 74° 8′, leaves 73° 58′ W=the longitude at the first observation.

=

Example 3. The angle BAC CAS+BAN 180° 51° +45° 180° 84°. By Trig. 143, 10 : sin CAB::8 : sin CBA; that is, 10: sin 84°::8: sin CBA; and, by the tables, 10.000000 : 9.997614 :: 0.903090: 9.900704 43' CBA. Now the angle CBS' the course=CBA— ABS=(Euc. I. 29.) NAB; therefore, the course=52° 43′ -45°-7° 43'.

52°

Example 4. BAS=38°, BAN=142°, BAC = 142° 21°=121°, DAB=142°—47°=95°, CBA = 38°—5° = 33°, DAB=38° + 13°=51°, BCA=121°+33° ☛ 180° = 26°, BDA=95°+51° ☛ 180°=34°, and CAD=47°—21° =26. By Trig. 143, sin BCA : R:: sin CBA : AC ; that is, sin 26° : 1:¦sin 33° : AC; and, by the tables, 9.64 1842: 10.000000 :: 9.736109: 1.094267=12.42. Again,

sin BDA: R::sin DAB: AD; that is, sin 34° : 1: :sin 51° : AD; and, by the tables, 9.747562: 10.000000::9.8905 03: 1.142941=13.89=AD. Now ACD+ADC=180°26° 154°; hence by Trig. 144, AC+AD : AC-AD:: tan (ACD + ADC): tan (ACD-ADC); that is, 26.31 : 1.47::tan 77° : tan (ACD-ADC); and, by the tables, 1.420121 0.167317::10.636636: 9.383832=13° 36′ 16′′. By Alg. 341, 77°+13° 36′ 16" 90° 36′ 16′′=ACD=the greater angle. By Trig. 143, sin ACD: AD::sin CAD : DC; that is, sin 90° 36′ 16′′ : 13.89: :sin 26° : DC; and, by the tables, 9.999976 : 1.142741::9.641842 : 0.784807 =6.093 miles 6 miles 29.76 rods.

CURRENT SAILING.

Example 2. Art. 88. Let C (Fig. 28) be the point which the boat is endeavoring to make; then CAO-22° 30'. Let AB=the current, then BAO=78° 45'. BAC=101° 15', CB=AD=5, and AB=3. By Trig. 143, CB : sin BAC ::AB: sin ACB; that is, 5 : sin 101° 15′::3 : sin BAC; and, by the tables, 0.698970: 9.991574::0.477121 : 9.7697 15=36° 03'. Now by Euc. I. 29, DAC÷CAO=ACB= 58° 33′ W=the course.

36° 03'+22° 30′

Example 1. Art. 89. By the Traverse Table, Dif. Lat. =39 S., and the departure 15.61 W. By Plane Sailing, Dif. Lat. R::Depart. : tan Course; that is, 39:1::15.61 : tan Course; and, by the tables, 1.591065: 10.000000:: 1.1934039.602338 : = 21° 48′ 58": the course. Again, R:Dif. Lat.::sec Course: Dist.; that is, 1:39:: sec 21° 48′ 58′′ Dist.; and, by the tables, 10.000000 : 1.591065:: 10.032754 1.623819-42.05=the distance.

Example 2. By the Traverse Table, Dif. Lat.=10.31 S, and Depart. 23.38 E. By Plane Sailing, Dif. Lat. : R:: Depart. tan Course; that is, 10.31:1::23.38 : tan Course; and, by the tables, 1.013259: 10.000000::1.368844 : 10.3 55585 = 66° 13' = the course. Again, R: Dif. Lat.::sec Course Dist.; that is, 1: 10.31::sec 66° 13′: Dist.; and, by the tables, 10.000000 : 1.013259::10.394394: 1.407653 =25.56 the distance.

3

E

SURVEYING.

Example 2. Art. 114. By the given courses the angle A=90°, the angle B =90° + 10° 30′ = 100° 30', the angle C=41°-10° 30′ 31° 30', the angle D=42° +58°=100°, and the angle E-58°: Hence by Trig. 153, BC+CD: BCCD:tan (CBD + CDB): tan (CBD — CDB); that is, 50.61 13.91::tan (180°—31° 30′)=tan † 148° 30′ : tan (CDB-CBD); and, by the tables, 1.704236 : 1.14 3327:10.549706 : 9.988797 = 44° 15′ 40′′ = 1⁄2 (CDB – CBD). By Trig. 143, sin DBC: DC:: sin BCD: DB; that is, sin 29° 59′ 20′′: 18.35: :sin 31° 30′ : DB; and, by the tables, 9.698824: 1.263636::9.718085 : 1.282897=19. 18=DB. By Mens. 9, R: sin BCD::BC×CD : 2 area BDC; that is, 1 sin 31° 30'::591.9710 : 2 area BDC; and, 10.000000: 9.971808: 2.772300: 2.490385 = 309. 3. Again, (Mens. 9.) R : sin ABD::AB×B×D : 2 area ABD; that is, 1 : sin 100° 30′— 29° 59′ 20′′=70° 30′ 40′′ : 505.2012; and, by the tables, 10.974376: :2.703464 : 2.6 77840 476.3 = 2 area ABD. Again, (Mens. 9.) R: sin AED::AE×ED : 2 area AED; that is, 1 : sin 53°::718. 5360: 2 area AED; and, by the tables, 10.000000 : 9.928 420: 2.856448: 2.784868=609.3 2 area AED. Twice the area of the three triangles ABD, BCD, AED=309.3+ 476.3+609.3=1394.9, which divided by 2=697.4 chains=

69.74 acres.

A

By addition, we find the sum of the North areas to be 643.4217, and the sum of the South areas to be 3553.6385. Their difference=2910.2168, which divided by 2=145.51084 the answer in miles.