Εικόνες σελίδας
PDF
Ηλεκτρ. έκδοση

82835.646007=442500= the semi-diameter of the sum, and 885000=his diameter.

1

PROMISCUOUS EXAMPLES.

Example 1. See Figure 7. By PROB. III. The angle PBC=46°, the angle A=31°, and their difference=15°= the angle ACB. By Trig. 150, sin ACB : AB: sin BAC BC; that is, sin 15° : 100::sin31° BC; and, by the tables, a. c. 0.587004; 2.000000::9.711839 2.298843= 199. Again, by Trig. 134, R: BC:: sin PBC: PC; that is, 1 199: sin 46° : PC; and, by the tables, 10.000000: 2.298843::9.856934: 2.155777=143.1.

Example 2. See Figure 8. By PROB. III. ACD=50° -32° 18°. PAC=50°, PAB=39°, ADC = 32°, and AD=150. By Trig. 150, sin ACD: AD:: sin ADC : AC; that is, sin 18° : 150::sin 32° : AC; and, by the tables, a. c. 0.510018: 2.176091::9.724210: 2.410319=25 7.34. Again, sin ABC: AC:: sin BAC: BC; that is, sin 29° 257.34::sin 11° : BC; and by the tables, a. c. 0. 109497 2.410319::9.280599: 1.800415=63.04. Again, by Trig. 134, R: AC::sin PAC: PC; that is, 1 : 257.34 ::sin 50°: PC; and, by the tables, 10.000000: 2.410319 ::9.884254 : 2.294573-197.05, from which subtract 63. 04, and there remains 134.01 the hight of the hill, and 63.04 the hight of the tower.

Example 3. See Figure 5. Making the base radius (Trig. 127.) AB: R::BC: tan A; that is, 4: 1 :: B : tan A; and, by the tables, 0.602060: 10.000000: :0.477121 : 9.87 5061-36° 52′ 15′′.

Example 4. h=√‡D2 +d2 −‡D : But D= 7940, whose square=63043600, which divided by 4=15760900. (1031)2 =106777, which added to 15760900=15771577.7777+, whose square root-3971.345; and, 3971.345-3970=1.3 45 miles 7101.6 feet.

Example 5. See Figure 7. Let AB the breadth of the river, and PC the hight of the rock=55; then, PCB= 90°—55° 54′=34° 6′, and PCA=90°—33° 20′=56° 40'. By Trig. 138, R: PC::tan PCB: BP; that is, 1:55:;

tan 34° 6' : BP; and by the tables, 10.000000 : 1.740363 ::9.830621 : 1.570984–37.3=the distance from the rock. Again, by Trig. 138, R: PC:: tan PCA: AP; that is, 1:55::tar 56° 40′ : AP; and by the tables, 10.000000 : 1.740363::10.181965 : 1.922328=83.62, from which sub. tract 37.3, and there remains 46.32=the breadth of the river.

Example 6. See Figure 15, Art. 30. By Trig. 137, R : EC: sin CEA: AC; that is, 1: 240000:: sin 15° 37' : AC; and, by the tables, 10.000000 : 5.380211::7.657100 : 3.037311=1090= the diameter, and, 2180=the diameter of the moon.

Example 7. See Figure 7. By PROB. III. The angle A= 30° 58′, PBC=36° 52′, AB= a mile=160 rods, ACB= 36° 52′-30° 58' 5° 54'. By Trig. 150, sin ACB : AB:: sin A: BC; that is, sin 5° 54′ : : :sin 30° 58′ : BC; and, by the tables, a. c. 0.988038: 1.698970 :: 9.711419: 0.3 98427 2.503. Again, R: BC: :sin PBC : PC; that is, 1: 2.503: sin 36° 52′; PC; and, by the tables, 10.0000 00: 0.398427::9.778119: 0.176546=1.501 miles.

Example 8. See Figure 7. By PROB. III. The angle A =32°, PBC=67°, AB= a mile=160 rods, ACB=67°— 32° 35°. By Trig. 150, sin ACB: AB:: sin A : BC ; that is, sin 35° : 160: :sin 32° : BC; and, by the tables, a. c. 0.241409: 2.204120 :: 9.724210 : 2.169739 = 147.83. Again, R: BC ¦ ¦ sin PBC : PC; that is, 1 : 147.83: :sin 67°: PC; and, by the tables, 10.000000: 2.169739::9.9 64026: 2.133765=136.06 rods=2244.990 feet.

NAVIGATION.

CASE I. Given the course and distance, to find the departure and difference of latitude.

Exam. 1. Making the distance radius, (Trig. 134.) See Fig. 20. R: Distance::sin Course: Departure; that is, 1: 38 ::sin 33° 45′ Departure; and, by the tables, 10,000000: 1.579784 9.744739: 1.324523 = 21.12. Again, R: Dist.cos Course: Difference of latitude; that is, 1 : 38:: cos 56° 15': Dif. Lat. ; and, by the tables, 10.000000 : 1.5 79784: 9.919846 1.499630-31.6. Therefore, the departure=21,12, and Dif. Lat.-31.6.

Example 2. R Dist.::sin Course: Departure; that is, 1:34: sin 29°: Departure; and, by the tables, 10.000000 : 1.531479: 9.685571 1.217050 = 16.49 = Departure: Again, R Dist.::cos Course: Dif. Lat.; that is, 1:34: cos 29° Dif. Lat. ; and, by the tables, 10.000000 : 1.5314 79: 9.941819: 1.473298=29.73-Dif. Lat.

CASE II. Given the course and departure, to find the distance and difference of latitude.

Example 1. Sin Course Departure::R Dist. ; that is, sin 37° 62::1: Dist.; and, by the tables, 9.779463: 1.7 92392::10.000000 ; 2.012929=103=the distance: Again, sin Course Departure::cos Course: Dif. Lat. ; that is, sin 37° 62: cos 37°: Dif. Lat.; and, by the tables, 9.779463 : 1.792392: 9.902349 1.915278-82.28 Dif. Lat.

=

Example 2. Sin Course Departure::R Dist. ; that is, sin 54° 74::1. Dist.; and, by the tables, 9.907958 : 1.869232::10.000000: 1.961274 91.47. Again, sin Course Departure cos Course: Dif. Lat. ; that is, sin 54°: 74::cos 54°: Dif. Lat.; and, by the tables, 9.907958 : 1.869232: 9.769219: 1.730493 = 53.765 0° 53'.8, which subtracted from 63°, leaves 62° 6'.8.

CASE III. Given the course and difference of latitude, to find the distance and departure.

Example. Cos Course: Dif. Lat.::R: Dist.; that is, cos 50°: 660::1: Dist. ; and, by the tables, 9.808067: 2.819 544: 10.000000: 3.011477 = 1026.77 the distance. Again, cos Course: Dif. Lat. :: sin Course: Departure; that is, cos 50: 660: :sin 50°: Departure; and, by the tables, 9.808067 : 2,819544::9.884254 : 2.895731=786.52 =the departure.

CASE IV. Given the distance and departure, to find the course and difference of latitude.

Example. Making the distance radius, (Trig. 135.) Dist. R: Departure: sin Course; that is, 400 1:180 : sin Course; and, by the tables, a. c. 7.397940: 10.000000: 2.255273: 9.653213=26° 44′ 20′′. Again, R: Dist.:cos

Course: Dif. Lat.; that is, 1 : 400 :: cos 26° 44′ 30′′ : Dif. Lat.; and, by the tables, 10.000000 : 2.602060;;9.950905 : 2.55 2965 = 357.3=5° 57′.3, which diminished by 3°=2° 57.3=the ship's latitude.

Case V. Given the distance, and difference of latitude, to find the course and departure.

Example. Making the distance radius, (Trig. 136.) Dist. : R:: Dif. Lat. : cos Course; that is, 66:1::50′ : cos Course; and, by the tables, a. c. 8.180456: 10.000000::1. 698970: 9.879426=40° 45′ the course. Again, R : Dist. ::sin Course: Departure; that is, 1: 66::sin 17° : Departure; and, by the tables, 10.000000: 1.819544::9.814753 : 1.634297=43.08=the departure.

CASE VI. Given the departure and difference of latitude, to find the course and distance.

Example. Making the Dif. Lat. radius, (Trig. 139.) Dif. Lat.: R::Depart. : tan Course; that is, 352:1::264 : tan Course; and, by the tables, a. c. 7.453457 : 10.000000:: 2.421604 9.875061 = 36° 52' the course. Again, R: Dif. Lat. sec Course: Dist.; that is, 1: 352::sec 36.°52′ : Dist.; and, by the tables, 10.000000: 2.546543::10.0968 90: 2.643433=440=the distance.

EXAMPLES FOR PRACTICE.

Example 1. CASE II. Sin Course: Depart. ::R: Dist. ; that is, sin 46° : 59::1: Dist.; and, by the tables, a. c. 0. 143066: 1.770852: 10.000000: 1.913918 82.02 = the distance. Again, sin Course: Depart.::cos Course: Dif. Lat.; that is, sin 46° : 59::cos 17° : Dif. Lat.; and, by the tables, a. c. 0.143066: 1.770852::9.841771 : 1.755687= 56.98 Dif. Lat.

Example 2. CASE IV. Dist.: R::Depart. : sin Course; that is, 68: 1:47: sin Course; and, by the tables, a. c. 8. 167491: 10.000000::1.672098 : 9.839589-43° 43′.5=the course. Again, R: Dist.::cos Course::Dif. Lat. ; that is, 1:68: cos 17°: Dif. Lat. ; and, by the tables, 10.000000 : 1.8325099.85900: 1.691512-49.15=Dif. Lat.

Example 3. CASE I. R Dist. sin Course Depart. ; that is, 1 57: sin 22° 30': Depart. ; and, by the tables, 10. 000000 1.755875::9.582846 1.338721=21.31=the departure. Again, R Dist.::cos Course Dif. Lat. ; that is, 1:57::22.° 30': Dif. Lat.; and, by the tables, 10.000000 1.755875: 9.965615 1.721490=52.66-Dif. Lat.

Example 4. CASE III. As the course is NW by N, it= 33° 45'. Dif. Lat.=2×60+36=156 miles. Cos Course :Dif. Lat. :: R; Dist. ; that is, cos 33° 45′ :156::1: Dist.; and, by the tables, a. c. 0.080154 2.193125::10.00 00002.273279=187.65=the distance. Again, cos Course Dif. Lat.sin Course Depart.; that is, cos 33° 45′ : 156::sin 33° 45′ : Depart.; and, by the tables, a. c. 0.0801 54 2.193125::9.744739 : 2.018018=104.24.

Example 5. CASE VI. Dif. Lat. R::Depart. tan Course; that is, 86: 1::92: tan Course; and, by the tables, a. c. 8.065502 : 10.000000 :: 1.963788: 10.029290 =46° 56′= the Course. Again, R: Dif. Lat.::sec Course Dist.; that is, 186 sec 46° 56′ Dist.; and, by the 1.934498: 10.165675 2.100173

tables, 10.000000

125.92 the distance.

Example 6. CASE V. Dist. R::Dif. Lat. cos Course; that is, 123: 1::96: cos Course; and, by the tables, a. c. 7.910095 10.000000::1.982271: 9.892366 = 38° 42′ = the Course. Again, R: Dist.::sin Course Depart.; that is, 1.: 123: sin 38° 42′ Depart.; and, by the tables, 10. 000000 departure.

2.089905::9.796049 1.885954 = 76.91 = the

PARALLEL SAILING.

Example 1. Com Lat,

[ocr errors]

90°-38° 52°. Dividing 117, the departure, into 78, and 39, under 52° and opposite 78 in Departure, will be found 99, opposite 39 will be found 491, which added to 99=148 miles or minutes-2° 28′ 30′′, and this subtracted from 16°, leaves 13° 31′ 30′′.

Example 2. R: cos 46°:: 210: Depart. or Dist.; and, by the tables, 10.000000 ; 2.322219::9.841771: 2.163990 =145.87 miles the departure or distance between the two places.

« ΠροηγούμενηΣυνέχεια »