1.7190000 (log. 0.5236)=11.4168196, and this subtracted from 17.8481700, leaves 6.15035041,413,680. .

Example 5. The area of the base =202 X.7854-314. 1600, the area of the top, 82 X.7854=50.2650, the square root of the product of these areas=125.6624, and the three added together=490.0880, which multiplied by 22 ( the hight)=10781.9360 the solidity of the whole tower, not allowing for the cavity. The area of the base of the cavi. ty=122 X.7854-113.0976, the area of the top of the cavity=42.7854=12.5664, the square root of the product of these areas 37.7000, and the sum of the three=163.3640, which multiplied by 22 3594.0080 the solidity of the cavity, which subtracted from the solidity of the whole tower leaves 7187.9220.

=

Example 6. Let x=the diameter; then r3 X.5236=.52 36x3231 the number of gallons, and xxx × 3.14159= 3.14159x2=the surface of the globe. By the conditions of the question, (.5236x3 ÷ 231)× $5,00 3.14159x2 X 20: Clearing of fractions 2.6180x3-14.4280055x2, and by reduction, x=55.43=the diameter.

Example 7. the diameter of the wire, and (2)2 × ́ .7854×1584000000=31101840000=the solidity of the wire. One mile=63360 inches, 25000 miles 1,584,000,000. 31 101840000 .5236=5940000= the cube of the diameter, whose cube root=181.1 inches=15 feet 1 inch nearly.

Ex. 8. Complete the cone and find its solidity: 21:37: 9 the hight of the whole cone, whose solidity=7.52 X.7854 x3=132.53625. To find the solidity of thefrustum, 7.52 X .7854-44,17875,52 X.7854-19.635, √44.17875 x 19.635 =29.45, and the sum of these three results 93.26375=the solidity of the frustum. The solidity of the top the frustum = 85.90435. By Euc. Sup. III. 11, 17 and 18, making a the hight of a cone which the solidity of the top part+the frustum, 132.53625:85.90435::93: x3; and x3=472.592, x=7.79, and 9-7.79=1.21 feet-14.52 inches.

Example 9. See Figure 15. CG=10, and GN=8; then 102-83-36=CN2, and CN=6, and Na=the hight of the

cylinder which perforates the sphere, and PD=4=the hight of the segments cut off by the cylinder. (162 ×.7854×2) +(43 x.5236)=435.6352=the solidity of one of the seg. ments, and 435.6352×2=871.2704-the solidity of the two segments. The solidity of the cylinder 162 X.7854x12 =2412.5088, which added to 871.2704-3283.7792=the whole solidity cut away by the cylinder. 20×3.14159 × 4 X2=502.6544 the area of the surface cut away by the cylinder.

Example 10. The diagonal of the cube=3, and 32=the square of a linear side of the cube+the square of the diagonal of one of its faces (Euc. I. 47.) the square of three linear sides; hence 32÷3=3=the square of a linear side. Log. 3.4771213, which divided by 2=0.2385606, and this into 3=0.7156818=5.196+.

Example 11. 162 ×.7854=201.0624, 82 X.7854-50.26 56, the product of these two results=10106.52217344 whose square root 100.5312, which added to 201.0624+50.2656 =351.8592, and this into 12=4222.3104.

Example 12. As the diameter of the sphere=16, and the hight of the segment 4, the diameter-the hight of the segment=12, and the radius=8. By Euc. VI. K, 4×12= the square of the radius of the base of the segment=48, the radius 6.9, and the diameter=13.8. 13.82.7854= 149.617430, and this into 2=299.234860, which added to 33.5104 (43.5236)=332.745260.

HIGHTS AND DISTANCES.

PROBLEM I. To find the perpendicular hight of an accessible object standing on a horizontal plane.

Example 1. See Figure 2. Making the hypothenuse radius (Trig. 131.) Cos BAC: AB::sin BAC : BC; that is, cos. 35° 30': 98::sin 35° 30': BC; and, by the tables, 9.910686 1.991226::9.763954: 1.844494-69.9 feet.

Example 2. R AB::tan A: BC; that is, 1: 163: tan 40°: BC; and, by the tables, 10.000000 2.212188::9.9 23813 2.136001-136.76.

Example. Art. 7. Making the base radius (Trig. 137.) Sin ABC::R; AB; that is, sin 3° 30′ ; 99::1': AB; and, by the tables, 8.785675 0.778151: 10.000000 : 1.9924 76 98.28 rods.

Example. Art. 8. Making the perpendicular radius (Trig. 138.) RBC::tan C: AB; that is, 1:66::tan 10° AB; and, by the tables, 10.000000 0.602060::10.7536 81:1.355741=22.68 rods.

PROBLEM II. To find the hight of an accessible object standing on an inclined plane.

Example. Art. 9. Sin C: AB::sin A; BC; that is, sin 33° 53′ : 76:: sin 44° 42′ ; BC; and, by the tables, 9.746 248 1.880814::9.8471991.981766-95.89 feet.

Example. Art. 10. See Figure 4. ab bc:: AB ; BC; that is, 3 :5::69; 115 feet.

PROBLEM III. To find the hight of an inaccessible object above a horizontal plane.

Example 1. See Figure 7. To find BC, sin ACB; AB ::sin BAC:BC; that is, sin 20°: 100:: sin 40° : BC ; and, by the tables, a. c. 0.465948: 2.000000::9.808067 2.274015=187.94. To find CP, R:BC:: sin PBC CP; that is, 1:187.94::sin 60° : CP; and, by the tables, 10.000000 ; 2.274015;:9.937531:2.211546=162.78.

Example 2. See Figure 7. sin ACB: AB::sin BAC : BC; that is, sin 8° : 120:: sin 68°; BC; and, by the tables, a. c. 0.856445: 2.079181::9.967166: 2.902792=79 9.45. To find CP, R ; BC: :sin PBC : CP ; that is, 1:79 9.45: 9.986904; 2.889696=775,7=2 miles 135.7 rods.

PROBLEM IV. To find the hight of any object, by observation

at two stations.

Example. Art. 13. See Figure 8. Sin ACD: AD: :sin ADCAC; that is, sin 34° : 83: :sin 51° : AC; and, by the tables, a. c. 0.252438: 1.919078::9.890503 : 2.0620 19 = 115.34. Again, sin ABD ; AD :: sin ADB : AB ;

that is, sin 26° 83: :sin 33°: AB; and, by the tables, a. c. 0.358158 1.919078: 9.736109: 2.013345 = 103.13. Again, AC+AB AC-AB::tan (ABC+ACB); tan (ABC-ACB); that is, 218.47: 12.21:: tan 77° : tan (ABC—ACB): and, by the tables, a. c. 7.660747: 1.086 960: 10.636636: 9.383743=13° 36'. Lastly, sin ACB : AB: sin BAC: BC; that is, (Alg. 341.) sin 63° 24′ : 103. 13: sin 26° BC; and, by the tables, a. c. 0.048588; 2. 013259: 9.641842 1.703588=50.54.

PROBLEM V. To find the distance of an inaccessible object.

Example 1. See Figure 9. By Trig. 150, sin C AB:: sin A: BC; that is, sin 48° 34′; 26.6::sin 38° 40′ ; BC; and, by the tables, a. c. 0.125097; 1.424882::9.795733 : 1.345712=22.17. Again, sin C:AB:¦sin B¦AC; that is, 48° 34' 26.6: :sin 92° 46': AC; and, by the tables, a. c. 0.125097 : 1.42882;:9.999493; 1.549472=35.44.

Example 2. See Figure 9. Sin C:AB::sin A; BC; that is, sin 78° 45′ 90: :sin 45°: BC; and, by the tables, a. c. 0.008426 1.954243: 9.849485 1.812154 == 64.9. Again, sin CAB::sin B: AC; that is, sin 78° 45' :.90 ::sin 56° 15′ : AC; and, by the tables, a. c. 0.008426 : 1. 954243: 9.919762: 1.882431=76.29.

Art. 15. To find CP, make BC (Fig. 9.) radius, (Trig. 134.) then, R: BC :: sin B: PC; that is 1 : 64.89:: sin 56° 15': CP; and, by the tables, 10.000000 : 1.812154:: 9.919762: 1.731916=53.94.

PROBLEM VI. To find the distance between two objects, when the pussage from one to the other in a straight line is obstructed.

=

the

Example. Art. 16. See Figure 10. Trig. 144 and 153. BC+AC=185, BC-AC-33, and the sum of the angles opposite AC, BC, 180°-101° 30' 78° 30', and sum=39° 15'. By Trig. 153, BC+AC; BC-AC::tan (A+B): tan (A-B); that is, 185: 33:: tan 39° 15': tan (A-B); and, by the tables, a. c. 7.732828: 1.5185 14: 9.912240: 9.163582-8° 18' the difference between the angles A, B, and the angle A is the greatest of the two

because it is opposite the greater side. By Alg. 341, the angle A=47° 32', and the angle B-30° 58': Hence, (Trig. 150.) sin A: BC: :sin C: AB; that is, sin 47° 32′ :109:: sin 101° 30′ : AB; and, by the tables, a. c. 0.132138 ; 2.0 37426: 9.991193: 2.160757=144.78.

PROBLEM VIII. To find the diameter of the earth, from the known hight of a distant mountain, whose summit is just visible in the horizon.

Example. Art. 20. 178.252=31773, and this divided by 4=7943.11, which diminished by 4=7939.17.=

PROBLEM IX. To find the greatest distance at which a given object can be seen on the surface of the earth.

Example. Art. 23. 7940×2=15880, and this added to 4=15884, whose square root=126.03=the answer.

Example. Art. 24. See Figure 13. 7940 × 320=2540 800. 132 feet 8 rods, and 33 feet 2 rods. 2540800×8= 20326400 which added to 64=20326464, whose square root =4508.04=distance from A to T. And, 2540800×2=50 81600 which added to 4=5081604, whose square root=22 54.41 = the distance from A to T. 4508.04+2254.41= 6762.45 rods=21.114 miles.

Example. Art. 25. 5 feet of a rod, and the diameter of the earth=2540800 rods, which multiplied by 3=84693 33, which added to 1=8469334, whose square root = 920.32 rods 2.85 miles.

PROBLEM X. To find the distance of any heavenly body whose horizontal parallax is known.

Example. Art. 28. See Figure 14. Making AC radius, RAC::cot AMC: CM; that is, 1: 3970: cot 0° 57′ : CM; and, by the tables, 10.000000: 3.598791::11.7803 595.359150-239,414.3 miles.

Example. Art. 30. See Figure 15. Making EC radius, REC::sin AEC; AC; that is, 1:95,000,000::sin 32′ 2" AC; and, by the tables, 10.000000 7.977724::7.66