In our preceding equations, sin. a, cos. a, etc., refer to natural sines; and by such equations we determine their values in natural numbers; and these numbers are put in Table II, under the heads of N. sine and N. cos., as before observed. When we have the sine and cosine of an arc, the tangent and cotangent are found by Eq. (3) and (6); thus, and the secant is found by equation (4); that is, sec. = R2 COS. For example, the logarithmic sine of 6° is 9.019235, and its cosine 9.997614. From these it is required to - find the logarithmic tangent, cotangent, and secant. The secants and cosecants of arcs are not given in our table, because they are very little used in practice; and if any particular secant is required, it can be determined by subtracting the cosine from 20; and the cose cant can be found by subtracting the sine from 20. The sine of every degree and minute of the quadrant is given, directly, in the table, commencing at 0°, and extending to 45°, at the head of the table; and from 45° to 90°, at the bottom of the table, increasing backward. The column having sine at the top has cosine at the bottom, and the opposite, because angles read from above are complementary to those read from below. The differences of consecutive logarithms corresponding to 10" are given for both sine and cosine, but the tangents and cotangents have the same column of differences for the reason that log. tan.+log. cot.=log. R2 and is therefore constant. Hence, by just as much as log. tan. increases, log. cot. de creases and the converse. As cosines and cotangents decrease when arcs increase, and increase when arcs decrease, the proportional parts answering to seconds for them must be subtracted. Example. Find the sine of 19° 17' 22". The sine of 19° 17', taken directly from the table, is for 6.02 x 22 = Hence, log. sine 19° 17′ 22′′ is 9.518829 132 9.518961 From this it will be perceived that there is no difficulty in obtaining the sine or tangent, cosine or cotangent, of any angle greater than 30'. Conversely: Given, the logarithmic sine 9.982412, to find its corresponding arc. The sine next less in the table is 9.982404, which gives the arc 73° 48'. The difference between this and the given sine is 8, and the dif ference for 1′′ is .61; therefore, the number of seconds corresponding to 8, must be discovered by dividing 8 by the decimal .61, which gives 13. Hence, the arc sought Is 73° 48′ 13′′. These operations are too obvious to require a rule. When the arc is very small,-and such arcs are sometimes required in Astronomy,-it is necessary to be very accurate; for this reason we omitted the difference for seconds For all ares under 30'. Assuming that the sines and tangents of ares under 30' vary in the same proportion as the arcs themselves, we can find the sine or tangent of any very small arc, with great exactness, as follows: The sine of 1', as expressed in the table, is 3.463726 Divide this by 60; that is, subtract logarithm 1.778151 The logarithmic sine of 1", therefore, is 4.685575 Now, for the sine of 17", add the logarithm of 17 1.230419 Logarithmic sine of 17", is 5.916024 In the same manner we may find the sine of any other small arc. For example, find the sine of 14' 211"; that is, 861.5". Two lines drawn, the one from the surface and the other from the center of the earth, to the center of the sun, make with each other an angle of 8.61". What is the logarithmic sine of this angle? One figure will be sufficient to represent the triangle in all of the following examples; the right angle being at B. PRACTICAL PROBLEMS. 1. In a right-angled triangle, ABC, given the base AB, 1214, and the angle A, 51° 40' 30", to find the other parts. B REMARK. When the first term of a logarithmic proportion is radius, the required logarithm is found by adding the second and third logarithms, rejecting 10 in the index, which is dividing by the first term. In all cases we add the second and third logarithms together; which, in logarithms, is multiplying these terms together; and from that sum we subtract the first logarithm, whatever it may be, which is dividing by the first term. To find this resulting logarithm, we subtracted the first logarithm from the second, conceiving its index to be 13. Let ABC represent any plane triangle, right-angled at B. 2. Given, AC 73.26, and the angle A, 49° 12′ 20′′; required the other parts. Ans. The angle C, 40° 47′ 40′′; BC, 55.46; and AB, 47.86. 3. Given, AB 469.34, and the angle A, 51° 26′ 17′′, to fnd the other parts: Ans. The angle C, 38° 33'43"; BC, 588.7; and AC, 752.9. 4. Given, AB 493, and the angle C, 20° 14'; required, the remaining parts. Ans. The angle A, 69° 46′; BC, 1338; and AC, 1425.5. 5. Let AB 331, and the angle A = 49° 14'; what are the other parts? = Ans. AC, 506.9; BC, 383.9; and the angle C, 40° 46'. 6. If AC=45, and the angle C'=37° 22′, what are the remaining parts? Ans. AB, 27.31; BC, 35.76; and the angle A, 52° 38 7. Given, AC=4264.3, and the angle A= 56° 29′ 13′′, to find the remaining parts. Ans. AB, 2354.4; BC, 3555.4; and the angle C, 33° 30′ 47′′, = 8. If 4B 42.2, and the angle A = 31° 12′ 49′′, what are the other parts? Ans. AC, 49.34; BC, 25.57; and the angle C, 58° 47′11′′. 8372.1, and BC = 694.73, what are the 9. If AB other parts? Ans. = AC, 8400.9; the angle C, 85° 15' 23"; and the angle A, 4° 44′ 37′′. 10. If AB be 63.4, and AC be 85.72, what are the other parts? Ans. BC, 57.7; the angle C, 47° 42'; and the angle A, 42° 18'. 11. Given, AC = 7269, and AB = other parts. Ans. 3162, to find the BC, 6545; the angle C, 25° 47' 7"; and the angle A, 64° 12′ 53′′. 12. Given, AC other parts. = 4824, and BC 2412, to find the The angle A= 30° 00', the angle = 60° 00', 13. The distance between the earth and sun is 91,500,000. miles, and at that distance the semi-diameter of the sun subtends an angle of 16'. What is the diameter of the sun in miles? Ans. 851,659. E In this example, let E be the center of the earth, S that of the sun, and EB a tangent to the sun's surface. Then the ▲ EBS is right-angled at B, and BS is the semi-diameter of the sun. The value of 2BS is required. |