serted in the table, because they may be easily computed from the cosines and sines. Thus: (Art. 33.) hence, R2 sec. COS. log. sec. = 20 — log. cos. = for log. of the quotient difference of logs. (art. 49 ;) and log. of the square of a number equal to twice the log. of the number, (Art. 54,) and log. of R = 10. To obtain the log. secant, therefore, we have this To obtain log. cosec, subtract log. sine from 20. Required log. cosec. 35° 27' 24". log. sin. 35° 27′ 24′′\ = log, cosec. 35° 27′ 24′′ 9.763493 = 20-9.763493 = 10.236507 A method of finding with greater accuracy the sine and tangent of a very small arc, or the cosine and cotangent of one near 90°, is pointed out at Art. 123. To find the trigonometrical lines of arcs greater than 90°, observe the rule at Art. 17. SOLUTION OF RIGHT ANGLED TRIANGLES, WITH THE AID OF LOGARITHMS. EXAMPLE. 61. Referring to the example of Art. 39, where the hypothe employing 1010 as R, instead of 1, because the tables which we are about to use, are constructed with that value of R, we have, by the rules for multiplication and division of logarithms Subtract log. sin 35° 9.758591 Remainder 1.417500=log. of 26.15= the hyp. EXAMPLE II. 30° 200 feet. Referring to Art. 41 where the height of the tower c is c = tan 30° × 200 which becomes, when the radius 1 in the first term of the proportion is changed into 1010 62. It is customary, where the subtraction of logarithms corresponding to the division of numbers is to be performed, to change this operation into addition by means of what is called the arithmetical complement of each subtractive logarithm. The arithmetical complement of a logarithm is the remainder after taking the logarithm from 10; thus the arithmetical complement of the logarithm 2.322447, is 7.677553. 10.000000 2.322447 7.677553 It may be formed most conveniently, instead of beginning at the right and subtracting each figure from 10 and carrying one each time throughout, by beginning at the left and subtracting each figure from 9 till you come to the last figure which subtract from 10. When we have a logarithm to subtract, we shall obtain the same result by adding its arithmetical complement and afterwards subtracting 10. Which may be proved as follows: By the definition arith. comp. log. b 10 log. b. Now add this arith. comp. to some other log. as log. a, the result will be log a 10 log b subtract 10 and there remains log a log b The same result as would have been obtained by subtracting log. b from log. a. Hence to perform operations containing a number of multiplications and divisions, by means of logarithms, we have the following RULE. Write the logarithms of all the multipliers, and the arith. complements of those of the divisors in a column. Add up the whole, and reject as many times 10 from the characteristic of the sum as there have been arith. complements employed. Applying this rule to the last example but one (art. 61) we have log. 10° 10.000000 log. 15 = 1.176091 arith. comp log. sin 35° = 0.241409 Sum rejecting 10 from characteristic 1.417500=log. 26.15 Applying it to the last example of Art. 61 we have log. tan. 30° log. = 9.761439 2002.301030 *ar. comp log. 1010 0.000000 sum rejec. 10 = 2.062469 = 115.4 Applying logarithms to the example in art. 40, we have 122 + 162 =400 log. of 400 = 2.602060 divide by 2, quot. 1.301030= log. v400 = From the tables 1.301030 is found to be the log. of 20 the value of the hypothenuse required. 63. We shall finish the subject of right angled triangles by presenting a case of their practical application which is likely often to occur. 77°30 1000 At the top of a mountain whose height was known by the barometer or otherwise to be 1000 feet above the level of the sea, a ship was observed through the tube of the instrument described at Art. 10, and the number of degrees between the tube and a plumb line from the centre of the circle was found. to be 77° 30'; required the distance of the ship. A right angled triangle is here formed, in which are given the perpendicular and angle at the vertex, and the base is required. Referring to the rule R : tan. of one of the acute angles :: side adjacent side opposite, we have in this case 1010 tan. 77° 30':: 1000: dist. required. * When arith. comp. of log. of R is used, it need not be written, since it is nothing To find the last term of a proportion, add the logarithms of the second and third terms which are the multipliers, and the arithmetical complement of the logarithm of the first term, which is the divisor. arith. comp. log. 101o 0.000000 log. 1000 = 3.000000 log. tan. 77° 30' = 10.654245 3.654245 = log. 4510.71 feet. Therefore the distance of the ship is 4510.71 feet, or a little over of a mile. In the first of these, by a proportion previously demonstrated (Art 38,) we have whence R: Sin B::AB: AD RXAD Sin B XA B Again, in the right angle triangle A C D, we have whence R: Sin C:: AC:AD RXAD Sin CX AC The first members of the two equations above, being the same, the second members are equal, hence sin BX AB sin C XA C Turning this equation into a proportion, by making the first product the extremes, and the second the means, we have sin B: sin c :: AC:AB |