The tables of Callet not containing the logarithmic secants and cosecants, the calculation of this example from his tables would be as follows, by the rule at Art. 60. log. cos 48° 35′ 27′′ = 9.82049 log. sec 48° 35′ 27′′ = (20-9.82049) = 10.17951 4. Required log. cosec 35° 27′ 24′′. Ans. by Tab. XXVII., 10•23651 or log. sin 35° 27′ 24′′ = 9.76349 log. cosec 35° 27' 24" 20-9-76349 10.23651 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. 8 App. I. To find the trigonometrical lines of arcs greater than 900, observe the rule at Art. 17. SOLUTION OF RIGHT ANGLED TRIANGLES, WITH THE AID OF LOGARITHMS. EXAMPLE. 63. Referring to the example of Art. 39, where the hypothenuse 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 350 = 9.75859 Remainder 1.41750= log. of 26.15 the hyp. Referring to Art. 41 where the height of the tower ċ is c=tan 300 X 200* which becomes, when the radius 1 in the first term of the proportion is changed into 101o 64. I. Another definition besides that given at Art. 14 for the sine of an angle, is the ratio of the hypothenuse to the side opposite the angle in any right angled triangle of which the angle forms one of the elements. For from formulas (1) Art. 38 may be obtained. II. A corresponding definition for the cosine of an angle is the ratio of the hypothenuse to the side adjacent the angle in any right angled triangle of which it forms an element. For, form. (3) Art. 38 It is evident that radius must be understood in the second member of this expression, because a line e cannot be equal to the rectangle of two lines. (Art. 31.) + The hypothenuse, which is the greater side, is evidently the denominator of the ratio, as the sine and cosine to radius unity are always fractions. III. In a similar manner the tangent of an angle is the ratio of the side adjacent to the side opposite the angle in any right angled triangle of which the angle forms an element. For Art. 41, (3) and (4) From these definitions the following consequences flow. IV. The hypothenuse multiplied by the sine of one of the acute angles of a right angled triangle will give the side opposite to the angle. For from (1) above clearing of fractions we have V. The hypothenuse multiplied by the cosine of the angle will give the side adjacent. For from (2) above VI. The side adjacent multiplied by the tangent will give the side opposite. For from (3) above VII. To find the hypothenuse when a side and angle are given, we must still use the sine and cosine of the angle, but as a divisor; the sine when the side opposite is given and serves for a dividend, the cosine when the side adjacent. For from formulas (4) and (5) above may be obtained VIII. When one of the perpendicular sides is given with an angle to find the other, the tangent of the angle is the multiplier of the side adjacent the angle to find the side opposite, and is the divisor of the side opposite, to find the side adjacent. See (6) above from which also may be obtained * When the hypothenuse is given with an acute angle it will be necessary always to multiply the hypothenuse by either the sine or cosine of the acute angle to obtain the other side; by the sine when it is the side opposite which is required, and by the cosine when it is the side adjacent. The memory will be aided by observing that for radius unity the sine and cosine are always fractions, and therefore the hypothenuse, which is the larger side, must evidently be multiplied by these in order to obtain the other sides. IX. Radius unity has no effect either as a multiplier or divisor, nevertheless when using the tables in which radius is 100 or 10000000000, it is necessary to know how the radius enters into the products or quotients formed by the above rules. And the following consideration will always. show, viz. that two equal quantities must be homogeneous. A line cannot be equal to a surface or the rectangle of two lines. Therefore all the products formed by the above rules must be understood to be divided by radius. Neither can a line be equal to the ratio or quotient of two lines, for this is an abstract number. Therefore all the quotients formed by the above rules must be understood to be multiplied by radius. In the use of logarithms, therefore, for the solution of right angled plane triangles, in every case two logarithms only will have to be employed, and their sum or difference taken according as the rule which applies to the case requires multiplication or division. When the sum of the two logarithms is taken, 10 must be rejected from the characteristic, which is in effect dividing by radius, and when the difference is taken, 10 must be added to the characteristic of the minuend, which is in effect multiplying by radius. All this will be easily comprehended from the following examples. The sum of the logarithms 2.69346 and 9.95960 which is 12-65306 after rejecting 10 from index will be the logarithm of 449.8 as may be seen by inspecting the tables, and this number is the value of the required side c. EXAMPLE II. The same things being given to find the side b. By Rule V. above Both the above operations may be conveniently connected together by preparing first a blank form thus: in which the arguments or elements of the triangle occupy the first column, the trigonometrical functions of these arguments necessary for the calculation of the side c the second, and those necessary for the calculation of the side b the third. This form is then to be filled up as follows The log. of 493 when found in the tables is written in two columns, and the log. sin and log. cos of 65° 40' can be taken out at one opening of the tables. EXAMPLE III. Let b73.94, c-57° 20', to find a by VII., subtracting the logarithms for division, adding 10 to the index of the minuend (see IX.) before subtracting, the calculation will be as follows: |