at the bottom of the page, and the minutes from the right hand column on the same horizontal line with the first. Therefore, sine, at the top of the page, should correspond with cosine, at the bottom; cosine with sine, tang with cotang, and cotang with tang, as in the tables (Art. 12). If the angle is greater than 90°, we have only to subtract it from 180°, and take the sine, cosine, tangent, or cotangent of the remainder. The column of the table next to the column of sines, and on the right of it, is designated by the letter D. This column is calculated in the following manner. Opening the table at any page, as 42, the sine of 24° is found to be 9.609313; that of 24° 01', 9.609597: their difference is 284; this being divided by 60, the number of seconds in a minute, gives 4.73, which is entered in the column D. Now, supposing the increase of the logarithmic sine to be proportional to the increase of the arc, and it is nearly so for 60", it follows, that 4.73 is the increase of the sine for 1". Similarly, if the arc were 24° 20′, the increase of the sine for 1", would be 4.65. The same remarks are applicable in respect of the column D, after the column cosine, and of the column D, between the tangents and cotangents. The column D, between the columns tangents and cotangents, answers to both of these columns. Now, if it were required to find the logarithmic sine of an arc expressed in degrees, minutes, and seconds, we have only to find the degrees and minutes as before; then, multiply the corresponding tabular difference by the seconds, and add the product to the number first found, for the sine of the given arc. Thus, if we wish the sine of 40° 26' 28". The decimal figures at the right are generally omitted in the last result; but when they exceed five-tenths, the figure on the left of the decimal point is increased by 1; the logarithm obtained is then exact, to within less than one unit of its right hand place. The tangent of an arc, in which there are seconds, is found in a manner entirely similar. In regard to the cosine and cotangent, it must be remembered, that they increase while the arcs decrease, and decrease as the arcs are increased; consequently, the proportional numbers found for the seconds, must be subtracted, not added. 3. Find the cotangent of 87° 57′ 59′′. Ans. 9.884592. Ans. 8.550356. CASE II. To find the degrees, minutes, and seconds answering to any given logarithmic sine, cosine, tangent, or cotangent. 19. Search in the table, in the proper column, and if the number is found, the degrees will be shown either at the top or bottom of the page, and the minutes in the side column either at the left or right. But, if the number cannot be found in the table, take from the table the degrees and minutes answering to the nearest less logarithm, the logarithm itself, and also the corresponding tabular difference. Subtract the logarithm ciphers to the remainder, and then divide the remainder by the tabular difference: the quotient will be seconds, and is to be connected with the degrees and minutes before found to be added for the sine and tangent, and subtracted for the cosine and cotangent. EXAMPLES. 1. Find the arc answering to the sine 9.880054 Tabular difference, 9.879963 1.81)91.00(50". Hence, the arc 49° 20′ 50′′ corresponds to the given sine 9.880054. 2. Find the arc whose cotangent is Tabular difference, 10.008688 4.21)97.00(23". Hence, 44° 26' - 23" 44° 25′ 37′′ is the arc answering = to the given cotangent 10.008688. 3. Find the arc answering to tangent 9.979110. The sides of a plane triangle are proportional to the sines of their opposite angles. 21. Let ABC be a triangle; then will For, with A as a centre, and AD equal to the less side BC, as a radius, describe the arc DI: and with B as a centre and the equal radius BC, describe the arc CL, and draw DE and CF perpen A EI L F AD : DE :: CF is the sine of B, to the same radius AD or BC. But by similar triangles, But AD being equal to BC, we have AC : CF. In any triangle, the sum of the two sides containing either angle, is to their difference, as the tangent of half the sum of the two other angles, to the tangent of half their difference. 22. Let ACB be a triangle: then will A AB+AC: AB-AC:: tan (C+B): tan (C– B). With A as a centre, and a E radius AC, the less of the two given sides, let the semicircumference IFCE be described, meeting AB in I, and BA produced, in E. Then, BE will be the sum of the sides, and BI their difference. C B FGH Draw CI and AF Since CAE is an outward angle of the triangle ACB, it is equal to the sum of the inward angles C and B (Bk. L., Prop. XXV., Cor 6). But the angle CIE being at the circumference, is half the angle CAE at the centre (Bk. III., Prop. XVIII.); that is, half the sum of the angles C and B, or equal to (C+B). The angle AFC=ACB, is also equal to ABC+BAF; therefore, BAF-ACB-ABC. But, ICF=(BAF)=3(ACB — ABC), or 1⁄2(C– B). With I and C as centres, and the common radius IC, let the arcs CD and IG be described, and draw the lines CE and IH perpendicular to IC. The perpendicular CE E D since the right angle ICE must be inscribed in a semicircle. But CE is the tangent of CIE =(C+B); and IH is the tangent of ICB=4(C-B), to the common radius CI But since the lines CE and IH are parallel, the triangles BHI and BCE are similar, and give the proportion, BE : BI :: CE: IH, or by placing for BE and BI, CE and IH, their values, we have AB+AC : AB-AC :: tan (C+B): tan 1(C– B). THEOREM III. In any plane triangle, if a line is drawn from the vertical angle perpendicular to the base, dividing it into two segments: then, the whole base, or sum of the segments, is to the sum of the two other sides, as the difference of those sides to the difference of the segments. 23. Let BAC be a triangle, and AD perpendicular to the base; then will the squares of two lines is equivalent to the rectangle contained by their sum and difference (Bk. IV., Prop. X.), we have, and AC2- AB2=(AC+ AB) (AC-AB) CD2- DB2 = therefore, (CD+DB). (CD — DB)=(AC+AB).(AC— AB) |