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 the 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. 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 taken from the table from the given logarithm, annex two 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 10.008688 cot, 44° 26', next less in the table Tabular difference, 10.008591 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 CB : CA :: sin A : sin B. 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 D A B EI L F dicular to AB: now DE is the sine of the angle A, and CF is the sine of B, to the same radius AD or BC. But by similar triangles, AD DE :: AC : OF. But AD being equal to BC, we have By comparing the sides AB, AC, in a similar manner, we should find, AB : AC :: sin C sin B. THEOREM II. 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 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. D A B FGH Draw CI and AF Since CAE is an exterior angle of the triangle ACB, it is equal to the sum of the interior angles C and B (Bk. I., 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 Cand B, or equal to 1⁄2(C+B). The angle AFC= ACB, is also equal to ABC+BAF; therefore, BAF = ACB- ABC. But, ICF = (BAF)=(ACB-ABC), or (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 IO. The perpendicular CE will pass through E, the extremity of the diameter IE, 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=1(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 }(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 But since the difference of B D 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 (CD+DB).(CD — DB) therefore, (CD+DB). (CD — DB) = (AC+AB). (AC— AB) hence, CD+DB: AC+ AB :: AC-AB: CD-DB. THEOREM IV. In any right-angled plane triangle, radius is to the tangent of either of the acute angles, as the side adjacent to the side opposite. 24. Let CAB be the proposed triangle, and denote the radius by R: then B From the similar triangles CDG and CAB, we have, OD : DG :: CA : AB; hence, Ꭱ : tan C :: CA : AB. In every right-angled plane triangle, radius is to the cosing of either of the acute angles, as the hypothenuse to the sid adjacent. 25. Let ABC be a triangle, right-angled at B: then R : cos A :: AC : AB. D A EF -B For, from the point A as a centre, with a radius AD=R, describe the arc DF, which will measure the angle A, and draw DE perpendicular to AB: then will AE ba the cosine of A. The triangles ADE and ACB, being similar, we have, AD : AE :: AC AB: that is, К : COS A :: AC : AB. REMARK. The relations between the sides and angles of plane triangles, demonstrated in these five theorems, are |