10. Prove that if cos (AC) cos B = cos (A - B+C); tan A, tan B, and tan C are in harmonical progression. 11. Prove that if cos A = cos B cos C, then chd 72° = 2 cos 54°, and sec 72° = sec 60° + sec 36°. 15. Prove that vers 45° is an arithmetic mean between sec 60 ̊ and sec 225, and that cosec 150 is a geometric mean between cosec 105° and cosec 165°. ON THE USE OF THE TRIGONOMETRICAL TABLES. 19. It will now be necessary to give a description of the tables of sines, tangents, etc., the method of constructing such being reserved for a subsequent part. In the Table of Natural Sines, etc., the radius is unity, and therefore the sines and cosines of all angles are either unity or decimal fractions less than unity. The decimal point, however, is sometimes omitted. The same is the case with the tangents of angles less than 45 ̊. The Table of Log. Sines, etc., is formed by taking the logarithms of the numbers in the corresponding Table of Natural Sines, etc.; and, to avoid negative indices, 10 is added to each, so that log sin A = log nat sin of A+ 10, and so on. (A.) To find the sine, cosine, etc., of any angle less than 90° expressed by degrees and minutes. If the angle be less than 45°, find the degrees at the top, and the minutes on the left-hand side, of the page. In the same line with the minutes, and under the proper name at the top (sine, cosine, etc.), take out the number required. Thus the natural sine and tangent of 27° 16′ are 4581325 and 5154019 respectively. Also the log sine and tangent of the same are 9.6609911 and 9.7121461. For angles between 44° and 90°, find the degrees at the bottom, and the minutes on the right hand side, of the page. In the same line with the minutes, and above the proper name at the bottom, take out the number. Thus the natural cosine of 56' 7' is 5575036. (B.) To find the sine, cosine, etc., of any angle less than 90°, expressed by degrees, minutes, and seconds. Find the sine, etc., of the next less and next greater angles in the table, and take the difference of these; then because the difference of any two trigonometrical functions of two angles (except in extreme cases) is as the difference of the angles themselves when the difference of the angles is less than one minute, we have 60" given number of seconds :: diff. thus found correction for seconds. It must be kept in mind that the sine, tangent, secant, when the angle is under 90, increase as the angle increases; but the cosine, cotangent, and cosecant decrease as the angle increases. Hence the correction must be added in the case of the sine, tangent, and secant, but subtracted in the case of the cosine, cotangent, and cosecant. 1. Let it be required to find the sine and cosine of 59° 14′ 16′′. pro. pts. for 16" = sin 59° 14′ 16′′ = ·8592973 The log sines, cosines, etc., of angles and seconds, are found in the same way. to the nearest unit; and hence, if the greater than 5, we must add 1 to the last figure, as in the preceding examples. (C.) To find the angle corresponding to any given sine, cosine, etc., to the nearest second. If the function be a sine, tangent, or secant, find the next less to the given one; but if a cosine, cotangent, or cosecant, the next greater, and take out the corresponding number of degrees and minutes. Then having found the difference between the next less or next greater, as the case may be, and the given one, we have this proportion : tabular diff. diff. found :: 60" seconds, to be added to the degrees and minutes already found. This proportion will be obvious from what has been stated. If the "tabular diff." be not given in the tables, it will be found by taking the next less from the next greater in the tables. 2. Given sin A = ·5432107, to find A. sin A5432107 sin 32° 54′ (next less) = ·5431744 * Hutton's Tables are referred to, but other tables are used in a similar way. 2138 (tab. diff.): 1423 :: 60": 40" nearly, The method for log sines, etc., is exactly similar. Scholium. For angles greater than 90°, and less than 180°, the functions of π — A may be substituted for them, subject to the changes of sign as indicated in Art. 15. Thus sin Asin (180° - A), cos A= =- tan (180° - A), etc. Hence, sin 91° 11' 12" cos 91° 11′ 12′′ = = = - cos (180- A), tan A sin (18091° 11′ 12′′) = sin 88° 48′ 48", cos 88° 48′ 48′′, etc. PROPERTIES OF PLANE TRIANGLES, WITH NUMERICAL EXAMPLES.* THE RIGHT-ANGLED TRIANGle. 20. Let A B C be a triangle, right-angled at C. Denote the angles by A, B, C, and the sides opposite to these, respectively, by a, b, c. Then by Art. 14, 1. In the triangle A B C, right-angled at C, there are given BC = 142, and the angle A = 26° 17' 19", to find the other parts. The angle B being the complement of A, we have also, and B=90 A 63° 42′ 41′′; by the preceding expressions (A) and (C). Calculation of b and c, by Natural Tangents, etc. (p. p. in the work means proportional parts.) A triangle consists of six parts, viz., three sides and three angles; and if three of these be given (one being a side), the triangle is completely defined. 2. Given BA = 467·817, and AC = 328 914, to find the other parts. By (B) of the preceding formulæ, bc sin B; this will find the angle B, as b and c are both given. Then A = 90° - B, and a = b tan A, or we may find a from the relation (Euc. i. 47), a = √ (c2 — b3). By Logarithms.* To find the angle B, we have log sin B = log b log c + 10. (Here 10 is added to the right-hand member of this equation, because in the Tables of Log. Sines, etc., 10 is added to each, to avoid negative indices, as explained in Art. 19.) Hence B = 44° 40′ 29′′ and A = 45° 19′ 31", the complement of B. a = √(c2 = b2) = √ (c + b) (c − b); In similar cases the equation ab tan A. 138 9032·1427116 log a = 2.5220116, or a= 332 668. result may be verified by the solution of the 3. Given c=176 21 (C being again the right angle), and B = 32° 15' 26"; to find A, b, and a. - The angle A = 90° - B = 57° 44′ 34", and be sin B, or log b = log c +log sin B 10. We take away 10 in this case, because the tables give log sin B + 10 for log sin B, as explained in the preceding solution. 176 212 2460306 = 9.7272276 Now, log P. p. for 26": = 868 (Taking 10 away) 1·9733450 log 94 0471.9733449 or b = 94.047. Again, a = c sin A, which gives a = 149 014. *Logarithms are employed in trigonometry, as in other branches of mathematics, merely to facilitate the calculations when the numbers are large. The results are obtained in the preceding example without the use of logarithms. The same result would be obtained by adding the quantities if the arithmetical complement of the log of c were employed. 4. Given A C=56.7816, B = 36° 7′ 18′′, C90°, to find the other parts. Ans. BC=77·8052, A B = 96 3213, A = 53° 52′ 42′′. 5. Given A B=12, BC = 15, B = 90°, to find AC and the angles without logs. Ans. A = 51° 20' 24" 68, C = 38° 39' 35" 32, A C 19.2093727. 6. Given a 101, b = 103, C = 90°; to find c and the angles A and B. Ans. c = = 144·257, A = 44° 26' 17" 8, B = 45° 33′ 42′′ 2. 7. Given a 379 628, A = 39° 26' 15", C=90°; to find the other parts. Ans. B = 50°33′ 45", b = 461.5504, c=597.6171. 8. In the triangle ABC right-angled at C, there are given AB 170 235, A = 44° 1' 10"; to find A C, C B, and the angle at B. Ans. A C = 122.416, B C = 118.297, B=45° 58′ 50′′. 9. Given BA 402.015, B 56° 7' 18" and C 90° to find the other parts. Ans. B C 224.0957, A C=333 7621, A = 33° 52′ 42′′. 10. Given b=31·76, A = 17° 12′ 51′′ and C=90, to find a, c, and Ans. a 9.8399, c= 33 249, B=72° 47′ 9′′. = B. = = = = THE OBLIQUE-ANGLED TRIANGLE. 21. To investigate a relation between the sides of a plane triangle and the angles opposite to them. Let A B C be a plane triangle; A, B, C as usual the angles, and a, b, c, the sides respectively opposite to these. Draw CD perpendicular to AB or A B produced; then by (B) of Art. 20, and (13) of Art. 15, A B sin B = sin A These properties are employed in the solution of a plane triangle, when two angles and a side, or when two sides and an angle opposite to one of them, form the data of the problem. If in the latter case the given angle be opposite to the less side, there are two triangles which fulfil the conditions of the problem Thus if C B, C A, and the angle A be given, then it will be obvious that when CB is less than CA, CB will meet A B in two points B and B', to the left of A, so that the angle A will be common to the two triangles A BC, A B'C. But if CB be greater than C A, B' will be to the right of A, and hence the angle A will not be common to B both triangles. B' This double solution is also indicated by the formula for the particular case. For since two sides and an angle opposite to the less are given, we find the sine of the angle opposite to the greater side, and as the sine of an angle is equal to the sine of its supplement, there is no reason, without other considerations, to prefer the acute angle found by the tables to the supplement of this angle. |