Ex. Find log 7994.3726– M. of log 79943 = .9027804 381 1'i , 6 = 32 :: M. of log 79943726 = .9027843 Hence, log 7994-3726 = 3.9027843. 28. Having given the logarithm of a number to find the number. After the explanations of Art. 26, the method of working the following examples will be easily understood : Ex. 1. Find the number whose logarithm is 1.9030173. Taking from the tables the mantissæ next above and below, we have •9030194 = M. of log 79987 54 = D. Again,-9030173 = M. of log N..................... Hence, subtracting (1) from (7) 33 = d, the difference between the logarithms of the required number and the next lower. Now * = •61, the difference between the next lower number and the required number. Hence •9030173 = M. of log 79986.61 ; ... 7.9030173 = log •7998661 ; .. •7998661 is the number required. Ex. 2.* Find the value of (1.023)3 + (500123)* (1.32756)* We have— log N = 3 log 1.023 + 1 log •00123 – 4 log 1.32756. Now, 3 log 1.023 = 3 x .0098756 = .0276268 1 log •00123 = (3.0899051) = 1 (4 + 1.0899051) = 1.2724763 * The logarithms used in this example are taken from the tables. :: adding, 3 log. 1.023 + 1 log .00123 = 1:3001031 Again, M. of log 13275 = .1230345 and diff. for •6 = 196 ... 4 log 1.32756 = 4 x •1230541 = 4922164 Then, subtracting, log N = 2.5078867 Hence we have, -5078867 = M. of log N, and •5078828 = M. of log 32202; .. 39 = d. ,39 ... 2.5078867 = log •03220229. Trigonometrical Tables. 29. We use trigonometrical tables much in the same way as we do tables of ordinary logarithms of numbers. Tables have been formed of natural sines, cosines, &c., and also of logarithmic sines, cosinės, &c. It is with the latter only we shall now deal, though many of our remarks apply equally to the former. As the values of the natural sines and cosines of all angles between 0° and 90° are (Art. 11) less than unity, it follows (Art. 21) that their logarithms are negative. To avoid, however, printing them in a negative form, and for other reasons, it is usual to add 10 to their real value, and hence in using them we must allow for this. The same thing is also done in the case of logarithmic tangents, cotangents, secants, and cosecants. We generally express the true logarithmic sine by log sin, and the tabular logarithmic sine by L sin. Hence, we have, log sin A = L sin A - 10, log cos A = L cos A - 10, &c. It must be remembered in using the tables that, although (Art. 11) the sine, secant, and tangent of an angle increase as the angle increases from 0° to 90°, yet the cosine, cosecant, and cotangent diminish as the angle increases. Hence, when any angle is not exactly contained in the tables, we must add the difference in the case of a sine, secant, or tangent; but subtract it in the case of a cosine, cosecant, or cotangent. And, conversely, when the given logarithm is not contained exactly in the tables, we must in the case of the sine, secant, or tangent take out the next lower tabular logarithm as corresponding to the angle next lower; but in the case of a cosine, cosecant, or cotangent, we must take out the next higher tabular logarithm as corresponding to the angle next lower in the tables. We shall assume that small differences in the angles are proportional to the corresponding differences of the logarithmic trigonometrical ratios Ex. 1. Find L sin 56° 28' 24". = 9.9209393 Tab. diff. for 60" or D = 836 ... diff. for 24" or d = 24 : L sin 56° 28' 24" = 9.9209727 Ex. 2. Find L cos 29° 31' 28". Now L cos 29° 31' = 9.9396253 Tab. diff. for 60” or D = - 716 ... diff. for 28" = - 20 716 = - 334 TO X 836 334 .. Lcos 29° 31' 28" = 9.9395919 Ex. 3. Find the angle A, when I tan A = 9.8658585 We have 9.8658585 = L tan A. 883 = difference or d, 8 x 60" = 20". Hence, 9.8658585 = L tan 36° 17' 20". And And Ý x 60" = 2648 x 60" = 883 Ex. 4. Find the angle A, when L cot A = 10·0397936. We have, 10.0397936 = L cot A, 1834 = difference or d, no 1834 x 60" = 43". And 6 x 60" = 2537 * Hence, 10-0397936 = L cot 42° 22' 43". Ex. V. . 1. Given log 47582 = 4.6774427, and log 47583 = 4:6774518, find log 47.58275. 2. Given log 5•2404 = .7193644, and log 524.05 = 2:7193727, find log ·5240463. 3. Given log •56145 = 1.7493111, and log 56.146 = 1.7493188, find log 7.05614581. 4. Given log 61683 = 4.7901655, and log 616-84 = 2.7901725, find the number whose logarithm is 2.7901693. 5. Find the value of (1:05)15, having given log 1.05 = •0211893, log 20789 = 4:3178336, and log 20790 = 4:3178545. 6. Find the compound interest of £120 for 10 years at 4 per cent. per annum, having given log 1.04 = .0170333, log 14802 = 4:1703204, and log 14803 = 4.1703497. 7. A corporation borrows £8,630 at 41 per cent. compound interest, what annual payment will clear off the debt in 20 years? Log 1.045 = .0191163, log 4.1464 = .6176712, and log 4.1465 = .6176817. 8. Find the value of (1•032)10 x 337.62 76347275)6 , having given Log 1•032 = .0136797, log 34722 = 4.5406047. 9. Find L sin 32° 28' 31", having given L sin 32° 28' = 9.7298197, and L sin 32° 29' = 9.7300182. 10. Find L coseo 43° 48' 16'', having given L sin 43° 48' = 9.8401959, and L sin 43° 49' = 9:8403276. 11, Required the angle whose logarithmic cotangent is 10.1322449, having given L cot 36° 25' = 10.1321127, L cot 36° 26! = 10:1318483. 12. Construct a table of proportional parts, having given 163 as the tabular difference. 13. In what time will a sum of money double itself at 5 per cent. per annum, compound interest? 14. Find a when 1.034 = 1.2143, having given that log 1.03 = .0128372, and log 12143 = 4:0843260. 15. Solve the equation 22–1 – 40 = 9.2%, having given log 2 = •3010300. 16. Given L cos 32° 45' = 9.9341986, D = 752, find L cos 32° 45' 12", and L sec 32° 45' 20". 17. Given L tan 28° 38' = 10.2628291, D = 3003, find L tan 28° 37' 15", and L cot 28° 38' 42". 18. Find the angle whose logarithmic cosine is 9.9590635, having given L cos 24° 29' = 9.9590805, CHAPTER VII. PROPERTIES OF TRIANGLES. 30. The sines of the angles of a triangle are proportional to the opposite sides. We shall designate the sides opposite to the angles A, B, C, by the small letters a, b, c, respectively. |