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′′. Tab. diff. for 60" or D = 836 == 9.9209393 .. Lcos 29° 31 28/ Ex. 3. Find the angle A, when L tan A = L-tan A. Ex. 4. Find the angle A, when L cot A - = L cot A, 10.0397936. Next higher, 10-0399770 = L cot 42° 22′, 1834 difference or d, = Also, Hence, 10-0397936 = L cot 42° 22′ 43′′. = = 3. Given log 56145 17493111, and log 56.146 = 1.7493188, find log 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 104 0170333, log 4.1703204, and log 14803 4.1703497. 14802 = = = 7. A corporation borrows £8,630 at 41 per cent. compound interest, what annual payment will clear off the debt in 20 years? 9. Find L sin 32° 28′ 31′′, having given Lsin 32° 28' 9-7298197, and L sin 32° 29' = 9-7300182. 10. Find L cosec 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 x when 1-03 1-2143, having given that log 1.03 = 0128372, and log 12143 = 4.0843260. 15. Solve the equation 2a—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 Ltan 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 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. Draw AD perpendicular to BC, or to BC produced. It follows, therefore, from the symmetrical nature of this Taking the figures of the last article, we have (1.) When C is an acute angle By Euc. II., 13, AB2 CD = BC2+ AC2 - 2 BC. CD. Now = cos C, or CD = AC cos C. AC Hence we have, AB BC2 + AC2 2 BC. AC cos C, or c2 = a2 + b2 = 2 ab cos C. (2.) When C is an obtuse angle, as in the second figure— By Euc. II., 12, AB2 = BC2 + AC2 + 2 BC. CD ; and CD AC cos ACD = = AC cos (180° C) BC2 + AC2 + 2 BC ( = a2 + b2 2 ab cos C. a2 + b2 c2 2 ab = AC cos C. AC cos C); as before. 32. To express the sine of any angle of a triangle in terms (a+b+c) (b + c − a) (a + c − b) (a + b −c). (2 bc)2 Hence, taking the square root, and taking the positive sign, because (Art. 11) the sin A is always positive when A is the angle of a triangle, we have— √(a + b + c) (b + c − a) (a + c b) (a + b −c). Let a + b + c = 2 s, then b + c a = 2 (s − a), b), a + b c = 2 (sc). 2 (8 - b), a + b 1 2 bc 2 bc √28.2 (sa).2 (s - b). 2 (s — c) a) (s – b) (s – c). |