5. In the same manner, from (44) we obtain 7. Again, dividing the expressions for the sine in (46), the corresponding expression for the cosine in (45), we ain the following expressions for the tangents of half = angles: viz. 48. Whence Tan.¿A=√/}(S—a)(S—b)(S—c). 49. From the expressions in (48), is easily derived the following very convenient rule for calculating the three angles of a triangle when the three sides are given. RULE. Add the three sides together, and take half their sum; from half the sum subtract each side separately, then subtract the logarithm of the half sum from 20, and under the remainder write the logarithms of the three remainders; half the sum of these will be a constant, from which, if the logarithms of the three remainders be successively subtracted, the new remainders will be the logarithmic tangents of half the angles of the triangle. NOTE 1. The above rule has this advantage over all others, that the three angles are obtained with little more labour than one, and when the three angles are thus found independently, if their sum be 180°, the calculation is correct; if not, it must be examined till it prove. NOTE 2. In order to know which of the angles we have obtained, it is necessary to observe, that the constant -(S-a), gives tan. A, the constant (S-b), gives tan. B, and the constant (S—c), gives tan. C. NOTE 3. The angles might also be calculated from the expressions in (Arts. 44, 45, 46, or 47); but those in (44) are not suited to logarithmic calculation, those in (45 and 46) do not give the angles with the same accuracy in particular cases, since for a small angle the cosine varies slowly, and for an angle nearly 90° the sine varies slowly, and those in (47), as well as all the others, require an independent calculation for each of the three angles. EXAMPLE. Given the three sides of a triangle ABC; viz. AB 340, BC 380, and AC 360, to find the three angles. AB, c= 340 340 B α b 360 380 BC, a= 380 AC, b= 360 2/1080 Tan. A 32° 50′ 31-7"-9-8098945 { cont. - Log.(S-a)} Tan. B 29° 50′ 46′′ =9-7587415 {cont. - Log.(S-b)} Tan. C 27° 18' 42-3"-9-7129845 con'.-Log.(S-c)} Therefore A 65° 41′ 3·4", LB 59° 41′ 32", and C 54° 37' 24-6", the sum of which is exactly 180°. EXERCISES. 1. What are the three angles of the triangle ABC, AB being 100 yards, BC 150 yards, and AC 120 yards? Ans. LA 85° 27′ 34′′, ¿B 52° 53′ 28′′, and ≤C 41° 38′ 58′′. 2. The three sides of a triangle are AB 470, BC 398, and AC 420; what are the angles? Ans. LA 52° 45′ 49′′, LB 57° 9′ 22′′, and LC 70° 4′ 49′′. 3. The three sides of a triangle are AB 260·1, BC 140·4, and AC 190.5; required the angles? Ans. LA 31° 47′ 31′′, LB 45° 37′ 46′′, and ДC 102° 24′ 43′′. 4. The three sides of a triangle are AB 562, BC 320, and AC 800; required the angles? Ans. A 18° 21′ 24′′, LC 33° 34′ 47′′, LB 128° 3′ 49′′. 50. It is required to find expressions for the sine and cosine of the sum and difference of two angles, in terms of the sine and cosine of the angles themselves. Let CAD-a, BAC=b, then BAD=a+b, and it is required to find expressions for the sine and cosine of (a+b), in terms of the sine and cosine of a and b. From B draw BCLAC, and BEAD, and through C draw CF BE, and CDAD, then A FEDC will be a parallelogram, and' B F a E D FE will be=CD, and FC will be =ED; also since the right angled triangles AOE, BOC, have the right angles AEO, BCO, equal to one another, and the angles at O= being vertically opposite, the remaining angles OAE and OBC are equal, or the [CBF=a. Now, (6.), Sin. (a+b)= BE CD+BF CD BF = + AB AB AB AB = X + X ; hence (51.) Sin. (a+b)= sin. a cos. b+cos. a sin. b. Also cos. (a+b)= AE AD-CF AD = X = AB AC AB BC AB CF AB hence D 52. Cos. (a+b) cos. a cos. b — sin. a sin. b. Let now CAE=a, and CAB=b, then BAE will =(a—b); draw BC LAC, CD and BE each C AE, and F B BF to CD; then since DAC+ = is art, take away the common LACD, and there remains the LBCF the A4 CAD; .. the angle BCF=a, also Sin. (a-b)= BE CD-CF CD AB A Ο CF AB AB X ; therefore AB AC AB CB AB 53. Sin. (a—b)= sin. a cos. b— cos. a sin. b. AE AD+FB AD FB Also cos. (a—b)= = AB = + AB AD AC FB BC E = X + X ; therefore AC AB BC AB 54. Cos. (a-b)= cos. a cos. b+ sin. a sin. b. Collecting now these four results for convenience of reference, we have the expressions sought. 51 53 52 a Sin. (a+b)= sin. a cos. b+ cos. a sin. b. Sin. (ab) sin. a cos. b— cos. a sin. b. Cos. (a+b) cos. a cos. b— sin. a sin. b. 54 Cos. (a-b)= cos. a cos. b+ sin. ɑ sin. b. (51)+(53) |55| Sin. (a+b)+ sin. (a—b)=2 sin. a cos. b. (51)-(53) 56 Sin. (a+b)— sin. (a—b)=2 cos, a sin. b. (54)+(52) 57 Cos. (a-b)+cos. (a+b)=2 cos. a cos. b. (54)–(52) 58 Cos. (a-b)-cos. (a+b)=2 sin. a sin. b. Let now a+b=S, and a-b-d, then a=(S+d), and b=(S-d). Substituting these values in the last four, they become The last four expressed in words are as follow:59. The sum of the sines of two angles, is equal to twice the sine of half their sum, into the cosine of half their difference. 60. The difference of the sines of two angles, is equal to twice the cosine of half their sum, into the sine of half their difference. 61. The sum of the cosines of two angles, is equal to twice the cosine of half their sum, into the cosine of half their difference. 62. The difference of the cosines of two angles, is equal to twice the sine of half their sum, into the sine of half their difference. NOTE. Since S and d are any two angles, we may write instead of them a and b, by writing at the same time (a+b) for (S+d), and (a—b) for (S-d); or instead of S write A, and instead of d write B, then (S+d)=4(A+B), and §(S—d)=1(A—B); whence sin. A+ sin. B 2 sin. (A+B)cos. (A—B) (59)÷(60) gives 63 tan. (A+B) = tan. (A—B)* sin. A- sin. B 2 cos. (A+B)sin. (A-B)* By dividing both numerator and denomi nator of the second side by 2 cos. (A+B) cos.¿(A—B). In Arts. (51, 52, 53, 54), let the angles a and b be equal, then a+b=2a, and a—b—0. sin. 2a 2 sin. a cos. a. |