34. Suppose A + B + C = 90°, then tan c = cot 1 or, tan a tan c + tan в tan c = tan A tan в (by mul.) or, 1 tan A tan B+tan A tan c + tan в tan c (by trans.) and cot A Cot B cot ccot A + cot B + cot c.... (6.) And this formula will obtain, in like manner, so long as A + B + C = (2n + 1) 90o. 35. Substituting for tan, its value preceding, it becomes sin COS in equa. (4) sin A sin B sin c + + COS B COS A whence, taking away the denominators, there results, sin A COS B COS c + sin B COS A COS C .... (7.) + sin C COS A COS B = sin A sin в sin c 36. By adding 3 sin A sin B sin c to both sides of equa. (7) we have B - sin ccOS (B — A) sin (B+C—A) sin (CB+A) sin (B+C—A) 4 sin A sin в sin csin A COS B COS C + sin A sin B sin c + sin B COS A COS C + sin в sin A sin c + sin c cos A COS B + sin c sin a sin B sin A cos (c. - B) + sin B COS (CA) + sin (A+C-B) + sin (A ·C + B) + + sin (B-C+A) + sin (C+B — A) + = sin (A+C—B) + sin (A+B − C) + =sin 2B + sin 2c + sin 2a .... (8.) Let A': = a + 90°, b′ = B + 90°, c′ = c + 90°, then we shall have - 4 cos A' COS B' Cos c' = cos 2A+ COS 2B + cos 2c.. (9.) The formulæ 7, 8, 9, are due to M. Mello: the last applies to the case where A' + B′ + c′ = (2n + 1) 90o. 37. We may now add all that is requisite in the solution of rectilineal triangles: and shall first take the case. where, instead of having two sides a, b, and their in cluded angle c given, the logarithms of those sides are given; as frequently happens in geodesic operations, and in astronomical tables for the distances of the planets from the sun. Here if a and b are regarded as the sides of a right angled triangle, in which a denotes the angle opposite to the side a, we shall have tan &= But, since a is supposed greater than b, this angle will be greater than half a right angle, or will be measured by an arc greater than 45°. a Hence, because tan (a 45°) = -- tana- tan 45° 1+tan a tan 450 and because tan 45° R 1, we have a tan (∞ — 45°) = (— — 1) Consequently, from equa. (3) - b tan (A — B) tan (A — B) a+b tan (A + B) cot c tan (AB) cot ctan (a- 45°).... (10.) Hence results the following practical rule: Subtract the less from the greater of the given logs, the remainder will be the log. tangent of an angle: take 45° from that angle, and to the log. tangent of the remainder add the log. cotangent of half the given angle, the sum will be the log. tangent of half the difference of the other two angles of the plane triangle. 38. Let us next return to the equation (1) in art. 11 of this chapter; namely, a = b cos c + C COS B b a cos c + C COS A ca cos B + b cos A The first of these equations being multiplied by a, the second by b, the third by c, and each of the equations thus obtained, being taken from the sum of the other two, there will arise b2 + c2—a2=2be cos A, whence cos A = b2 + c2-a2 2bc a2+ c2—b2 a2+c2-b2=2ac cos B, a2+b2-c2=Zab cosc, COS B 2ac a2+b2-c2 (11.) These equations are, obviously, equivalent to prop. 17, chap. ii. They may be expressed in a form rather better suited to computation, thus: Where, since the expressions are perfectly symmetrical, those for cos B, cos c, need not be put down. 39. Substituting in equa. (12), for cos a its equivalent 2 cos2 A cos A = 1, we have bc If, in equa. (11) we substitute 1-2 sin2 A for Cos A, we shall have If, again, equa. (14) be divided by equa. (13), there will result, tan A = ..(15.) 40. Of the rules for the determination of the angles of a plane triangle when the three sides are given, comprehended in the formula, marked 11, 12, 13, 14, 15, the last three are the best in practice, except the sides are integers and lie within the compass of a table of squares. When the angle sought is small, it is usually better to employ the method of equa. (14) than that of equa. (13). The method of equa. (15) is tolerably commodious, and very correct, except when a is either very small or near 180°. In some cases where great accuracy may be required, the student may wish to obviate the uncertainties that would arise from the use of some of these rules. For this purpose Dr. Maskelyne has given, in the introduction to Taylor's Logarithms, the following rules in reference to the sines and tangents of very small arcs. 1: To find the sine. To the logarithm of the arc reduced into seconds, with the decimal annexed, add the constant quantity 4.6855749, and from the sum subtraet one third of the arithmetical complement of the log. cosine, the remainder will be the log. sine of the given arc. 2. To find the tangent. To the log. arc and above constant quantity, add two-thirds of the arithmetical complement of the log. cosine, the sum is the log. tangent of the given arc. 3. To find the arc from the sine. To the given log. sine of a small are add 5.3144251, and of the arith. comp. of log. cosine: substract 10 from the index of the sum, the remainder will be the logarithm of the number of seconds and decimals in the given arc. 4. To find the arc from the tangent. To the log. tangent add 5.3144251, and from the sum subtract of the arith. comp. of log. cosine; take 10 from the index, and there will remain the logarithm of the number of seconds and decimals in the given arc. 41. Having at the end of the 2d chapter adverted briefly to the method of constructing a table of natural sines and tangents, the subject need not be resumed here. We shall merely subjoin the values of a few of the sines, cosines, &c in surd expressions, and a few formulæ of verification, such as may readily be reduced to identical equations. Now, to radius 1, we have sin 15° sin 18° sin 22° = = cos 75° = {√2 = √3 = {( √6 − √2) =cos 72° = (−1 + √5) cos 6702-√2 sin 30° cos 60° sin 72° = cos 18° = sin 75° = cos 15° = √2 + √2 10+2√5 √2 + √3 = { ( √6 + √2). From these, by means of the formula (G), (1), (N), &c. the sines of any other arcs proposed may be computed: from these, also, may be deduced the tangents and secants. The secants may be √6-√2 =2+ √3, computed from either the tangents or from the sines. Thus, from the tangents, since sec=√(tan2 + rad2), we have sec 15° (2√3)2+12= √8—4√3=√6−√2. sec 75° (2+ √√3)2+12 = √8+4√3=√6+ √2. =√(2+ 2 sec 671°=√(√2+1)2+12 = √4+2√2• |