L. cos a............8.0982928 L. sin .........9.9999563 L. sin (C4) 9.6998211 C — 4 = 33° 40′ 54′′ . 5 ....99 9 45.5 C=132 50 0.0 We could not take the supplement of 33° 40′ 54′′. 5 as the value of C; because this would have given us C200. It is plain, therefore, that the proposed problem admits only of one solution.* Such as wish to know the most useful applications of Trigonometry, cannot do better than consult M. Puissant's Traité de Topographie, d'Arpentage et de Nivellement (Treatise on Topography, Surveying and Levelling), Paris, 1807. END OF THE TRIGONOMETRY. APPENDIX, Containing the Solution of various particular Cases in Trigonometry. XCVI. THE manner of solving triangles, which has now been explained, leaves nothing to desire in regard to the universality of its application. There are some cases, however, in which it might be advantageous to employ particular solutions in place of the general ones, either for the purpose of abridging the calculation, or of rendering the results more accurate and more independent of errors in the tables. We shall now exhibit a few of these cases, selecting such as are of most frequent use, or as lead to the most remarkable formulas. We shall, as formerly, designate the angles of the proposed triangle, spherical or rectilineal, by A, B, C ; the sides which lie respectively opposite them by a, b, c. We shall likewise suppose the radius of the tables=1, an hypothesis, which cannot affect the universality of the results. The angles A, B, C are expressed in calculation, either by the degrees or the absolute lengths of the arcs which measure them, these arcs being taken in a circle whose radius is 1. If an angle or arc x is very small, then in place of sin x, and cos x, their values ex pressed in series may be employed, namely, sin x = x cos x = 1 3 x 1.2.3 + &c. 1.2+ &c; but in that case, a must be expressed in parts of the radius. An arc being found in parts of the radius, to obtain its value in minutes, it must be multiplied by the number of minutes contained in the radius; which number is logarithm being 3.80388012297. 20000 =6366.1977237, its § I. Of rectilineal triangles, two of whose angles are very small. XCVII. Suppose two angles A and B to be very acute, and consequently C to be very obtuse; we may put sin AA-A3, sin B= B-B3, and sin C = sin (A+B)=A+B-(A+B). Hence if the side c with the adjacent angles A and B are known, the other c sin A two sides may be found by the formulas a = c sin B sin (A+B)' b = sin (A+B which, by substituting the preceding values, and reduc and from this, results a + b. cc AB. These values are accurate, excepting such terms as contain four dimensions in A and B. XCVIII. In the second place, suppose that we have the two sides a and b given, with the contained angle C4, being very small. We shall first have c2=a+b2+2ab cos = a2+b2+2ab (1 — 102) = (a+b)2 ab 02; hence Next, the angle A will be found by the equation sin A== sinC= a sin e, from which, by substituting the value of c, and that of sin e, we α C we might deduce the value of B, by interchanging the letters a and b: but A being known, we have immediately B-A. If is given in minutes, in order to obtain A likewise expressed in minutes, we must A @ in place of A substitute, in the preceding formulas, the ratios and, R being the number of minutes contained in the radius. We shall thus have XCIX. To give an example of these formulas, put a = 1000 yds., b=2400 yds., C = 199° 32′ or ◊ = 68′; we shall have a+b➡c= = 0.037806, whence c = 3399. 962194 yds. Then by a first approximation, we have A = a @ =20', and B='—A= 19'. 99988946, and consequently B=48'. 00011054, values which are correct to the last decimal. § II. Solution of the third case of rectilineal triangles, by the method of series. C. The two sides a and b, with the included angle C being given, to find the angle B, we have the proportion b: a:: sin B: sin (B+C), which gives a sin Bb (sin B cos C + cos B sin C), and consequently sin B b sin C cos B a - b cos C = If in this equation, instead of the sines and cosines, we insert their values in imaginary exponentials (Art. 35.), we shall have Taking the logarithms of each member, and expanding the second into a common series by the formula e Hence, dividing by 2/-1, and observing that mC-1 -mc√1 -e =2√ 1 sin m C, we shall have This is the value of the angle B, expressed in parts of the radius, by a series, the law of which is very simple, and which will converge the faster, the less b is in proportion to a. The value just found must likewise satisfy the equation, tang (B+} C ) = a+b tang C, which is the same as CI. The angle B being known, we shall have the third angle A= 2000 -B-C. As for the third side c, it depends on the equation c2 = a2 - 2ab cos c+b2, which by extracting the root gives -b cos C+ b2 2 a - sin2 C+ b3 2a2 sing C cos C - &c. But this series has not a regular movement, and cannot be continued at will. On the other hand, a very simple series may be found for the value of the hyperbolic logarithm of c Thus, it is manifest that the quantity a2—2ab cos C+b2 = (a—be©√ —1) (a—be C—~—1), C for the developed product of these two factors gives a2 · ab (e C√ — 1 + e−C√−1)+b2, or a2 — 2 ab cos C+b2. Hence we have c2 = : ( a − b e C N − 1 ) ( a − be — C √ − 1). Taking the logarithm of each member, there will result Hence, by again reducing, according to the formula a series not less elegant than the one which gives the value of B. If the logarithms are required to be those of the common tables, we have only to multiply the different terms by the modulus 0.43429448. § III. Solution of the third case of spherical triangles by means of series. CII. In the preceding paragraph, we have shewn, that the value of x drawn from the equation tang x = tang C may be expressed m+ n Now, in a spherical triangle, when the two sides a and b, and the in |