By the method, however, of right-angled trigonometry alone, it would be necessary always to calculate the perpendicular; but this unnecessary calculation is avoided by employing the analogies given in the table, p. 71, vol. II. Geometry, the application of which is explained in the following articles: Or, and The Third Case. 361. Let the given parts be a, b, and C; and let a perpendicular BD be drawn from angle B upon the side b; let be the segment of b that is nearest to C, reckoning from C towards A. To find angle A. Tantan a cos C, and tan A = Or, and tan a : tan ... rad: cos C sin (b — ◊) : sin = tan C : tan A Find by [1], and then A by [2.] When 6, the perpendicular is without the triangle, and b ◊ is negative; and hence also its sine is so (328); consequently, if tan C is negative, then tan A will be positive (336), and conversely, according to the rule in article. 336. - If b exceeds 90°, still the calculation is similarly performed. But as there are two perpendiculars (370) that fall without the triangle, the segment between C and the other perpendicular may be taken, and the value of 0 is found in the same manner, only for C must then be substituted TC in the two formulas, and b + taken instead of b—0. Or, as would now be reckoned in an opposite direction, if it be considered negative, the formulas above will apply without alteration, observing the remark in 328, and also that b- becomes b + 0. To find the side c. Using the same notation, tantan a cos C, and cos c = tan C sin sin (b-9) R: cos C tan a : tan ... These are exactly the analogies given in the table, p. 71, vol. II. Geom., and may be derived as there shown; or they may be proved by applying the analogies (335) of rightangled trigonometry to the two right-angled triangles formed by the perpendicular from B on b. Let BD = h, and suppose R = 1, then, by article 335 in the triangle BCD, cos C cot a• tan tan tan a cos a = cos • cos h, cos a and cos h = cos a cos ' cos The first and last of these formulas are just the formulas [3] and [4] given above; the second formula containing h is used merely for eliminating it from the third. The formulas for A are similarly found by means of the two triangles. 362. It is easily known, after 0 is found, if the perpendicular falls without the triangle, for then is b. The required parts could be found by merely using the principles of right-angled trigonometry; but then the perpendicular h would always require to be calculated, in the first triangle, before the parts of the other could be found; which would generally be an unnecessary trouble. But by eliminating the perpendicular h from the analogies thus obtained, as is done above, the preceding formulas are found. The angle B can be found exactly in the same manner as A, by supposing the perpendicular to be drawn from A upon the side a. The formulas for this purpose are easily obtained from those for A, by merely changing A into B, a into b, and b into a. 363. It frequently happens in the applications of this subject that only one of the unknown parts of a triangle is required, and much unnecessary computation is prevented by having formulas for finding the required part in terms of the given parts; such as those given above, by means of which any of the unknown quantities A, B, c, can be found independently of each other. Several other formulas are used for finding the side c directly; but that given above is as simple and concise as any of them. The Fourth Case. 364. Let the given parts be A, B, c, and let a perpendicular be drawn from B upon b; and let angle ABD = 0, then, Or, since Or, and Or, and Cos ccot cot A, tan a = 1 tan A' : cos ctan A: cot → [1] cos (B-): cos = tan c : tan a [2] By exactly similar formulas the side b is to be found. Cottan c⚫ tan A, cos C = sin (B-) sin : sin (B-) = cos A: cos C To find the side a. cot A = R: tan c⚫ cos cos (B cos c⚫tan Acot → ... ... sin A, or sin a: sin 6 ... The Fifth Case. 365. Let a, b, and A, be given, and let a perpendicular CD be drawn from C upon c; let the segment of c next to A be denoted by 0, and the opposite angle ACD by ; then, To find the angle B. [3] [4] sin b Sin B= sin A: sin B...[1] sin a When a is nearer to 90° than b, B has only one value, which is of the same species as b. When a differs more from 90° than b, then B has two supplementary values (358). To find the side c. Tantan bcos A, and cos (c— 0) = and R: cos Atan b: tan cos b: cos a = cos : cos (c―0) 366. The value of found above shows whether the perpendicular is within or without the triangle, as in article 361. Also the species of A and B determine this circumstance; for the perpendicular falls within or without the triangle, according as angles A and B are of the same or of different affection." are found, then c is known, for When and c c = c — 0 + 0. Or, and cos a cos b To find the angle C. Cot = cos b⚫tan A, and cos (C-☺) = tan a : tan b = cos : cos (C-) ... To find the side b. ... tan b tan a cos 0. [2] ... • cos →. [4] [5] 367. When B has two values, one of them, for instance its acute value, should first be taken, and the side c, and angle C of the triangle to which it belongs, are then to be calculated; then its other value being taken, the side c and angle C of the triangle to which it belongs are to be calculated. The Sixth Case. 368. Let A, B, and a, be given, and let a perpendicular be drawn as in the preceding case; then, sin a, or sin A: sin B sin a: sin b [1] sin B Sin b = sin A When A is nearer to 90° than B, 6 has only one value, and of the same species as B; in any other case, b has two supplementary values (359). ... Or, and To find angle C. Cotcos b⚫tan A, sin (C— 0) = Or, and R: cosb tan A : cot = cos A : cos B = sin ℗ : sin (C — ~) cos B cos A ... ... The species of C is not thus determined, as its sine is the fourth term; but those of B and a being known in triangle BCD, that of DCBC is known (329, 330). To find the side c. sin . tan A tan B [2] sin 0. R: cos A tan 6 : tan tan B: tan A = sin : sin (c—0) 0 The species of c- is not determined, as its sine is the last term; but in triangle BCD, the species of (c—0) is known from those of a and B. [4] [5] 369. When b has two values, one of them, for instance its acute value, can first be taken, and the angle C, and side c of the triangle to which it belongs, are then to be calculated; its other value being next taken, the angle C, and side c of the triangle to which it belongs, are to be computed. 370. The rules given in articles 358, 359, for determining the species of the parts, are proved in some treatises on analytical trigonometry, and they can be very easily proved by spherical geometry, with the aid of these simple theo rems: 371. The sines of an angle and its supplement are equal. When two arcs are each less than a quadrant, the sine of the greater is greater than the sine of the other; but when each of the arcs exceeds a quadrant, and is less than a semicircle, the sine of the greater is less than the sine of the other. When the magnitude of an arc is intermediate between that of another arc and its supplement, its sine is greater than that of either of the latter.' 372. All the forms of a triangle that can exist when two |