PROB. II. Fig. 16. 17. In an oblique angled plane triangle, of the three sides, and any two of the angles, any three being given, to find the other two and the remaining angle. The four cases of this problem may be resolved by means of the foregoing propositions, as in the following table; observing, that when two angles of any plane triangle are given, the third is also given, being the supplement of the sum of the two given angles; and if one of the angles is given, the sum of the other two is also given. When the three angles only are given, the sides cannot be found from them, but their ratios may, being the same with those of the sines of their opposite angles. Case. Given. Sought. Solution. 1. 2. A, B C=180°-(A+B) AC sin C: sin A :: AB: BC (2.) sin C: sin B :: AB: AC. с AC: AB :: sin B : sin C (2.) A=180°—(B+C) BC sin B:sin A:: AC: BC. B AB+AC: AB-AC :: tan AC С {(C+B) : tan {(C-B) (3.) A BC then by Lem. 3. C=} (C+B) Two sides Two angles + (C- B); and B = and the in-and the in- (C+B)—!(C—B) cludedangle.cluded side. Cos }(C—B): cos }(C+B) :: AB+AC: BC (4.) Or, sin 1 (C—B): sin A (C+B) :: AB-AC: BC. 3. Case. Given. Sought. Solution. 4. AB A Let P=AB+AC+ BC; then AC B {P.(1 P-- BC):({P-AB). BC С (1 P-AC) : : rad> : tana The three The three BAC (7.) sides. angles. or, AB.AC : {P.({P-BC) : : rad? : cosa _BAC (8.) (P -- AB) : : rad" : sin? Other solutions of the last two cases are sometimes used. For example, Case 3. may be resolved by Prop. V, (fig. 18.) Thus : BA: AC:: rad: tan ABC. Rad: tan (ABC-45°):: tam A(B+C): tam A(BẠC.) Now, since }(B-C) is thus found, and 1(B+C) is given, B and C may be found by Lemma III. Then, sin C: sin A :: AB: BC. And Case 4. may be resolved by Prop. X, (fig. 12. 13.) Thus: BC: CA+AB::CA-AB: F. Then F is either the difference or sum of CD, DB, the segments of the base, according as F is less or greater than BC. In either case, the sum of CD, DB, and their difference being given, CD and DB may be found by Lemma III. Then, CA: CD:: rad: cos C. BA: BD:: rad: cos B. Thus C and B are found, and consequently A. * SECT. III. THE CONSTRUCTION OF THE TRIGONOMETRICAL CANON. The Trigonometrical Canon is a set of tables exhibiting the lengths of the sines, tangents and secants of every arch of a quadrant, containing an exact number of minutes, (or seconds,) in parts of the radius, the radius itself being 1: so that the sine, tangent and secant of any arch or angle may be found directly from such tables; and conversely, any sine, tangent and secant being given, the arch or angle which it expresses may also be readily found. The trigonometrical canon may be constructed by means of the following propositions: PROP. I. THEOR. Fig. 1. Any two sides of a right angled triangle being given, the other side may be found. For, since (47. 1.) AC=AB2+ BC; therefore AC— BC2 = AB”, and AC — AB’ = BC? ; and by extracting the square root, AC will be = v(AB+ + BC%); AB=v(AC?BC%); and BC= V(AC-AB-). The sine (DE) of an arch (DB) and the radius (CD) being given, to find the cosine (DF). Since, in the right angled triangle CDE, the two sides CD and DE are given, the other side CE, (which is equal to DF the cosine,) will, by the preceding proposition, be = N(CD: -DE-). EXAMPLE Given the radius CD=1, and DE, the sine of an arch of 30°=\, to find DF, the cosine of 30°. DF=CE=v(CD:-DE)= V(1—25)=V75= .8660254 =cos 30o. The sine (DE) of any arch (DB) being given, to find (DM or BM) the sine of half that arch. Since DE and the radius DC are given, CE may be found, (by Prop. 2.) and also EB, which is the difference between the radius and the cosine CE. Then, in the right angled triangle DBE, the two sides DE and EB being given, the other side DB may be found, (by Prop. 1.); the half of which DM or BM is the sine of the arch DL or LB=) half the given arch BD. |