DETERMINATION OF THE SPECIES, 1. AD is acute or obtuse, according as Ac is of the same, or of different species with the Z.A. B 2. CD is of the same species as the ZA; therefore, according as the ZA and DB are of the same, or of different species, Bc is acute or obtuse. SOLUTION. 1. In the triangle ADC, find AD. + Thus, rad X cos ZA-cot Acx tang AD. Cot Ac : rad.: cos Z A : tang AD=74°.40' acute. Then AB-AD=46°.7'-DB acute. (G) Case IX. Given two angles, and the side adjacent to both of them, to find a side opposite to one of the given angles. DETERMINATION OF THE SPECIES. 1. The ZACD is acute or ob- C. tuse, according as the Z A is of s 3...o.......” the same, or of different species %. # with the side Ac. A. I. To . ...’ AN i. 2. The ZA is of the same species as CD; therefore, according as the ZA and the Z. BCD are of the same, or of different species, Bc is acute or obtuse. (H) CASE X. Given two angles, and the sides adjacent to both of them, to find the other angle. DETERMINATION OF THE SPECIES. 1. The Z. AcD is acute or obtuse, according as the Z. A is of the same, or of different species with the side Ac. - Ti, - IB .2. The angles A and B are of the same, or of different species, according as the ZACD is less or greater than the ZAcB. (I) CASE XI. Given the three sides, to find the angles. The side Ac- 80°. 19' • Given & The side BC= 63°.50' Required the angles A, - B, and c. The side AB = 120°.47' The sum of the sides Ac and Bc being less than 180°, the perpendicular falls nearest to the less side BC; therefore if it falls without the triangle, the rule determines ED in fig. 2. tang # AB = tang 60°.23.30"= - - 1024544 : tang # (Ac--Bc)=tang 72°.4.30"- 10:49016 :: tang # (Ac-Bc)=tang 8° 14'.30"= 9:16090 : tang ED=tang 14°.16.35"= - - 9:40562 Here ED is less than half the 4... ...Q....D. base, therefore the perpendi- ‘. ÖN } cular falls within the triangle, ‘. - A and consequently the angles A - & AS 2p and B are acute. & B DTE, F. The segment Ad=# AB+ ED=74°.40', and DB-46°.7'. SECONDLY. In the triangle ADC, the segment AD and the hypothen use Ac are given to find the ZA-51°.31', and the Z. Acd. . • OR Thus, Let Bc be considered as the base. Then, the sum of the sides AB and Ac being greater than 180°, the perpendicular falls nearest to the greater side AB; R * therefore if it falls without the triangle, the rule determines Ed in fig. 2. tang # Bc-tang 31°.55'.- - - 9°79438 : tang (AB+Ac)=tang 100°.33'= 10.72992 :: tang # (AB-Ac)=tang 20°. 14'- - - 9:56654 : tang Ed or Ed=tang 72°.32'- - 9:50208 Here the fourth propor- A tional is greater than half the base, therefore the perpendicular falls without the triangle. SECON DI.Y. In the triangle AdB, or ADB, the Z B may be found=59°.17'. And in the triangle Acd, or Aca, the z c may be found=131°.28'. w - o / Given #: 3. Required the sides Ac, BC, - - and AB. The Z. C-131°.30' Let the angles A and B, which are of the same species, be considered as base angles, then the perpendicular will fall within the triangle. And since the sum of the two angles A and B is less than 180°, the sum of the sides Ac and Bc is less than 180°, therefore the perpendicular falls nearest to B, the greater angle. Or Thus, Let Bc be considered as the base. . . The base angles being here of different species, the perpendicular falls without the triangle. And because the sum of the two angles B and c is greater than 180°, the perpendicular falls nearest to the less angle B, or greatest side AB, consequently the rule determines the Z EAd. Cot # (h+c)=cot 95°.23'-, - 897421 d.................A......D. : tang # (c.—B)=tang 36°.7'- 9-86312 : “; ::tang # Z BAc=tang 25°45'-9"68336 %. : tang Z. EAd=tang 75°.1'= 10:57227 B CHAP. IX. PRACTICAL RULES For SOLVING ALL THE DIFFERENT CASES OF OBLIQUE-ANGLED spher ICAL TRIANGLES, WITHOUT A PERPENDICULAR, witH THEIR APPLICATION BY LOGARITHMS. CASE I. (L) When two sides and an angle opposite to one of them are given, to find the rest. RULE. 1. To find the other opposite angle. Sine of the side opposite to the given angle, Is to sine of its opposite angle. |