As in the Oblique angled Triangle ABC, whose Angles at B Fig. and Care both Acute, the Perpendicular AD shall fall within XXII. the Triangle: For, if it fall not within, it must be the fame with one of the Sides, or elle it must fall without the Triangle: If it be the fame with either of the Sides, then the angle at Bor C must be a Right Angle; which is contrary to the Propofition: Ifit fall without the Triangle, as fuppofe at E, then the Angle AEB shall be a Right Angle: But the Angle AB E is Obtufe, for it is the Complement of the doute Angle ABC, and therefore the Side A E is greater than a Quadrant: And the Angle ACE be- Fig. ing Acute, A E thall be alfo less than a Quadrant: But, that XXII. the fame Side Thould be both More and Less than a Quadrant, is abfurd: And therefore, in this Cafe, the Perpendicular fhall fall within the Triangle.. But, In the Triangle A E B, Obtufe-angled at B, and Acute at E, the Perpendicular AD shall fall without the Triangle upon! the Side EB, contrived: Or, if otherwise, it must be the fame with one of the Sides, or fall within the Triangle: It cannot be the fame with either of the Sides, for then the angle at B or E fhould be a Right Angle: And it cannot fall within the Triangle, because then the Angles at B and E must either be both Obtufe, or both Acute, as hath been already proved. If therefore the Angles at the Bafe be of different Affections, the Perpendicular shall fall without; as was to be proved. However this Perpendicular falleth, it must be always oppofite to a known Angle; and for better Direction herein, take this General Rule. RULE II. Let your Perpendicular fall from the End of a Side given, and adjacent to an Angle given. As in this Triangle A BC, if there were given the Side A B, and the Angle at A; by the former, and this, Rule, the Perpen dicular must fall from B upon the Side A. C. But if there were given the Side A C, and the Angle at A, the Fig. Perpendicular must fall from C, upon the Side AB, continued XΧΙΙΙ. to E-And to know whether the Side upon which the Perpen dicular shall fall, must be continued or not, is no more than to ask whether the Perpendicular must fall within or without the Triangle. But, if the former Directions be not fufficient, the Calculation will determine it. For, RULE 70 1 Janua Mathematica. XXIII. Fig. RULE III. If the Ark found at the first Operation (whether be of a Side or an Angle) be more than the Arch given; the Perpendicular shall fall without; if less, within the Triangle. And this will plainly appear in the Solution of the fol lowing Cafes. General Rules to be observed in the Second Operation of the Solution of Oblique angled Spherical Triangles, when they are reduced into Two Right-angled Triangles. XXIV. After the first Operation, whereby either the Segments of the Base, or the Angle at the Cathetus, or Perpendicular, is found; a diligent Care being had to the Addition or Substraction of them: The second Operation will be performed by one of the Four following Rules. Fig. RULE I. The Sines of the Complement of the Hypotenuses, to the Sines of the Complement of the Bases, are in dirett Proportion. So, csBA:csDA::csBC: cs DC. RULE II. The Sines of the Bases, to the Tangents of the Angles at the Base, are in Reciprocal Proportion: So, SBA:SDA::ctB:ctD::B: D. RULE III. The Sines of the Complement of the Angles at the SBCA:SDEA::csB:cs B: cs D. RULE IV. The Tangents of the Hypotenuses, to the Sines of the Complement of the Angles at the Cathetus (or Perpendicular) are in Reciprocal Proportion: So, csBCA: ESDCA::ctBC:DC:::DC:BC. CHAP. CHAP. V. The Solution of Oblique-angled Spherical Triangles: By letting fall a Perpendicular, whereby the Oblique Triangle is Reduced into Two Right-angled. CASE I. In the Oblique-angled Spherical Triangle A BC, there is given, the Two Sides A B and BC, with the Anglè C, oppofite to BA; to find the Angle at B: If it be also known whether the enquired Angle be Acute or Obtufe. Analogy. AssBA:SBC::SC:s A. For by the Universal Proposition, (1.) As Rad. SAB::SA:s BD. S (2.) As Rad.; BC::SC:sBD. Therefore, Ass AB:SBC::SC:SA. CASE II. Two Angles, A and C, and the Side BC, opposite to the Angle A, given; To find the Side BA: If it be known whether the enquired Side be more or less than a Quadrant. Analogy. Ass As C::$BC:s BA.. CASE III. In the Oblique-angled Spherical Triangle A CD, there is given, the Two Sides AD and A C, with the Angle DAC, contained by them: To find the third Side DC. Fig. XXV. In this Cafe the Perpendicular may fall from the Extremity of Fig. either Side, but oppofite to the Angle given. Analogies. (1.) Asct. AC: Rad. ::cs. DAC: 1. AB. There XXVL |