Prob. 61. To draw a leffer Circle, e f g, parallel to the right Circle BD, at 53° from it. Take 53° from the Line of Chords, and lay them from B to e, and from D to g; then lay a Ruler from B to g, and it will cut the right Circle A C in f; thro' the three Points e, f, and g, by Prob. 17, draw e f g, the Parallel required. Prob. 62. To draw a lesser Circle, abc, parallel to an oblique Circle, Be D, at 22° diftant from it. Find p, the Pole of the oblique Circle Be D, by Prob. 43; a Ruler, being laid from D to p, will cut the primitive Circle in a; fet off 68°, the Complement of the given Diftance, A from a, both Ways, to d and e; a Ruler, being laid from D to d and e, will interfect the right Circle AC in b and g; then, by Prob. I, bifect bg in d B D p b, the Center, on which, with the Radius hb, or hg, draw the Arch a be, the Parallel required. Cafe 2. When the given Point, as A, is in the Periphery of the primitive Circle, fet off 90° from A, both Ways, to B and D; then draw the right Circle BOD, the Circle required. Cafe 3. When the given Point, as a, is neither in the Center or Periphery of the primitive Circle, thro' a and the Center of the primitive Circle draw a right Circle, A C, and another at right Angles to it, as BD; a Ruler, being laid from B.to a, will cut the primitive Circle in c; fet off 90° from c to d, and lay a Ruler from B to d, which will cut the right Circle AC in e; then thro' the three Points B, e, and D, by Prob. 17, draw the Circle Be D, which will be the required oblique Circle to the Pole a. Spherical TRIGONOMETRY. LAIN or right-lined Triangles are formed by the Interfection of right Lines: Spherical Triangles are formed by the Interfection of Circles: The Sides of these are Parts of those Circles; and fpherical Angles are measured by the Arch of a great Circle, intercepted between the Arches which conftitute the Angle., The Properties of Spherical Triangles are as follows: 1. The Sum of the three Angles of any spherical Triangle are greater than two right Angles, and lefs than fix right Angles. 2. The Side of a fpherical Triangle is less than a Semicircle. 3. The three Sides of a fpherical Triangle are less than a Circle or 360°. 4. The greateft Side always fubtends the greatest Angle. 5. In a right-angled fpherical Triangle, if the Legs or Sides containing the right Angle are, each of them, greater or lefs than 90°, then will the Hypothenufe be less than 90°. 6. But, if the two Legs are, one greater, and the other lefs, than 90°, then will the Hypothenufe be greater than 90°. 7. In a right-angled fpherical Triangle, if the Legs containing the right Angle are, each of them, greater than 90°, each of the Angles oppofite will be greater than 90°; but, if cach each of the Legs are less than 90°, each of the oppofite Angles will be lefs than go°. 8. If the Hypothenufe be more than 90°, the Sides.containing the Angles will be, one lefs, the other more, than 90°; and the Angles oppofite to thofe Sides will be, one lefs, the other more, than 90°. 9. If the Hypothenufe is lefs than 90°, the other Sides will be each more than 90°, or each lefs than 90°; and the Angles oppofite thofe Sides will also be each more, or each lefs, than 90°, as were the Sides. The Solution of right-angled Spherical Triangles. The Lord Nepier, the noble Inventor of Logarithms, difcovered a Method for folving right-angled spherical Triangles, which is called the Catholick Propofition: This being beft adapted for the Eafe of the Memory, I fhall endeavour to explain it fo that the Reader may meet with as little Difficulty as may be in this Part of the Work. The Hypothenufe, Bafe, Perpendicular, and the other two Angles, that is, the five Parts of a spherical right-angled Triangle, exclufive of the right Angle, are called the five circular Parts. Of these five circular Parts, if the two of them which are given, and the third which is to be found, are fo fituated that no other Part of the Triangle lies between them, it is called an Extreme conjunct: The right Angle is not reckoned as any Part, it not being one of the five circular Parts: And the Part that then lies in the Middle of these three Parts is called the middle Part, and the other two the Extremes conjunct. But, if between the two Parts that are given, and the third which is to be found, there lies any other Part except the right Angle, it is called an Extreme disjunct; and that Part, between which and the other Parts there lies either a Side or an Angle, is called the middle Part; and the two Parts lying together are called the Extremes disjunct. This Diftinction I advife the Reader to be careful to obferve. Rule 1. When it is an Extreme conjunct, and the middle Part is to be found: As the Radius Is to the Tangent of one Extreme, So is the Tangent of the other Extreme To the Sine of the middle Part. M m Note Note, That, if the middle Part, or either of the Extremis conjunt, be the Hypothenufe, or either of the oblique Angles, then, in the above Proportion, inftead of the Tangents and Sines, ufe the Co-tangents and Co-fines; but, if the Bafe, or Perpendicular, fhould be the middle Part, or the Extremes conjunct, then ufe only the Tangents and Sines. Rule 2. When it is an Extreme conjunct, and one of the Extremes is to be found: As the Tangent of the given Extreme So is the Sine of the middle Part To the Tangent of the required Extreme. The fame Note must be obferved here as in the firft Rule. Rule 3. When it is an Extreme disjunct, and the middle Part is to be found: As the Radius Is to the Co-fine of one Extreme, So is the Co-fine of the other Extreme To the Sine of the middle Part. Note, That, if either of the Extremes disjunct be the Hypothenufe, or either of the oblique Angles, in the above Proportion, instead of the Co-fine, use the Sine; but, if the Bafe, or Perpendicular, fhould be the Extremes disjunct, then ufe the Co-fine. Rule 4. When it is an Extreme disjunct, and one of the Extremes is to be found: As the Co-fine of the given Extreme Is to the Radius, So is the Sine of the middle Part To the Co-fine of the required Extreme. The fame Note is to be observed here as in the third Rule. Prob. 64. In the right-angled spherical Triangle ABC, right-angled at B, given the Hypothenufe AC 75°. 20′, and the Angle BCA 57°. 16', to find the Perpendicular BC. Conftruction. With the Chord of 60°, draw the primitive Circle EDBC; draw the oblique Circle CAD, by Prob. 48, fo fo as to make an Angle at C of O P E D In the spherical Triangle A B C there is given, befides the right Angle at B, the Hypothenufe A C 75°. 20, and the Angle at C 57°. 16', to find the Perpendicular BC: Thefe Parts having no Parts betwixt them, it is an Extreme conjunct, and AC and BC are the two Extremes, and the Angle at C is the middle Part; hence the prefent Cafe is an Extreme conjunct, and one of the Extremes is to be found: Therefore, to find the Perpendicular B C, by the fecond Rule, and the Note, as the middle Part is an oblique Angle, and one of the Extremes is the Hypothenufe, Co. Ar. 0.582158 IO. As the Co-tangent of A C, 75°. 20', The Perpendicular B C, being Part of the primitive Circle, may be measured with a Line of Chords, by Prob. 41; by which the Learner will fee the Agreement between the Conftruction and Calculation. Prob. 65. In the right-angled spherical Triangle ABC, right-angled at B, given the Hypothenufe A C 75°. 20', and the Angle at C 57°. 16', to find the other Leg, or Bafe, A B. The fame Things being given as in the laft Problem, the Conftruction is the fame: And here I advife the Learner to conftruct the Figure by the Directions there given, and to mark the Parts given with a-, and the Part required with an o; the Reader will find the Advantage of doing this at every Problem. The |