XXVI. As cs. AB:cs,AC;;cs.BD:cs.DC. CASE IV. Two Sides AC and DC, with the Angle ADC, opposite to AC given; To find the third Side A D. In this Cafe, Let the Perpendicular fall from the Concourse of the given Sides, upon the Side enquired, continued, if need be. (1.) As ct. CD: Rad.::cs. ADC: t. BD. (2.) As cs. BD:cs.CD:: Rad.: cs. CВ. (3.) As cs. CB: Rad.::csAC:csAB. Therefore, cs.CD:cs.BD:;cs. AC:cs. AB. } Triangle. CASE V. Two Sides C A and DA, and their contained Angle In this Cafe the Perpendicular may fall from the Extremity of either of the Sides, oppofite to the Angle given. (1.) Asct. AC: Rad.:: Cs. DAC: 1. AB. (2.) Asct. CAD: S.AB: Rad.: t. B С. Therefore, } Triangle. AssAB: ct. CAD: : SBD: ct. AD C. CASE VI. Two Angles, ADC and C A D, and the Side AC, opposite to one of them being given, To find the Side between them. If it be known whether the Side fought, or the Side oppofite to the other given Angle, be Acute or Obtuse. In this Cafe, Let the Perpendicular fall from the Extremity of the given Side, on the Side enquired, continued, if need be. Analogies. Analogies. (1.) As ct. AC: Rad. :: cs. DAC: t. A B. Therefore, As ct. CAD:S.AB::ct.ADC: s. B D. But, D B --- AB = A D in the 2d } Triangle. Fig. XXVI. CASE VII. Two Angles D AC and ADC, and the Side Fig. In this Cafe, Let the Perpendicular fall from the Angle en- Analogies. (1.) As ct. CAD: Rad.::cs. AC:ct. ACB. (3.) As cs. BC: Rad.::cs.BDC:s.B C D. Therefore, Ascs. CAB:s. ACB::cs.BDC:s. BCD. } Triangle. CASE VIII. Two Angles, ACD and CAD, with the Side Fig. between them, AC, being given; To find the third Angle D. XXVL In this Cafe, Let the Perpendicular fall from the Extremity of CajeVIII. the given Side, and oppofite to the Angle enquired. Analogies. ACB=BCD in the ist (1.) As ct. CAB: Rad.cs. AC: ct. ACB. And, ACD --- A C B = B C D But, ACD+ACB=BCD in the 2d (2.) As s. ACB:cs.CAB:: Rad. :cs. В С. (3.) As Rad.: cs.BC::s.BCD:cs. CDB. L Triangle. There Fig. XXVI. Cafe IX. Therefore, As S. ACB:cs.CAB::s.BCD:cs. CDB. CASE IX. Two Sides, A C and CD, with the Angle ADC, opposite to one of them, (viz. A.C) being given; To find their contained Angle AC D. If it be known, whether the enquired Angle, or the Angle opposite to the other given Side, be Acute or Obtufe. In this Cafe, Let the Perpendicular fall from the Angle ens quired. Analogies. (1.) As ct. CDB: Rad. ::cs.CD:ct. BCD. (2.) As ct. CD:cs.BCD: : Rad. : t. B С. (3.) As Rad.: t. BC::ct. AC:cs. ACB.. Therefore, As ct. CD:cs.BCD::ct.AC: cs. ACB. But BCD ACB ACD in the 2d } Triangle.. Fig. CASEX. Two Angles, DAC and ACD, with the Side beXXVI. tween them, A C, being given; To find either of the other Cafe X. Sides, DC or AD. In this Cafe, Let the Perpendicular fall from the Concourse of the Side given and fought, on the third Side, continued, if need be. Analogies. (1.) As ct. CAB: Rad. :: cs.AC:ct. A C B. Therefore, } Triangle. As cs. BCA: ct. AC::cs. BCD:ct. CD. CASE * CASE XI. The Three Sides being given ; To find an Angle. Fig. For the refolving of this Problem, there must be fome Prepara- Cafe XI. ANALOGY. As the Tangent of half the Base, Is to the Tangent of half the Sum of the other Two Sides; So is the Tangent of half the Difference of those Sides, To the Tangent of half the Difference of the Segments of the Base. Thus then, In the Oblique-angled Spherical Triangle ACD, there is given, Analogies. As the Tangent of half A D, Is to the Tangent of half A C and AD; So is the Tangent of half the Difference of A C and A D, To the Tangent of half A E. And half AD+ half AE = AB in the ist But half A E-half AD = BE or B D in the 2d Hence, to find the Angle at A. As Rad.: ct. AC::t. AB:cs.CA В. Fig. }Triangle. XXVII. CASE XII. The Three Angles being given; To find a Side. Fig. XXVI. This Cafe is but the Converse of the last beforegoing, and is to Cafe XII. be Solved after the fame manner: If fo be that we convert the An- L2 And Fig. And for the Proportions or Analogies for the Resolving of this XXVII. Cafe there are also Two Ways proposed, to which I referr you. CHAP. VI. The foregoing XII Cafes, of Oblique-angled Spherical Triangles, Geometrically Demonstrated, and Resolved, with... out letting fall a Perpendicular. F : OR the Performance of what is here promised, these Six THEOREM I. In any Spherical Triangle, whether Right or Oblique-angled.. The Sines of the Angles are Proportional, to the Sines of their op-pofite Sides: & contra. This is demonftrated in Seft. 1. Chap. 4. of Right-lined Triangles, and is the fame in Spherical Triangles, whether Right or Oblique Angular. THEOREM II. In all Oblique-angled Spherical Triangles, whose Three Sides together are less than a Semicircle, or 180 Degrees. As the Sine of half the Sum of the Angles at the Base, Is to the Sine of the half Difference of those Angles; So is the Tangent of half the Base, To the Tangent of half the Difference of the Sides.. And also, As the Co-fine of half the Sum of the Angles at the Bafe, So is the Tangent of half the Base, To the Tangent of half the Sum of the Sides. DE |