Laftly, in the rightangled fpherical Triangle ABC, let there be given the Hypothenufe AC = 60°, and the Angle A = 23° 29'; to find the Base and Perpendicular. Then (by Theor. 1. p. 23.) the Operation will be as follows. Having exhibited the Manner of refolving all the common Cafes of plane and fpherical Triangles, both by Logarithms and otherwife; I fhall here fubjoin a few Propofitions for the Solution of the more difficult Cafes which fometimes occur; when, inftead of the Sides and Angles themselves, their Sums, or Differences, &c. are given. PROPOSITION I. The Sine, Co-fine, or verfed Sine of an Arch being given, to find the Sine and Co-fine, &c. of half that Arch. From the two Extremes of the Diameter AB, let the Chords A E and BE be drawn, and let the Radius CQ bifect AE,perpendicularly, in D (Vid. 1. 3.); then will AD be the Sine, and CD the Co-fine, of the Angle ACD, or ACE. = But 4AD2 A E2 (by Cor. 1. to 5.2.) = A B x AF (by Cor. to 11.4.) = 2AC x AF; whence AD2 =AC x AF: Alfo 4CD BE AB × BF = 2 AC x BF; whence CD'ACxBF. From which it appears, that the Square of the Sine of half any Arch, or Angle, is equal to a Rectangle under balf the Radius and the verfed Sine of the Whole; and that the Square of its Co-fine is equal to a Rectangle under balf the Radius and the verfed Sine of the Supplement of the whole Arch, or Angle. PROP. II. The Sines and Co-fines of two Arches being given, to find the Sines, and the Co-fines, of the Sum and Difference of thofe Arches. Let AC and CD (=BC) be the two propofed Arches; let CF and OF be the Sine and Cofine of the greater AC, and let m D (Bm) and Om, be thofe of the leffer CD (or BC): Moreover, let DG and OG be the Sine and Co-fine of the Sum AD; and BE and OE, thofe of the Difference AB. Draw mn parallel to CF, meeting AO in n; alfo draw mu and and BH parallel to AO, meeting GD in v and H: Then it is plain, because DmBm, that Dv is = Hv, and munG= En; and that the Triangles OCF, Omn and mDv are fimilar; whence we have the following Proportions, Now, by adding the two first of these Equations together, we have mn+Dv x OC (DG × OC) = OmxCF+DmxOF; whence DG is known, Moreover, by taking the latter from the former, we get mn-Dv x OC (BE x OC) Om x CF — Dm x OF; whence BE is known. In like manner, by adding the third and fourth Equations together, we have On+mv x OC (OEx OC) = Om × OF + Dm × CF; and, by fubtracting the latter from the former, we have On-mv × OC = OČ (OG x OC) Om × OF Dm x CF; whence OE and OG are alfo known. 2. E. I. COROLLARY I, Hence, if the Sines of two Arches be denoted by Sands; their Co-fines by C and c; and Radius by R; then will COROLLAR Y 2. Hence, the Sine of the Double of either Arch 2CS (when they are equal) will be = and its Co fine= C2 -S R : Whence it appears that the Sine of the Double of any Arch, is equal to twice the Rectangle of the Sine and Co-fine of the fingle Arch, divided by Radius; and that its Co-fine is equal to the Difference of the Squares of the Sine and Co-fine of the fingle Arch, alfo, divided by Radius. COROLLARY 3. X Moreover, because Dm x CFOC xmv (OC XEG=+OC × OE-OG); and Om x OF=OC x On (OC x 2On OC x OE+OG), it follows, that the Rectangle of the Sines of any two Arches (AC, CD (BC) is equal to a Rectangle under half the Radius, and the Difference of the Co-fines, of the Sum and Difference of thofe Arches; and that the Rectangle of their Co-fines is equal to a Rectangle under half the Radius, and the Sum of the Co-fines, of the Sum and Difference of the fame Arches. PROP. III. The Tangents of two Arches being given, to find the Tangents of the Sum, and Difference, of those Arches Let the firft Cafe, and the Tangent of their Difference, in the fecond, and let CF, perpendicular to the Radius DN, be drawn: Then, because of the equiangular Triangles BAD and BFC, we fhall have BD x CFDA × BC? BD BF BAX BC by 18. 3. Take each of the last equal Quantities from BD3, and there will remain BD BD × BF (BD × DF) = BD - BA× BC: Now BD × DF (BD2- BÁ x BC): BDxCF (DA × BC) :: DF: CF :: DN (DA): NE :: DA2: DA x NE; whence, alternately, BD-BA x BC: DA (:: DA × BC: DA * NE) :: BC: NE. But the first Term, of this Proportion, because BD DA+BA", will alfo be expreffed by DA+ BA-BA x BC, or by DABA-BA x AB+AC; or, laftly, by, DA2 + BAXAC: Therefore, the three firft Terms of the Proportion being known, the fourth NE will likewife be known. 2. E. 1. = COROLLAR Y. Hence, if Radius be fuppofed Unity, and the Tangents of two Arches be denoted by Tand t, it follows that the Tangent of their Sum will be= T+t and the Tangent of their Difference = T -T |