AC. 3. To find the Perpendicular A C. The Analogy, s C: Rcs: cs AG. Thus you fee all the Variety of Stating Analogies, for the Cafes of Right-angled Spherical Triangles, performed by the Inftrument above defcribed, whereby you may also observe how expeditiously, as well as infallibly, the whole Bufinefs is difpatched. This Inftrument fupplies every way the Deficiency of Memory for who, without conftant Practice, or a prodigious Memory, can always think of the Analogy for the Solution of any Cafe which may accidentally oc cur unless this Inftrument be at hand. The Use of which (efpecially if furnished with Artificial Sines and Tangents, as before advised) will fufficiently recommend it to all the ingenious Students in this moft curious Part of Mathematical Science. CHA P. XII. The fourth Method of Solving Right-angled Spherical Triangles by the Sector. T HIS admirable and universal Mathematical Inftrument I have already described, with the Conftruction of the various Lines thereon, fo far as they concern this Art, in Vol. 1, of Plain Trigonometry. I have alfo there largely fhewn its Ufe in folving Plain Triangles, which being nearly the fame in folving Spherical Triangles; I shall be much more brief in fhewing its use here. In order to which, the Reader must be here again informed, that the Lines of Sines and Tangents, on each Leg of the Sector, are call'd Lateral Sines aud Tangents; and are thus marked, || Sine, and || Tangent, or | S, || T. But thofe Sines or Tangents which correfpond to each other cross-wife from Leg to Leg, or lie parallel to each other, are called Parallel Sines and Tangents ; and are thus marked Sine, Tangent; or = S, T. But because the Line of Tangents on either Leg, runs but to 45 Degrees (that being equal to Radius ;) and it is inconvenient to manage the Operation by the Sector, when the Analogy contains a Tangent (or Cotangent) greater than 45°; therefore the best way will be to change the greater Tangent into its Complement; which is eafily done; for fince Radius is a Mean Proportional between the Tangent, and the Tangent Complement of an Arch; therefore 'tis but inverting the Order of Radius and the greater Tangent, ard you will have Radius and the leffer Tangent in its ftead; for inftance, in this Analogy, Rcs BC: t C: ct B. R: tC ctC: R; ctC: cs BC:: R:ct B. Rct B: t B: R; CtCcs BC: 1B: R; That is, : By Inverfioncs B ctC: R t B. Now this laft Analogy finds B as well as the firft; but in the First Analogy, fuppofe the t C and ct B were both greater than 45°, as in the Triangle in the foregoing Synopfis; then in this laft Analogy, we have got et C, and t B both leffer of Confequence; and and the doing this is eafy to thofe who are skilled in thefe Affairs. The like is to be done with all other Analogies of a like Nature, as in the following Operations. One thing more the Reader muft note, and that is, That the fecond Term, Sine or Tangent, is to be taken lateral-wife from off the Leg, and made a Paralkel on the Degrees of the firft Term, whether Sine or Tangent; fo fhall the Parallel of the third Term, applied laterally on the Leg, fhew the Quantity of the fourth Term fought. Except when Radius is the fecond Term, and then proceed with the Third and First, and Second and Fourth, as before directed; this finds the Fourth Term on the Leg, which is more certain and easy, than to find it by a Parallel Application from Leg to Leg, which is done in a Method juft the Reverse of what is How delivered. Here follows the Solutions of the Six Cafes and all their Varieties, by the Sector. 1. To find the Leg B A. The Analogy, R: SC:: SBC: s BA. With the Compaffes take the make this a Radius in the Sine of the Sine of 44° 52′ fhall give the For the Leg B A. Sine of 56° 57'; 90 and 90; then Sine of 36° 15′ 2. To find the other Leg AC. The Analogy, RtBC cs C:t AC. : Thus take the Tangent of 44° 52′, and make it a Radius; then fhall the Sine of 33° 03′ give the Tangent of 28° 30′ for the Quantity of the Side 4 C. 3. To find the Angle B The Analogy, RcsBCt Cot B; Therefore make the Tangent of 3o 03′ a = Sine of 45° 08′; then fhall the Radius give the Tangent of 42° 34′ for the Meafure of the Angle B. Cafe 2. Given { The Hypothenuse BC= 44 52 and one of the Legs, as BA = 36 15 1. To find the other Leg AC. The Analogy, csBA: R::cs BC cs AC. Set the Sine of 45° 08′ to the Sine of 53° 45′B B fo fhall the Radius give the Sine 61° 30′, the · Complement of 28° 30, for the Leg AC. 2. To find the Angle B. The Analogy, BC RtBA: cs B. ! Make the IT of 36° 15 a = 2 of 44° 52 fhall the R give the ] S of 42° 34′ for the An gle B. 3. To find the Angle C. The Analogy, s BC: R:: sBA: sC. Make the S of 36° 15′ a S of 44° 52'; fo Radius give the S of 56° 57' for the Cafe 3. One Leg, as BA 36 15 1. To find the Hypothenufe B C. The Analogy, Rcs B: ct AB: ctBC. For the Sector, thus, cs Bt AB RtBC. Make the IT of 36° 15′ a = S of 47° 26'; then fhall the Radius give the T of 44° 52'. for Hypothenuse BC. 2. To find the Leg A C.; The Analogy, R: SBA::t Bt AC. Make the S of 36° 15' a Radius; then the || = T of 429 34' will give the || T of 28° 30' for the Leg AC. 3. To find the other Angle C. The Analogy, Rcs BA:: s B: cs C. Make the S of 53° 45' a = Radius; then shall the S of 47° 26' give the || S of 33° 03' the Complement of C. Cafe |