Imágenes de páginas
PDF
EPUB

3. To find the perpendicular A C.
The Analogy, s C: R :: € s 13:05 AG:

Thus you see all the Variety of Stating Analogies, for the Cafes of Right-angled Spherical Triangles, performed by the Instrument above described, whereby you may also observe how expeditiously, as well as infallibly, the whole Business is dispatched. This Instrument supplies every way the Deficiency of Meinory : for who, without conftant Practice, or a prodigious Memory, can always think of the Analogy for the Solution of any Case which may accidentally occur, unless this Instrument be at hand. The Use of which (especially if furnished with Artificial Sines and Tangents, as before advised ) will fufficiently recommend it to all the ingenious Students in this most curious Part of Mathematical Science.

CH A P. XII.

The fourth Method of Solving Right-angled

Spherical Triangles by the Sector.

T

HIS admirable and universal Mathematical
Instrument I have already described, with the

Construction of the various Lines thereon, so far as they concern this Art, in Vol. 1, of Plain Trifonometry. I have also there largely shewn its Use in folving Plain Triangles, which being nearly the fame

in solving Spherical Triangles ; I shall be much more brief in Thewing 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 RS, || T. But those Sines or Tangents which correspond to each other cross-wise 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 easily done ; for fince Radius is a Méan Proportional between the Tangent, and the Tangent Complement of an Arch ; therefore 'tis but inverting the Order of Radius and the

greater

Tanyou

will have Radius and the lesser Tangent in its stead; for instance, in this Analogy,

gent, ard

[ocr errors]

R:csBC ::tc:ct B.
But it is R: C::ctC: R ;
Therefore ctC:(SBC::R:ct B.

R:ct B::tB:R;
Therefore C+C : CSBC ::tB:R; That is,
By Inversion c s B :ct C::R:t B.

But again

[ocr errors]

Now this last Analogy finds B as well as the first but in the First Analogy, suppose the t C and c t B were both greater than 45o, as in the Triangle in the foregoing Synopsis ; then in this last Analogy, we have got cic, and t B both lesser of Consequence ;

and

and the doing this is easy to thofe who are skilled in Defe 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 must note, and that is, Tivat the fecond Term, Sine or Tangent, is to be Taken lateral-wife from off the Leg, and made a Paral. lel on the Degrees of the first Term, whether Sine or Tangent ; fo shall the Parallel of the third Term, applied laterally on the Leg, shew the Quantity of the fourth Term fought.

Except when Radius is the second 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, luch is done in a Method just the Reverse of what is KÓW delivered. Here follows the Solutions of the Six Caf:s and all their Varieties, by the Sector.

Case 1.

Given {

The Hypothenuse
and an Oblique Angle

BC=44 52
C=56 57

1. To find the Leg B A. The Analogy, R:sC:: SBC:s B A. With the Compasses take the || Sine of 56° 57' ; make this a = Radius in the Sine of go and go ; then a = Sine of 44° 52' fhall give the | Sine of 36° 15' to the Leg B A.

2. To find the orber Leg A C. The Analogy, R :BC ::cs C:1 10. Thus take che || Tangent of 44° 52', and make it a = Radius ; then shall the = Sine of 33° 03' give the Tangent of 28° 30' for the Quantity of the Side A C.

3. To find the Angle B. The Analogy, Rics BC ::00: át B ; But for the Sector, csBC:ctc :: R:1B.

Therefore make the 11 Targent of 3.0 ' a = Sine of 45° 08' ; then shall the = Radius give the y Tangent of 42° 34' for the Measure of a Aligle B.

Cafe 2.

[blocks in formation]

The Hypothenuse

BC = 44 52 and one of the Legs, as BA = 36 75

:

1. To find the other Leg AC. The Analogy, csBA: R::15 B C :CS AC.

Set the || Sine of 45° 08' to the = Sine of 53° 45'; so shall the = Radius give the | Sine 61° 30', che · Complement of 288 zo', for the Leg AC.

2. To find the Angle B.
The Analogy, i BC:R: IBA:Cs B.

! Make the | T of 369 15 a -7 of 44.052?; lo shall 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 will the = Radius give the | S of 56° 57' for the Angle C.

Cafe 3.

Given {

One Leg, as
and an adjacent Angle

BA = 36 15
B = 42 34

1. To find the Hypothenuse B C. The Analogy, R:0 B.::ct AB : Ct-BC: For the sector, thus, csB: TAB :: R: BC.

Make the 1 T of 36° 15' a = S of 47° 26'; then shall the = Radius give the || T of 44° 52'. for Hypothenuse B C.

2. To find the Leg A C. The Analogy; R:s BA :: 1 B : 1 AC. Make the || S of 369 15' a = Radius ; then the = T of 429 34' will give the || T of 289 30' for the Leg AC.

3. To find the other Angle C. The Analogy, R:csBA:: SB:cs C. Make the | S of 53° 45' a = Radius ; then shall the = S of 47° 26' give the lI S of 330 oz' the Complement of C.

Cafe

« AnteriorContinuar »