Page images
PDF
EPUB

Make the Equinoctial Colure to contain, with the Brafs Meridian, an Angle of 57° 221, and there keep the Globe faft; then move it up or down, till, by the Quadrant of Altitude, you find the Equinoctial Colure and the Horizon contain an Angle of 57° 32′; the Globe being fixed in this Pofition, you have the Triangle POS formed thereon; in which the CoDeclination PS, will be found 66° 57'; the Latitude PO, will be seen to be 50° 56′ on the Meridian. ; and the Amplitude of the Sun's Rifing from the North SO, will appear on the graduated Horizon to be 51° 48'.

Thus I have briefly fhewn the Manner of folving Right-angled Spherical Triangles by the Globe or Sphere; I could indeed have been more large on this Head, but I abhor Tautology, and an unneceffary Prolixity; for as it is an Affront to the Ingenious, (a Word to the Wife being enough) fo it is a fruitless Labour to the Dull and Brainless; whofe pitious Fate is, always to be ignorant and indocible by any Means whatever. I am well affured thofe who have but a tolerable Genius, and know how to apply particular Rules and Examples to general Ufes, will, by what is here faid, understand what ever is unfaid on this Matter; for Example, they know how the Angle PSO may be measured, as was the Angle S PO in any Cafe; viz. by inverting the Triangle. They also know that the laft Cafe might have been refolved by changing the Angles into Sides, according to Theorem 13, and its Corollary, with fome other things of this kind, which I thought not needfull to insist upon.

СНАР.

CHAP. XVI.

The eighth Method of folving Right-angled. Spherical Triangles, by the Stereographic Projection.

T

HIS Way is (generally speaking) more artful than useful; not but that to a Perfon well verfed in Spherics, it is of peculiar Ufe and Service; for this Method difpels all Ambiguity, and Errors, which attend the Solution by moft other Methods; and by a little Ufe, is very practicable and cafy. So that if the Ingenuity, Certainty, Eafe, and Expeditiousness, of any Method, be fufficient to recommend it, this cannot fail of Acceptance with all thofe who have the leaft Genius and Tafte for this Science. The Problems I fhall all along refer to, are thofe of the Stereographic Projection in the Chapter beforegoing.

As in every Cafe, the Triangle may be made either at the Center or at the Circumference of the Primitive Circle, I fhall accordingly prefent the Reader with two Schemes, the one having the Triangle at the Center, the other at the Circumference of the Primitive Circle; with the Manner of conftructing them, and of folving the Triangles in all the fix Cafes, as follows.

Given

Cafe 1.

The Hypothenufe

[ocr errors]

BC 44 52

and one of the Oblique Angles C = 56 57

[merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][ocr errors]

fo will the Triangle

be made as required.

1. To find the Leg A B.

ACB,

In Fig 1, lay a Ruler from the Pole a of the Oblique Circle E AG, to the Point B, and it will cut the Primitive in b; measure Fb on the Line of Chords, and you will find it 36° 15'. The fame you will find it in Fig. 2, if A B be measured on the Half-Tangents from 90 downwards.

2. To find the Leg A C..

In Fig. 1, this is found on the Half-Tangents to to be 28° 30'; and in the Fig. 2, A C measured on the Chords will be found the fame.

3. To find the Angle B.

In either Figure, lay a Ruler from the Angular Point B, and the Pole of the Oblique Circle a, and it will cut the Primitive in m; then the Arch M m measured on the Line of Chords will be found to be 429 34'.

Cafe 2.

The Hypothenuse

BC 44 52

Given and one of the Legs, as 4B = 36 15

[blocks in formation]

Set the Half-Tangent of 56o 15' from

and draw the Oblique Circle

then will the Triangle

be that which is required.

[merged small][ocr errors][merged small][merged small][merged small]

B to A, DAG;

ABC,

365

P

a

[blocks in formation]
[ocr errors]
[blocks in formation]

53° 45′

[ocr errors]

and draw the Paral

[ocr errors][merged small][ocr errors][merged small]

With the Half-Tan

gent of defcribe the Parallel

then through the Points

ول

draw the great Oblique Circle I
alfo through the Point
draw the Right Circle

thus you have the Triangle
made as required.

I

VOL. II.

BKL; CBF, CQF; B, ABG; ACB

X

I. Ta

« PreviousContinue »