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Find m the pole of cae, and from c draw cmw; and through P, the pole of the primitive, draw cpe, then we applied to a scale of chords gives the measure of oce. To measure OCP, the angle formed by the right circle coe, and the oblique circle cae; through m, the pole of cal, and r, the pole of cpe, draw cw, and cr; then wr applied to a scale of chords is the measure of OCP.

Or, The angle oce may be measured by the line of semitangents; thus, let oe be applied to the scale of semi-tangents from 90 towards the left hand, the number of degrees contained between the points of the compasses, will be the measure of oce; and po, applied to the scale of semi-tangents, from the beginning of the scale towards 90, will give the measure of oøp.

III. If the angle be formed by two oblique circles, as cae and BAD.

Find m the pole of cae, and n the pole of BAD (N. 159.) From the angular point A, through m and n, draw ams and Anv, cutting the primitive in s, and v; then sv, applied to a scale of chords, will give the measure of the angle Bac, or of its equal ean.

CHAP. III.

INVESTIGATION OF GENERAL RULES FOR CALCULATING THE

SIDES AND ANGLES OF RIGHT-ANGLED SPHERICAL TRIAN

GLES.

PROPOSITION XXI.

(H) In any right-angled spherical triangle.

Radius, is to the sine of any side ; as the tangent of the adjacent angle, is to the tangent of the opposite side. DEMONSTRATION. Let ABC be a

F spherical triangle, right-angled at A.

rad : sine ab:: tang 2 B : tang AC.

For, let D be the centre of the sphere, and draw the radii DC, DA, and DB; also from the right-angle A, in the plane DBA, draw as perpendicular to DB, and it will be the sine of the arc AB. At the point E, in the plane DBC,

E draw EF perpendicular to do, then

B will the angle Fea be the inclination

Q. E. D.

of the planes DBA and DBC, and consequently (D. 133.) equal to the spherical angle CBA.

Draw ar from the point A, a tangent to the arc Ac, and produce the radius DC to F; then since the arc ac is perpendicular to the plane DBA (for by hypothesis it cuts the arc AB at rightangles) AF will be perpendicular to AE.

Because ac is less than a quadrant, and DA, DB, and AE, are in the same plane DBA, and that Er is at right-angles to DB (by construction); Dc cannot likewise be at right-angles to DB. Therefore pc and Ef being in the same plane DBC, and not parallel to each other, must meet.

The plane triangle FAE is right-angled at a; the perpendicular AF is a tangent to the spherical perpendicular Ac; the base AE is the sine of the spherical base AB; and the angle FEA is equal to the spherical angle ABC.

Hence, rad : AE:: tang FEA : AF.
That is, rad : sine AB::tang ABC : tang AC.

(I) SCHOLIUM. The foregoing proposition is of considerable importance in spherical trigonometry, and therefore ought to be clearly understood.-It will perhaps be

D of advantage to some students, to illustrate the nature of it in a mechanical way; in order to render the preceding

B в demonstration more perspicuous.

Upon a piece of pasteboard, with any radius describe a circle, make ac of any length less than a quadrant; draw ay perpendicular to the radius AD, and draw DCF through c; let AB be of any length less than a semi-circle, and draw DB. From a draw AEF cutting BD at right-angles,, make of equal to DCF, and draw BG a tangent to BD.

Then cut out the figure FABGFDCF, next cut the back of the line Ad half through, do the same with the line DB, and raise AB till it is perpendicular to the plane ADB; then raise up the plane DEBGF, so that the points F may coincide, and you will have a plane triangle Fae right-angled at A, which will shew the nature of the problem as clearly as possible.

PROPOSITION XXII. (K) In any right-angled spherical triangle.

Radius is to the sine of the hypothenuse, as the sine of any angle is to the sine of its opposite side.

DA.

A

H

DEMONSTRATION. Let abc be a spherical triangle, right-angled at A.

rad : sine bc::sine LB : sine AC.

For, let d be the centre of the sphere, draw the radii DC, DB, and

1. In the plane DBC draw cg perpendicular to DB, and it will be the sine

B of the hypothenuse BC.

From the point g in the plane , draw gh perpendicular to do, and join ch; then (H. 165.) Ghc will be a rightangle, and the angle cou will be equal to the spherical angle

The plane triangle Ghc is right-angled at H, the hypothenuse Gc is the sine of the spherical hypothenuse Bc; the perpendicular ch is the sine of the spherical perpendicular ac, and the angle com is equal to the spherical angle ABC.

Hence,
Radius : gc::sine cch : ch; that is,
Radius : sine of the hypoth. Bc::sine of the angle CBA :
Sine of the perpendicular ac.

Q. E. D.

ABC.

(L) SCHOLIUM. This proposition may perhaps be rendered more familiar to the understanding by a mechanical illustration. Upon a piece of pasteboard, with

any radius describe a circle, make BC, the hypothenuse of any length less than a qua

H) drant; and draw the radii BD, CD; from DA c draw cg at right-angles to BD; set off

KG AB of such a length that when ch is

produced it may cut AD somewhere, as in H; • at h erect the perpendicular hc, and let pc be drawn.

Cut out the figure CDCABC, then cut the lines Ad and bd half through, as in the xxist proposition, raise the plane adc perpendicular to the plane ADB, and the plane BDC, till the points c coincide; you will then have a right-angled plane triangle cug, from which the whole proposition will appear exceedingly simple and easy.

PROPOSITION XXIII.

(M) In any two right-angled spherical triangles, having tk same acute angle at the base.

The sines of their bases have the same ratio E to each other as the tangents of their perpendiculars; and, the sines of their hypothenuses have the same ratio to each other as the sines of their perpendiculars.

DEMONSTRATION. Let the right-angled spherical triangles ABC and AHG have the

A acute LA common.

B
Radius : sine AB::tang. A : tang. Bc. (H. 165.)

Radius : sine ar::tang. A : tang. Hg.(H. 165.)
Therefore, sine AB : sine Ah:.tang. BC : tang. HG.

Radius : sine ac::sine A : sine Bc. (K. 166.)

Radius : sine ar::sine a : sine hg. (K. 166.) Hence, sine ac : sine AG:;sine BC : sine HG.

Q. E. D.

SCHOLIUM.

(N) The different cases or varieties that may happen in the solution of right-angled spherical triangles, wherein two things, together with the right-angle, are always given to find a third, are in all sixteen, and from this proposition alone (by producing the sides of the triangle to quadrants (L. 151.) the whole may be solved. The five following corollaries, or cases, include the whole practice of right-angled spherical triangles, and are the foundation of Baron Napier's rules. CASE I. Sine ah : sine AB::tang GH : tang BC (M. 167.)

Viz. rad: sine AB::tang A : tang BC (L. 151.)
But rad : tang::cot : rad (Z. 103.)
.: Cot a : rad::sine AB : tang BC

Also, rad : sine ac::sine c: sine AB (K. 166.)
CASE II. Rad : sine sc:: tang c : tang AB (H. 165.)

But, rad : tang::cot : rad (Z. 103.)
.:: Cot c : rad::sine Bc : tang AB
Sine ag : sine ac::sine hg ; sine BC (M. 167.)

Viz. rad : sine ac::sine a : sine BC (L. 151.)
Case III. Sine Ei : sine Ed::tang IG : tang DF (M. 167.,

Viz. rad : cos c::tang AC : tang BC (L. 151.)
But, rad : tang::cot: rad. (Z. 103.)
.: Cot ac : rad::cos c: tang
Sine dc : sine Id::sine cF : sine FG (M. 167.)
Viz, rad : sine c::cos BC : cos A (L. 151.)

Or, rad : sine A::cos AB : cos c
Case IV. Sine fh : sine rg::tang bh : tang cg (M. 167.)

Viz, rad : cos A:: cot AB : cot ac (L. 151.)

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BC

But, rad : cot::tang: rad (Z. 103.)
:: Tang AB : rad::cos a: cot ac
Sine ci : sine cd::sine fg : sine id (M. 167.)

Viz. cos BC : rad::cos A : sine c (L. 151.)
CASE V. Sine cg : sine cı:.tang Fg : tang di (M. 167.)

Viz, cos ac : rad::cot A : tang c (L. 151.)
But, rad : tang::cot : rad (Z. 103.)
:: Cos Ac : cot A::cot c : rad
Sine Bh : sine cy::sine FB : sine cf (M. 167.)

Viz. cos AB : cos ac:: rad : cos BC (L. 151.) (O) If the extremes and means of each case be multiplied together, we shall obtain the same equations as are produced by Baron Napier's rules.

I.

Rad x sine AB=tang BC X Cot A

Rad x sine AB=sine ac x sine c
II. Rad x sine sc=tang AB x cot c

Rad x sine Bc=sine ac x sine A
III. Rad x cos c =cot AC X tang BC

Rad x cos c =sine A X COS AB
IV. Rad x cos A =cot AC X tang AB
Rad x cos a

=cos BC x sine c
V. Rad x COS AC =cot A x cot c

Rad x cos AC = COS AB X COS BC (P) If we use the notation of Legendre", and call the hypothenuse b, the base c, and the perpendicular a, and their opposite angles B, C, and A, the equations may be thus expressed. 1.

Rad x sine c=tang ax cot a=sine b x sine c II. Rad x sine a=tang cx cot c=sine b x sine A III. Rad x cos c=cot b x tang a=cos cx sine a IV. Rad x cos A=cot bx tang c=cos a x sine c V.t Rad x cos b=cot A Xcot c=cos cxcos a

Any two of these quantities, together with the right-angle, being given, the rest may be found.

* Éléments de Géométrie, page 380.

+ A late writer on trigonometry, after telling us in his preface that the wellknown rules of Napier, called the Five CIRCULAR Parts, are “too artificial and restricted to be generally employed in the present advanced state of the science;" gives the rules above, and actually solves all his cases of right-angled spherics by them, though they are exactly the same as those of Napier, but less commodiously expressed. And whatever improvements may have been made in trigonometry, either by the English or foreign mathematicians, it is a certain truth that “ Ma. thematical science cannot boast of a neater compendium of results, or a more valuable aid to the memory of the student," than Napier's rules. Vide Monthly

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