Upon a piece of pasteboard, with any radius describe a circle, make AC of any length less than a quadrant; draw AF 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 DF 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 AF 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 the sine of its opposite side. 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

DA.

:

In the plane DBC draw CG perpendicular to DB, and it will be the sine of the hypothenuse BC.

From the point & in the plane DBA, draw GH perpendicular to DB, and join CH; then (H. 161.) GHC will be a right-angle, and the angle CGH will be equal to the spherical angle ABC.

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 CGH is equal to the spherical angle ABC.

Hence,

Radius: GC:: sine CGH: CH; that is,

Radius: sine of the hypoth. BC:: sine of the angle CBA:
Sine of the perpendicular AC.

Q.E.D.

(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 quadrant; and draw the radii BD, CD; from c draw CG at right-angles to BD; set off AB of such a length that when CG is produced it may cut AD somewhere, as in H;

C

at H erect the perpendicular HC, and let DC be drawn.

G

Cut out the figure CDCABC, then cut the lines AD and BD half through, as in the xxIst problem, 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 CHG, from which the whole proposition will appear exceedingly simple and easy.

PROPOSITION XXIII.

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

The sines of their bases have the same ratio 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

acute A common.

E D

Radius: sine AB :: tang. A : tang. Bc. (H. 161.)
Radius: sine AH::tang. A: tang. HG. (H. 161.)
Ex æquali, sine AB: sine AH:: tang. BC: tang. HG.
Radius sine AC:: sine A: sine BC. (K. 162.)
Radius: sine AG :: sine A: sine HG. (K. 162.)
Ex æquali, sine AC: sine AG :: sine BC : sine HG.

:

SCHOLIUM.

F

G

Q.E.D.

H

(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 produ

cing the sides of the triangle to quadrants (L. 147.) 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. 163.)
Viz rad: sine AB:: tang A: tang BC (L. 147.)
But rad: tang:: cot: rad (Z. 97.)

.. Cot Arad::sine AB : tang BC

Also, rad: sine ac :: sine c: sine AB (K. 162.) CASE II. Rad: sine BC :: tang c: tang AB (H. 161.) But, rad: tang:: cot: rad (Z. 97.)

... Cot c: rad:: sine BC: tang AB

Sine AG sine AC :: sine HG : sine BC (M. 163.)
Viz. rad sine AC:sine A: sine BC (L. 147.)

CASE III. Sine EI: sine ED :: tang IG : tang DF (M. 163.)
Viz. rad: cos c:: tang AC: tang BC (L. 147.)
But, rad: tang:: cot: rad. (Z. 97.)

... Cot AC: rad :: cos c:tang BC.

Sine DC: sine ID :: sine CF : sine FG (M. 163.)
Viz. rad sine c:: cos BC: COS A (L. 147.)

:

OR, rad : sine A:: cos AB :cos C

CASE IV. Sine FH: sine FG :: tang BH: tang CG (M. 163.) Viz. rad: cos A :: cot AB: cot AC (L. 147.)

But, rad: cot :: tang: rad (Z. 97.)

.. Tang AB : rad: : cos a: cot AC

Sine CF: sine CD :: sine FG : sine ID (M. 163.)
Viz. cos BC: rad :: cos a: sine c (L. 147.)

CASE V.

Sine CG: sine CI :: tang FG: tang DI (M. 163.)
Viz. cos AC: rad: : cot A: tang c (L. 147.)
But, rad: tang:: cot: rad (Z. 97.)

... Cos AC: cot A :: cot c: rad

Sine BH : sine CG:: sine FB : sine CF (M. 163.)
Viz. cos AB : cos AC:: rad: cos BC (L. 147.)

(0) 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.

(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 ▲, the equations may be thus expressed. I. Rad x sine c=tang ax cot A =sine bx sine c II. Rad x sine a tang cx cot c sine b× sine ▲ III. Rad x cos ccot b x tang a=cos cx sine a IV. Rad x cos A cot bx tang c=cos a x sine c V.+ Rad x cos bcot A x cot c =cos c x cos a

=

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

BARON NAPIER'S UNIVERSAL RULES FOR SOLVING RIGHTANGLED SPHERICAL TRIANGLES.

PROPOSITION XXIV.

(Q) I. Radius x sine of the middle part=rectangle of the tangents of the extremes when conjunct.

And,

II. Radius x sine of the middle part=rectangle of the cosines of the extremes when disjunct.

Observing to use the complements of the hypothenuse and angles.

EXPLANATION OF THE RULES.

In every right-angled spherical triangle there are FIVE CIRCULAR PARTS, exclusive of the right-angle, which is not taken into consideration: and these five parts are the hypothenuse, the two legs or sides, and their opposite angles; they are called circular parts, because the measure of each of them is the arc of a circle.

Now in every case proposed for solution, there are three of these five parts concerned, viz. two given and a third required.

* 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 "Mathematical 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 Review, vol. LIII., page 280.

It is therefore obvious that when the three parts follow each other in a successive order, the middle one is the middle part, and the two others the extremes conjunct, that is, joined to the middle part.

Suppose in the triangle ABC, that the angles A and c, and the hypothenuse AC are the parts concerned, it is evident AC is the middie part, and that A and c are joined, or adjacent to AC, and therefore are conjunct.

D

E

B

Ω

H

Again, if the three parts do not follow each other in a successive order, that which is not connected with either of the other two is invariably called the middle part; and the other two, which are not connected with it, are called the extremes disjunct; that is, not joined to the middle part.

Suppose the hypothenuse AC, the base AB, and the perpendicular BC, the parts concerned; AC will be the middle part, for it is not connected either with AB or BC, the angle A intervening in the former case, and the angle c in the latter; the sides AB and BC are extremes disjunct, that is, not joined to AC, because of the intervention of the angles A and C, but the sides AB and c are considered as joined though the right-angle B is between them, for, as we have before observed, it is excluded, or not taken into consideration.

It must be remembered that, in speaking of the hypothenuse or either of the angles as a middle part, or extremes conjunct or disjunct, their complements are to be used; but when we speak of the sides or legs, their complements are not to be used, but the real sides or legs.

Hence the middle part must universally be either the base AB, the perpendicular BC, the complement of the angle c, the complement of the angle A, or the complement of the hypothenuse AC, that is, it must be some one of the five circular parts.

(R) CASE I. FIRST, let AB, BC, and the angle A be the parts under consideration, in the triangle ABC.

Here the three parts are joined together, because the right angle B is not regarded.—Therefore AB is the middle part, BC, and the complement of A, are the extremes conjunct.

Hence, rad x sine AB=tang BC X cot A.

SECONDLY. Let AB, AC, and the angle c be the parts under consideration, in the triangle ABC.

The