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BARON NAPIER'S UNIVERSAL RULES FOR SOLVING RIGHT

ANGLED SFHERICAL 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.

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

D the angles A and c, and the hypothenuse ac are the parts concerned, it is evident ac is the middle part, and that

T A and c are joined, or adjacent to AC,

P. and therefore are conjunct.

Again, if the three parts do not follow each other in a successive order, that which is not connected with

WB

H 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

BC 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 Be, 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 base AB is still the middle part, for it is not connected either with ac or the angle c, therefore these parts are the extremes disjunct ; the former being separated from AB by the angle A, and the latter by the side BC. Here we apply the second rule, recollecting that the cosine of the complement of an angle, or the hypothenuse, is the sine itself.

Hence, rad x sine AB=sine Ac x sine c. (S) CASE II. FIRST, Let AB, BC, and the angle c, be the parts under consideration, in the triangle ABC. This is exactly like the first part of the preceding case;

the three parts are connected, bc is the middle part, AB and the complement of c are the extremes conjunct.

Hence, rad x sine BC=tang AB X cot c. SECONDLY. Let BC, the hypothenuse ac, and the angle A, be the parts concerned in the triangle ABC.

This is similar to the second part of the preceding case; the perpendicular Bc is still the middle part, for the angle c separates it from Ac, and the side AB from the angle A: therefore ac and the angle A are extremes disjunct, that is, not joined to bc the middle part. Here we must apply the second rule, taking care to remember that the cosine of the complement of the hypothenuse, or an angle, is the sine itself.

(T) CASE III. First, Let BC, Ac, and the angle c, be the parts under consideration, in the triangle ABC.

The three parts follow each other, without the intervention of any other quantity; therefore the complement of c is the middle part, BC and the complement of ac are the extremes conjunct, that is, joined to the angle c.

Hence, rad x cos C=cot Ac x tang BC. SECONDLY. Let AB, the angle A, and the angle c; be the parts under consideration, in the triangle ABC.

The complement of the angle cis here the middle part, being separated from the angle a, by ac; and from the side ab, by the perpendicular Bc;, therefore, AB and the complement of the angle A, are the extremes disjunct, or not joined to c.

Hence, rad x cos c=sine a x Cos AB. (U) Case IV. First, Let AB, AC, and the angle A, be the parts under consideration, in the triangle ABC.

This is exactly similar to the first part of case 3d. The complement of the angle a is the middle part, AB and the complement of ac, are the extremes conjunct, that is, they are joined to the angle A.

Hence, rad x cos A=cot Ac x tang AB. SECONDLY. Let BC, the angle A, and the angle c, be the parts under consideration, in the triangle ABC. This is exactly of the same nature with the second part of the

The complement of the angle a is the middle part ; BC and the complement of c, are the extremes disjunct, the former being separated from the middle part a, by the base AB, and the latter by the hypothenuse ac.

Hence, rad x cos A=cos BC x sine c. (W) CASE V. First, Let the hypothenuse Ac, the angle A, and the angle c, be the parts under consideration, in the triangle

3d case.

ABC.

The three parts follow each other; therefore the complement of ac is the middle part, the complement of a and the complement of c are the extremes conjunct, that is, they are joined to the middle part ac.

Hence, rad x cos AC=cot A x cot c. SECONDLY. Let the hypothenuse Ac, the base AB, and the perpendicular Bc, be the parts under consideration, in the triangle ABC.

The complement of ac is here the middle part, being separated from AB by the angle A, and from Bc by the angle c; therefore AB and Bc are the extremes disjunct.

Hence, rad x cos AC=COS AB X COS BC.

SCHOLIUM.

The preceding cases include all the varieties that can possibly happen in the practice of right-angled spherical triangles.

Any of the equations may be turned into a proportion by putting the required term last, that with which it is connected first, and the other two in the middle in any order. These equations are exactly the same as those already given (O. 169.) and therefore Napier's rules are universally truc.

CHAP. IV.

INVESTIGATION OF GENERAL RULES FOR SOLVING THE DIF

FERENT CASES OF OBLIQUE SPHERICAL TRIANGLES, BY DRAWING A PERPENDICULAR FROM THE VERTICAL ANGLE, UPON THE BASE.

PROPOSITION XXV.

Shewing the manner of applying Baron Napier's rules to ob

lique spherical triangles, from which several useful corollaries are deduced.

(X) When the three given parts do not follow each other in a regular order, viz. when an unknown part intervenes, a perpendicular should always be drawn from the end of a given side, and opposite to an adjacent given angle.

But when the three given quantities follow each other without the intervention of an unknown quantity, the perpendicular should be drawn in such a manner, as to fall not only from the end of a given side and opposite to an adjacent given angle, but likewise from the end of a required sine, or opposite to a required angle, according as a side or an angle is the subject of enquiry."

(Y) Having drawn a perpendicular, agreeably to the foregoing directions; then, if the vertical angle of the oblique triangle be given or sought, find the vertical angle of that rightangled triangle wherein two things are given; but, if the base of the oblique triangle be given or sought, find the base of that right-angled triangle wherein two quantities are given.

Compare the perpendicular, the part given, and the part sought, in that triangle wherein only one quantity is given; find

* The reason of these rules for drawing a perpendicular is founded on practice and observation. For in every right-angled triangle there must be two given quantities, exclusive of the right angle, and it is obvious that the perpendicular must be so drawn as to form one right-angled triangle wherein two quantities are given.

the middle part, and make an equation agreeably to NAPIER's Rules, marking the term sought with an asterisk (*).

Compare the perpendicular and the similar parts in the triangle where two quantities are given, and make an equation, which place exactly under the former, and strike out such terms as are common to both the equations.

Turn the remaining quantities into a proportion; thus, if all the terms which are to be struck out of the equations be on the same side of both, put the required term last, that with which it is connected first, and the other two in the middle in any order.

But if the terms which are not struck out of the equations stand on both sides; one unmarked term, in the lower equation, is to the unmarked term exactly above it; as the other unmarked term in the lower equation, is to the marked or required term which stands above it.

(Z) Case I. Suppose two angles at the base of an obliqueangled triangle, and a side opposite to one of these angles were given, to find the vertical angle.

First, find the vertical angle BCD + in the triangle BDC, where two entire quantities are given.

D

с

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Here Bc is the middle part, and the other parts are the extremes conjunct. Hence,

rad X COS BC=cot LB X Cot BCD.
Therefore cot LB : rad::cos BC : cot BCD.
In the triangle ADC; rad x cos LA=cos DC x sin acD.
In the triangle BDc; rad X cos B=COS DC x sin BCD.
Therefore cos LB : cos LA::sine BCD : sine ACD.

Had the 2 c been given, and the L A required, we must have proceeded exactly in the same manner, only the L A must have been marked instead of act, and the proportion would have been thus; sine BCD : sine ACD::cos LB : cos LA.

The sum, or difference, of acD and BCD, gives ACB according as the perpendicular falls within, or without the triangle.

(A) COROLLARY I. The cosines of the angles at the base, are in proportion to each other as the sines of the angles at the

+ When both the angles at the base are given ; if the perpendicular fall without the triangle, it is the most convenient to draw it so that two known parts may come between the perpendicular and the given angle to which it is opposite, hence cd is the proper perpendicular in this case.

The terms which are to be struck out of the equations are printed in italics.

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