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 extreme 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 sin AB sin AC X sin c.

(388) 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 sin BC=tan AB X cot c.

SECONDLY. Let BC, the hypothenuse AC, and the angles, 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.

Hence, rad x sin BC=sin AC X sin a.

(389) 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 ACX tan 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 c is 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=sin a x cos AB.

(390) 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 III. 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 ACX tan 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 Case III. 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 sin c.

(391) CASE V. FIRST, Let the hypothenuse AC, the angle a, and the angle c, be the parts under consideration, in the triangle

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 AX 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 of any order. These equations are exactly the same as those already given (384), and therefore Napier's rules are universally true.

CHAP. IV.

INVESTIGATION OF GENERAL RULES FOR SOLVING THE DIFFERENT CASES OF OBLIQUE SPHERICAL TRIANGLES, BY DRAWING A PERPENDICULAR FROM THE VERTICAL ANGLE UPON THE BASE.

PROPOSITION XXV.

Showing the manner of applying BARON NAPIER's rules to oblique spherical triangles, from which several useful corollaries are deduced.

(392) 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 side, or opposite to a required angle, according as a side or an angle is the subject of inquiry.†

(393) Having drawn a perpendicular, agreeably to the foregoing directions; then, if the vertical angle of the obliquetriangle be given or sought, find the vertical angle of that right-angled 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 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.

+ 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.