The last table may also be conveniently expressed in the following manner, denoting the side opposite to the angle A, by a, to B by b, and to C by c; and also the segments of the base, or of opposite angle, by x and

y.

TO

SPHERICAL

TRIGONOMETRY,

CONTAINING

NAPIER'S RULES OF THE CIRCULAR PARTS.

THE

HE rule of the Circular Parts, invented by NAPIER, is of great use in Spherical Trigonometry, by reducing all the theorems employed in the solution of right angled triangles to two. These two are not new propositions, but are merely enunciations, which, by help of a particular arrangement and classification of the parts of a triangle, include all the six propositions, with their corollaries, which have been demonstrated above from the 18th to the 23d inclusive. They are perhaps the happiest example of artificial memory that is known.

DEFINITIONS.
I.

If in a spherical triangle, we set aside the right angle, and consider only the five remaining parts of the triangle, viz. the three sides and the two oblique angles, then the two sides which contain the right angle, and the complements of the other three, namely, of the two angles and the hypotenuse, are called the Circular Parts. Thus, in the triangle ABC right angled at A, the circular parts are AC, AB with the complements of B, BC, and C. These parts are called circular; because, when they are named in the natural order of their succession, they go round the triangle.

II.

When of the five circular parts any one is taken, for the middle part, then of the remaining four, the two which are immediately adjacent to it, on the right and left, are called the adjacent parts; and the other two, each of which is separated from the middle by an adjacent part, are called opposite parts.

Nn

Thus in the right angled triangle ABC, A being the right angle, AC, AB, 90°-B, 90° - BC, 90~—C, are the circular parts, by Def. 1.; and if any one as AC be reckoned the middle part, then AB and 90o -C, which are contiguous to it on different sides, are called adjacent parts; and 90°-B, 90° - BC are the opposite parts. In like manner

B

if AB is taken for the middle part, AC and 90°-B are the adjacent parts: 90° BC, and 90o — C are the opposite. Or if 90° BC be the middle part, 90-B, 90° – C are adjacent; AC and AB opposite, &c. This arrangement being made, the rule of the circular part is contained in the following

* PROPOSITION.

In a right angled spherical triangle, the rectangle under the radius and the sine of the middle part, is equal to the rectangle under the tangents of the adjacent parts; or to the rectangle under the cosines of the opposite parts.

The truth of the two theorems included in this enunciation may be easily proved, by taking each of the five circular parts in succession for the middle part, when the general proposition will be found to coincide with some one of the analogies in the table already given for the resolution of the cases of right angled spherical triangles. Thus, ́in the triangle ABC, if the complement of the hypotenuse BC be taken as the middle part, 90° - B, and 90°-C, are the adjacent parts, AB and AC the opposite. Then the general rule gives these two theorems, RXcos BC=cot BXcot C; and RXcos BC cos AB Xcos AC. The former of these coincides with the cor. to the 20th; the latter with the 22d.

and

To apply the foregoing general proposition, to resolve any case of a right angled spherical triangle, consider which of the three quantities named (the two things given and the one required) must be made the middle term, in order that the other two may be equidistant from it, that is, may be both adjacent, or both opposite; then one or other of the two theorems contained in the above enunciation will give the value of the thing required.

Suppose, for example, that AB and BC are given, to find C; it is evident that if AB be made the middle part, BC and C are the oppo

site parts, and therefore RXsin AB sin CXsin BC, for sin C=cos (90°C), and cos (90°-BC)=sin BC, and consequently

sin C

sin AB

sin BC

Again, suppose that BC and C are given to find AC; it is obvious that C is in the middle between the adjacent parts AC and (90°-BC), therefore RXcos C=tan ACXcot BC, or tan_AC=

cos C cot BC

cos C+

tan BC; because, as has been shewn above,

1

tan BC.

cot BC

In the same way may all the other cases be resolved. One or two trials will always lead to the knowledge of the part which in any given case is to be assumed as the middle part; and a little practice will make it easy, even without such trials, to judge at once which of them is to be so assumed. It may be useful for the learner to range the names of the five circular parts of the triangle round the circumference of a circle, at equal distances from one another, by which means the middle part will be immediately determined.

Besides the rule of the circular parts, Napier derived from the last of the three theorems ascribed to him above, (schol. 29.), the solutions of all the cases of oblique angled triangles. These solutions are as follows: A, B, C, denoting the three angles of a spherical triangle, and a, b, c, the sides opposite to them,

I.

Given two sides b, c, and the angle A between them.

To find the angles B and C.

Given the two sides b, c, and the angle B opposite to one of them.

To find C, and the angle opposite to the other side.