of the same denomination with it; that is, if the middle part is one of the sides, the opposite parts are the other two; and if the middle part is the supplement of one of the angles, the opposite parts are the supplements of the other two.

42. Mr. Walter Fisher has also given, in the Transactions of the Royal Society of Edinburgh, rules of easy recollection which will serve for the solution of all the cases of plane and spherical triangles. They are included in four theorems, which may be applied to plane triangles by taking, instead of the sine or tangent of a side, the side itself.

Theor. 1. Given two parts, and an opposite one.

S.AS.0::s.as.0.

Theor. 2. An included part given, or sought.

Theor. 4. Given the three sides or angles of an ob

Here м denotes the middle part of the triangle, and must always be assumed between two given parts. It is either a side or the supplement of an angle; and is sometimes given, sometimes not.

A and a are the two parts adjacent to the middle, and of a different denomination from it.

o and o denote the two parts opposite to the adjacent parts, and of the same denomination with the middle part.

l is the last, or most distant part, and of a different denomination from the middle part.

That these four theorems may be called to mind with the greater facility, the following four words formed by

abbreviating the terms of the respective analogies should be committed to memory, viz.

Sao, satom, tao, sarsalm.

SECTION VI.

On the Areas of Spherical Triangles and Polygons, and the Measures of Solid Angles.

Theorem I.

43. In every spherical triangle, the following proportion obtains, viz. as four right angles, (or 360°), to the surface of a hemisphere; or, as two right angles (or 180°), to a great circle of the sphere; so is the excess of the three angles of the triangle, above two right angles, to the area of the triangle.

Let ABC be the spherical triangle. its sides, as BC, into the circle BCEF, which may be supposed to bound the upper hemisphere. Prolong also, at both ends, the two sides AB, Ac, until they form semicircles estimated from each angle, that is, until BAE

Complete one of

q

a Dim

ABDCAF ACD=180°. Then will CBF 180° BFE; and consequently the triangle AEF, on the anterior hemisphere, will be equal to the triangle BCD on the opposite hemisphere. Putting m, m', to represent the surface of these triangles, p for that of the triangle BAF, q for that of CAE, and a for that of the proposed triangle ABC. Then a and m' together (or their equal a and m together) make up the surface of a spheric lune comprehended between the two semicircles ACD, ABD, inclined in the angle A: a and lune included between the semicircles CAF, CBF, making p together make up the the angle c:a and q together make up the spheric lune included between the semicircles BCE, BAE, making the angle B. And the surface of each of these lunes, is to

that of the hemisphere, as the angle made by the comprehending semicircles, to two right angles. Therefore, putting is for the surface of the hemisphere, we have 180°: A::s: a + m.

180°: B::: a + q.

180°: cs: a + p.

Whence, 180°: A+B+C::s:3a+m+p+q=2a+1s; and consequently, by division of proportion,

as 180°: A+B+C-180°:: $: 2a + s − s = 2a;

A+B+C-180°

or, 180°: A+B+C-180°::s: a=s.
ăs. 360°

Q. E. D. *

Cor. 1. Hence the excess of the three angles of any spherical triangle above two right angles, termed technically the spherical excess, furnishes a correct measure of the surface of that triangle.

Cor. 2. If = 3.141593, and d the diameter of the A+B+C 1800

sphere, then is d2

spherical triangle.

720°

= the area of the

Cor. 3. Since the length of the radius, in any circle, is equal to the length of 57-2957795 degrees, measured on the circumference of that circle; if the spherical excess be multiplied by 57 2957795, the product will express the surface of the triangle in square degrees.

Cor. 4. When a = 0, then A + B + C = 180°: and when a = s, then A + B + C = 540°. Consequently the sum of the three angles of a spherical triangle, is always between 2 and 6 right angles: which is another confirmation of art. 19, p. 83.

* This determination of the area of a spherical triangle is due to Albert Girard (who died about 1633). But the demonstration now commonly given of the rule was first published by Dr. Wallis. It was considered as a mere speculative truth, until General Roy, in 1787, employed it very judiciously in the great Trigonometrical Survey, to correct the errors of spherical angles. See Phil Trans. vol. 80, and chap. xii, of this volume.

Cor. 5. When two of the angles of a spherical triangle are right angles, the surface of the triangle varies with its third angle. And when a spherical triangle has three right angles its surface is one-eighth of the surface of the sphere.

Remark. The mode of finding the spherical excess, and thence the area when the three angles of a spherical triangle are given, is obvious enough; but it is often requisite to ascertain it by means of other data, as, when two sides and the included angle are given, or when all the three sides are given. In the former case, let a and b be the two sides, c the included angle, and E the spherical excess: then is cot E =

cota. cot b+cos c

sin c

When

the three sides a, b, c, are given, the spherical excess may be found by the following very elegant theorem, discovered by Simon Lhuillier:

The investigation of these theorems would occupy more space than can be allotted to them in the present volume.

Theorem II.

44. In every spherical polygon, or surface included by any number of intersecting great circles, the subjoined proportion obtains, viz. as four right angles, or 360°, to the surface of a hemisphere; or, as two right angles, or 180°, to a great circle of the sphere; so is the excess of the sum of the angles above the product of 180° and two less than the number of angles of the spherical polygon, to its area.

For, if the polygon be supposed to be divided into as many triangles as it has sides, by great circles drawn from all the angles through any point within it, forming

at that point the vertical angles of all the triangles. Then, by theor. 1, it will be as 360°: s :: A + B + C

180° its area. Therefore, putting P for the sum of all the angles of the polygon, n for their number, and v for the sum of all the vertical angles of its constituent triangles, it will be, by composition,

as 360° s :: P + V ~ 180°. n: surface of the polygon. But v is manifestly equal to 360° or 180° x 2. Therefore,

as 360°: :: P (n-2) 180°: s. s

area of the polygon. Q. E.D.

P (n-2) 180°

360°

the

Cor. 1. If and d represent the same quantities as in theor. 1, cor. 2, then the surface of the polygon will be P- − (n − 2) 180°

expressed by ad2.

7209

Cor. 2. If R° = 57.2957795, then will the surface of the polygon in square degrees be = R°. [P — (n − 2) 180°].

Cor. 3. When the surface of the polygon is 0, then P = (n 2) 180°; and when it is a maximum, that is, when it is equal to the surface of the hemisphere, then P = (n − 2) 180° + 360° = n. 180°: consequently 1, the sum of all the angles of any spheric polygon, is always less than 2n right angles, but greater than (2n- 4) right angles, n denoting the number of angles of the polygon.

Nature and Measure of Solid Angles.

45. A solid angle is defined by Euclid, that which is made by the meeting of more than two plane angles, which are not in the same plane, in one point.

Others define it the angular space comprised between several planes meeting in one point.

It may be defined still more generally, the angular space included between several plane surfaces or one or more curved surfaces, meeting in the point which forms the summit of the angle.