PROBLEM V.

To find a small arc from its log. sine or its log. tangent. 1. Let the log. sine be given; then n being the number of seconds in the arc, the expression (2), in problem Iv., gives

log.n log. sin. x -4.6855749 + arith. comp. log. cos. r

= log. sin. + 5.3144251 — 10+ arith. comp. log. cos. r;

therefore, to find the arc from the log. sine the rule is this. To the log. sine of the small arc add 5·3144251, and } of the arithmetical complement of the log. cosine; subtract 10 from the index of the sum, and the remainder will be the logarithm of the number of seconds in the

arc.

2. Let the log. tangent be given; then from the expression (3), last problem, we have

log. n log. tan. r

- 4.6855749

arith. comp. log. cos. a

=log. tan. x + 5·3144251 — 10 — arith. comp. log. cos. x;

=

that is, to the log. tangent of the small arc add 5.3144251, and from the sum subtract of the arithmetical complement of the log. cosine, take 10 from the index of the remainder, and we shall have the logarithm of the number of seconds in the arc.

Let us now apply each of the foregoing rules to an example.

1. Required the log. sine of 1′ 4·8754′′.

By the rule in problem iv. the process is as follows:

... log. sin. 1' 4·8754′ = 6·4917548 + ·8754 × ⚫0067334

... log. tan. 7'2·38′′ = 7·3108879 +38 0010279

7.3112785.

3. Required the arc whose log. sine is 6.4976550.

The proposed log. sine lies between log. sine 1′ 4′′ and log. sine 1' 5", and the difference between these logs. is 0067334; also the difference between the proposed log. and log. sine 1′ 4′′ is 59002; hence,

4. Required the arc whose log. tangent is 7-1644398.

The proposed log. is between log. tan. 5′ 1′′ and log. tan. 5′ 2′′; the difference of these logs. is '0014404, and the difference of the proposed and log. tan. 5′ 1′′ is '0002981.

190

CHAPTER II.

INVESTIGATION OF EXPRESSIONS FOR THE SURFACE OF A SPHERICAL TRIANGLE, AND FOR THE

SPHERICAL EXCESS.

(84.) It has been already shown (36) that two great circles always intersect in two points at the distance of a semicircle from each other. The space thus included by two great circles is called a lune, (see the fig. at page 66.)

The surface of a lune is to the surface of the whole sphere as the arc QQ', or as the angle P of the lune, is to the whole circumference IQHI. This is pretty obvious, but it may be rigorously proved in the same way as it is proved in plane Geometry, that in the same circle any sector is to the whole circle as its arc is to the circumference, (Geom. prob. 23, Book 6). Hence, if we call the surface of the sphere S, and the angle of the lune w degrees, the expression for its area will be

S -; or if, instead of degrees, w represent the absolute length of

360

those degrees to radius 1, then the expression may be written S-,

2

It can be proved, although not by the elementary principles of Trigonometry, that the surface of a sphere is equal to four times the area of one of its great circles; that is, r being the radius of the sphere S=4r2, so that the expression for the area of the lune is 2r2w. If we supposer to be unity, the surface will be expressed by 2w, that of the whole sphere being 4π.

* See "The Elements of the Integral Calculus," page 144.

SURFACE OF A SPHERICAL TRIANGLE.

191

PROBLEM I.

To express the area of a spherical triangle in terms of its three angles.

Let ABC be any spherical triangle, and produce the sides AC, BC, till they meet again in C', forming the lune CC'. The triangle CAB will be a portion of an opposite lune equal to the lune CC'; and this portion will obviously be equal to the portion C'A'B', provided the arcs CA, CB, are equal to the arcs C'A', C'B'. Now AA' is equal to CC', each being a semicircle; hence, taking from each the common part CA', we have CA — C'A'. In like manner CB C'B', and, therefore, the triangles ABC, A'B'C', are equal. Hence the surface of the hemisphere, whose base is AB'A'B', is equal to the sum of the three lunes AA', BB', CC', minus twice the triangle that is, calling the surface of this triangle Σ.

ABC;

S=22(A+B+C)-2

.. Σ= p2 (A + B + C) — 4S = r2 {(A + B + C) — π}.

where it must be observed that A, B, C, denote the lengths of the arcs which measure the angles of the proposed triangle to radius unity. But, if we take A, B, C, and π in degrees, then since

180° : π : : A + B + C − 180° : { A+B+C ~ 180°}

the expression for Σ will be

П

180°

If the radius of the sphere, on which the triangle is, be taken for unity, then calling the area in this case &, we have

which indicates that the area of a triangle, on the surface of a sphere, whose radius is unity, is equal to the excess of its three angles above two right-angles. This quantity is technically called the spherical excess, and the theorem (2) is known by the name of Girard's theorem.