METHOD OF COMPUTING A TABLE OF NATURAL SINES.

68. Since the length of the semi-circumference of a circle whose radius is 1, is equal to the number 3.14159265 if we divide this number by 10800, the number of minutes in 180°, the quotient, .0002908882..., will be the length of the arc of one minute; and since this arc is so small that it does not differ materially from its sine or tangent, this may be placed in the table as the sine of one minute. Formula (3) of Table II., gives,

Having thus determined, to a near degree of approximation, the sine and cosine of one minute, we take the first formula of Art. 67, and put it under the form,

and make in this, b = 1', and then in succession,

sin 3′ = 2 sin 2' cos 1'- sin 1'= .0008726646...

sin 4' 2 sin 3' cos 1' — sin 2′ =

.0011635526 ...

sin 5' =

&c.,

(

thus obtaining the sine of every number of degrees and minutes from 1' to 45°.

The cosines of the corresponding arcs may be computed by means of Equation (1).

Having found the sines and cosines of arcs less than 45°, those of the arcs between 45° and 90°, may be deduced, by considering that the sine of an arc is equal to the cosine of its complement, and the cosine equal to the sine of the complement. Thus,

sin 50° = sin (90° - 40°) = cos 40°, cos 50° = sin 40°,

in which the second members are known from the previous computations.

To find the tangent of any arc, divide its sine by its cosine. To find the cotangent, take the reciprocal of the corresponding tangent.

As the accuracy of the calculation of the sine of any arc, by the above method, depends upon the accuracy of each previous calculation, it would be well to verify the work, by calculating the sines of the degrees separately (after having found the sines of one and two degrees), by the last proportion of Art. 67. Thus,

sin 1° : sin 2o — sin 1° :: sin 2° + sin 1° : sin 3o ;

sin 2° : sin 3° - sin 1° ::

sin 3° + sin 1° : зin 4°; &c.

SPHERICAL TRIGONOMETRY.

69. SPHERICAL TRIGONOMETRY is that branch of Mathematics which treats of the solution of spherical triangles.

In every spherical triangle there are six parts: three sides and three angles. In general, any three of these parts being given, the remaining parts may be found.

GENERAL PRINCIPLES.

70. For the purpose of deducing the formulas required in the solution of spherical triangles, we shall suppose the triangles to be situated on spheres whose radii are equal to 1. The formulas thus deduced may be rendered applicable to triangles lying on any sphere, by making them homogeneous in terms of the radius of that sphere, as explained in Art. 30. The only cases considered will be those in which each of the sides and angles is less than 180°.

Any angle of a spherical triangle is the same as the diedral angle included by the planes of its sides, and its mea sure is equal to that of the angle included between two right lines, one in each plane, and both perpendicular to their common intersection at the same point (B. VI., D. 4).

The radius of the sphere being equal to 1, each side of the triangle will measure the angle, at the centre, subtended by it.

Thus, in the triangle ABC, the angle at A is

the same as that included between the planes AOC and

AOB; and the side α is the measure of the plane angle BOC, ( being the centre of the sphere, and OB the radius, equal to 1.

71. Spherical triangles, like plane triangles, are divided into two classes, right-angled spherical

B

triangles, and oblique-angled spherical triangles. Each class will be considered in turn.

We shall, as before, denote the angles by the capital letters A, B, and C, and the opposite sides by the small letters a, b, and c.

FORMULAS USED IN SOLVING RIGHT-ANGLED SPHERICAL

TRIANGLES.

72. Let CAB be a spherical triangle, right-angled at A,

and let O be the centre of the

sphere on which it is situated.
Denote the angles of the triangle
by the letters A, B, and C,
and the opposite sides by the
letters a, b, and c, recollecting
that B and C may change
places, provided that 6 and
b
change places at the same time.

B

P

A

Draw OA, OB, and OC, each of which will be equal to 1. From B, draw BP perpendicular to OA, and from P draw PQ perpendicular to OC; then join the points and B, by the line QB. The line QB will be perpendicular to OC (B. VI., P. VI.), and the angle PQB

will be equal to the inclination of the planes OCB and OCA; that is, it will be equal to the angle C.

From the right-angled triangles OQP and QPB, we

If, in (2), we change c and C, into b and B, we have,

sin b sin a sin B

(5.)

If, in (3), we change b and C, into c and B, we

have,

cos B tan (90°-a) tan c

(6.)

If, in (4), we change b, c, and C, into c, b, and B, we have,