« PreviousContinue »
hence from the equality of the second members of equations, (432) and (435),
Secondly. From triangle A'B'C' we have by (5) (437) and (428), and the fact that the angle B'A'C' is equal to the inclination of the two planes BOC and BOA,
cos. B'A'C' cos. A=
and, from triangles A'OC' and A'OB', by (5) and
cos. Atang. b. cotan, h.
Thirdly. Corresponding to the preceding equation between the hypothenuse h, the angle A, and the adjacent side b, there must be a precisely similar equation between the hypothenuse h, the angle B, and the adjacent side a; which is
cos. Btang. a cotan. h.
Fourthly. From triangles B'OC', B'OA', and B'A'C', by (5), (429), and (437),
Fifthly. The preceding equation between h, the angle A, and the opposite side a, leads to the following corresponding one between h, the angle B, and the opposite side b;
sin. b = sin. h sin. B.
Sixthly. From triangles C'OA', B'A'C' and B'OC', by (5), (429), and (437),
Seventhly. The preceeding equation between the angle A, the opposite side a, and the adjacent side b, leads to the following corresponding one between
the angle B, the opposite side b, and the adjacent
sin. a=cotan. B tang. b.
Eighthly. From (10),
which, substituted in (454) and (455), give s
Multiplying (458) by cos. b and (459) by cos. a,
sin. b cos. a=cotan. A sin. a.
The product of (460) and (461) is
(462) sin. a sin. b cos. a cos. b cot. A cot. B sin. a sin. b; which, divided by sin. a sin. b, becomes
cos. a cos. bcotan. A cotan. B.
But, by (436),
cos. h cos. a cos. b;
cos. hcotan. A cotan. B.
Ninthly. We have, by (436) and (449),
the product of which is by (10) and (11)
Hence, from the equality of the second members of
The preceding equation between the side a, the opposite angle A, and the adjacent angle B, leads to the following similar one between the side b, the opposite angle B, and the adjacent angle A;
cos. B cos. b sin. A.
6. Corollary. The ten equations, (436), (442), (443), (448), (449), (454), (455), (465), (470), and (472) (471), have, by a most happy artifice, been reduced to two very simple theorems, called, from their celebrated inventor, Napier's Rules.
In these rules, the complements of the hypothenuse and the angles are used instead of the hypothenuse and the angles themselves, and the right angle is neglected.
Of the five parts, then, the legs, the complement of the hypothenuse and the complements of the angles; either part may be called the middle part. The two parts, including the middle part on each side, are (473) called the adjacent parts; and the other two parts are called the opposite parts. The two theorems are as follows:
(474) I. The sine of the middle part is equal to the pro
duct of the tangents of the two adjacent parts.
(475) II. The sine of the middle part is equal to the product of the cosines of the two opposite parts.
Demonstration. To demonstrate the preceding rules, it is only necessary to compare all the equations which can be deduced from them, with those previously obtained (472).
Let there be the spherical right triangle ABC (fig. 2.) right-angled at C.
First. If co. h were made the middle part, then, by (473), co. A and co. B would be adjacent parts, and a and b opposite parts; and, by (474) and (475), we should have
tang. (co. 4) tang. (co. B),
sin. (co. h)
cos. h = cotan. A cotan. B,
which are the same as (465) and (436).
Secondly. If co. A were made the middle part; then, by (473), co. h and b would be adjacent parts, and co. B and a opposite parts; and, by (474) and (475), we should have