by R and the second by cos b. Performing this multiplication, the two equations become

and

R COS A sin b sin c R3 cos a R2 cos b cos c

cos b cos c sin a sin b R2 cos b cos c

Add these, cancelling the common term R2 there results

R COS Asin bsinc+cos bcos c sin a sin b―R3cosa

R COS a cos2 b

cos b cos c and

-RCOS a cos2 b substituting for cos2 b in the last term its value R2 — sin2 b (Art. 72) this result becomes

R COS Asin b sin c+cos b cos csin a sin br3cos a-R3 cosa+ R sin2 b cos a cancelling + R3 cos a and R3 cos a in the second member, and dividing the whole equation by sin b, we have

R COS A sin c + cos b cos c sin a = R sin b cos a Transposing the second term of the first member, this last equation becomes

R COS A sin c R sin b cos a cos b cos c sin a Had we set out with the value of cos в and combined it with that of cos c in the same manner that we have that of cos A, it is plain that we should have obtained a similar result which we may write without the trouble of repeating the operation, by changing a into в and a into b, and vice versa. We thus have

A

R COS B sin c = R sin a cos b cos a cos c sin b Adding this and the preceding equation together we have R sin c (cos A+COS B)=(R—COS C) (sin a cos b+sin b cos a) but (Art. 69)

sin a cos b+sin b cos a R sin (a + b)

Substituting this second member for the first in the preceding expression, and dividing the whole by R, we have

sin c (cos A+COS B)=(R-COS C) sin(a+b)... (1)

But (Art. 81)

sin

c

the denominators of which being made to disappear it be

sin c (sin a + sin B) = sin c (sin a + sin b) subtracting (3) from (2) there results

sin c (sin A sin B) = sin c (sin a

dividing (4) by (1) we have

sin A+ sin B

COS A+ COS B

sin asin b
sin (a + b)

sin b)

sin c

R-COS C

sin A sin B

COS A+ COS B

sin c

But it appears from equation (2) of the preceding article

that

sin A + sin B tan (A+B)

COS + cos B

R

from formula (3) of the same article that

from formula (7) of the same article that

from formula (5) of the same article that

sin a + sin b cos (a

sin (a+b) cos (a + b)

and from formula (6) of the same article that

substituting these values in equations (6) and (7) of the pre

sent article those equations become

and

tan (A B) =cotc

sin § (a — b)

sin(a+b)

These may each be written in the form of a proportion,*thus cos (a+b): cos(ab): : cot c: tan (A+B)...(S) and sin(a+b) sin (a - b) :: cot 1 c : tan (A — B) c: (AB)... (9) That is, the cosine of half the sum of two sides of a spherical triangle is to the cosine of half their difference, as the cotangent of half the included angle is to the tangent of half the sum of the other two angles.

The second may be repeated in a similar manner, changing cosine into sine and tangent of the half sum into tangent of the half difference of the other two angles.

When two sides and the included angle of a triangle are given, the first three terms of both these proportions will be known, and the fourth in both, or rather its logarithm, may be found by adding the logarithm of the second and third, and the arithmetical complement of the logarithm of the first. We shall then know half the sum (A + B) and half the dif ference (AP) of the two unknown angles. By adding and subtracting these results the angles themselves will be obtained; thus

From the proportions (8) and (9), by means of the property of polar triangles, another set may be derived, which shall be applicable to the solution of a spherical triangle, when two angles and the interjacent side are given. Thus in the proportions (8) and (9) change each of the sides and angles into 180° minus the side opposite in the polar triangle. cos (a + b) becomes cos(180o — A + 180° — B): = cos (180° — (A + B)) COS (A + B)

-

A

since the cosine of the supplement of an arc is equal to minus the cosine of the arc. (Art. 25). Cos } (a - b) becomes cos } (180° (180° — B))

cos } (- A + B) = cos } (A --- B) since (A + B) =-(A — B) and since the cosine of a negative arc is positive (Art. 27.) Cot fc becomes cot} (180°—c)= cot (90°— } c)=tan şc since the cotangent of the complement is the tangent. In a similar manner tan } (A + B) becomes -- tan } (a+b)

b) The two extremes of the proportion by these substitutions become negative; their product will still be equal to that of the means, if they be both made positive; proportion (8) thus becomes

cos } (A + B): cos } (A — B) :: tan c: tan } (a + b) By similar changes proportion (9) becomes sin } (A + B) : sin } (A — B) :: tan} c: tan } (a−b)

- } The last one may be translated into ordinary language thus:

The sine of half the sum of two angles of a spherical triangle is to the sine of half their difference as the tangent of half the interjacent side is to the tangent of half the sum of the other two sides.

The other may be repeated in a similar manner.

These two proportions, and the two from which they were derived, are known by the name of Napier's analogies, * having been first given by Lord Napier, who is celebrated for many useful inventions of a similar character, but chiefly for that of logarithms:

We shall now apply the first set to an

- )

a

EXAMPLE

The latitudes and longitudes of two celestial objects being given to find their distance apart.

Let p be the pole of the ecliptic, s and s' the places of the two celestial objects, then Ps and Ps' will be their coaltitudes, and the angle P will be the difference of their longitudes, since P will be measured by that arc at a quadrant's distance on the ecliptic. (Geom. B. 9, Prop. 6.) Let the latitudes of the two celestial objects be 51° 30' and 20°; and let

their difference of longitude be 31° 34' 26". Their colatitudes will be 38° 30' and 70°. Then we shall know in the above triangle the two sides opposite s and s' which we will calls and s', and the included angle P. The greater side s' =70°, s = 38° 30′ and P = 31° 34′ 26′′. Applying Napier's analogies with the use of logarithms, we have by the first ar.comp.log.cos.(s'+s)or54°15'=0.233401 2

log.cos.(s'-s)or15°45' 9.983381 log.cot. Por 15° 47′13′′ 10.548635

sum rejecting 10 = 10.765417=log.tan.(s'+s) (s' + s) 80° 15′ 41′′

Had (s's) been greater than 90°, its cosine must have been negative, and the first term of the proportion being negative the fourth must have been negative also, and (s' + s) would have been the supplement of the angle found in the tables, since the tangent of the supplement is equal to minus the tangent of an arc. (Art. 36.)

Applying the second proportion we have ar.comp.log.sin.(s'+s) or 54°15' 0.090672 log.sin.(s-s)or 15°45'-9.433675 log.cot.por 15°47′13′′=10.548635

Add

sum rejecting 10=10.072982-log.tan.(s'-s) (ss) 49° 47' 30"

sum

=

(s's')=80° 15′ 41′′

s' = 130° 3' 11"