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not to be used, as was stated in regard to the analogous formulas for a planè triangle (Part Ï. art. 213); the third may be used in any case. In this and the next case, 20 is to be subtracted from the index of the sum of the logarithms.

EXAMPLE. The sides of a spherical triangle are 143° 46′, 67° 24', and 132° 11'; find the angles (see fig. in art. 310). Let a 143° 46' To find angle A.

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b= 67 24
c=132 11

343 21

Cosec c

40

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then

s = 171 s-a 27 54

Ꭶ b=104 16

2)18.996226

s-c 39 29

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In a similar manner, the angles B and C can be found by any of the three rules (see p. 218).

EXERCISES.

1. Find the angles B and C of the triangle, whose sides are given in the preceding example.

Ans. Angle B = 111° 4', and C = 131° 30'. 2. The sides of a spherical triangle are 62° 54′ 4′′, 125° 20', and 131° 30'; what are its angles?

Ans. 83° 12′ 10′′, 114° 30′, and 123° 20'. 351. Case II.-When the three angles of a spherical triangle are given.

I. From half the sum of the three angles subtract the angle opposite to the required side; then add together the logarithmic cosines of the half sum and of this remainder, and the logarithmic cosecants of the other two angles; and half the sum will be the logarithmic sine of half the required angle' (see p. 218).

Let a be the required side, and S half the sum of the angles, then

Or,

cos S cos (S- A)

sin2a =

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={L cos S+L cos (S—A)+ L cosec B+L cosec C}

II. From half the sum of the three angles, subtract separately the angles adjacent to the required side, then add together the logarithmic cosines of the two remainders and the logarithmic cosecants of the other two angles; and half the sum will be the logarithmic cosine of half the required angle."

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={L cos (S-B)+Lcos (S-C)+L cosec B+L cosec C}

A rule may also be gous to that for tan tangent is

And

tan2 α =

given for the value of tana, analoA in 349. The expression for the

cos S⋅ cos (SA) cos (S-B) cos (S—C)*

L tana

={Lsec (S—B)+L sec (S—C)+Lcos S+Lcos (S—A)} 352. The remarks in 350 apply to these three formulas.

EXAMPLE. The angles of a spherical triangle are 114° 30', 83° 12', and 123° 20′; find the sides.

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In the same manner the other sides may be found.

EXERCISES.

1. The three angles are 143° 18', 111° 4′, and 131° 30′; find the sides.

Ans. The sides are 143° 46′, 67° 24', and 132° 11'.

2. The three angles A, B, C, of a spherical triangle are respectively 70° 39′, 48° 36′, and 119° 15'; what are the sides? Ans. The side a 89° 16′ 53′′·5, b = 52° 39' 4"-5, and c = 112° 22′ 58′′-6.

353. The two following cases can be solved by means of the analogies of the circular parts,* which are expressed in the following manner :—

"Let one of the six parts of a triangle be omitted, and let the part opposite to it, or its supplement when it happens. to be an angle, be called the middle part (M); the two parts next it, the adjacent parts (E, e); and the two remaining parts, the opposite parts (0, 0); then

sin(E+e): sin (E-e) tan M: tan (0-0), and cos (E+e): cos (E-e) = tan M: tan (0+0). 1⁄2 1 By means of these two analogies, half the sum and half the difference of O and o are found, and each of them is then found by the principle stated in article 211, Part I.

354. When E, e, O, and o, are given, M can be found from the first of these analogies, by placing it for the last term, and sin (E-e) for the first, and the other two indifferently for the second and third; thus,

=

sin(E-e): sin (E+e) tan (0-0): tan M. Or M can be similarly found from the second analogy.

355. Case III.—When two sides and the contained angle are given, as a, b, and C.

Omit the side c, and make the supplement (353) of C the middle part M; then the sides a, b, are the adjacent parts E, e; and the angles A, B, the opposite parts 0, 0;

and hence (353)
sin(a+b): sin (a~ b)

=

cot C: tan (A~ B), cos (a+b): cos (a ~ b) cot C: tan(A+B). The half sum and half difference of A and B being found by these two analogies, each of them is then easily found by the principle in article 211, Part I.

*These are called Napier's analogies, as they were discovered by

To find the side c.

Reject C, and make c the middle part; then c is M; angles A, B, are adjacent parts; and the sides a, b, are opposite parts; hence (354)

sin (A~B): sin 1 (A + B) = tan 1 (a ~6): tan c.

EXAMPLE. In a spherical triangle, two sides are 84° 14' 29", and 44° 13′ 45′′, and the contained angle 36° 45′ 28′′; required the remaining parts.*

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1

(a + b) = 64° 14′ 7′′, (a - b) = 20° 0′ 22′′;
and C18 22′ 44′′.

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By 211, Part I.,

A = 81° 15' 44"-41 + 48° 49′ 38′′ 130° 5′ 22′′-41,

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=

· 48° 49′ 38′′ — 32° 26′ 6′′·41.

* This example is taken from Woodhouse's Trigonometry.

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1. Given two sides and the contained angle equal to 89° 17', 52° 39', and 119° 15′; to find the other parts. Ans. The other side is 112° 23', and the other angles 48° 36′ and 70° 39'.

2. The sides a and b are 109° 21' and 60° 45', and angle C is 127° 20′ 55′′-5; find the other parts.

Ans. Angles A and B are 90° 43′ 6′′·5 and 67° 37′ 1′′·4, and the side c is 131° 24'.

356. Case IV.-When two angles and the interjacent side are given.

Let the angles A and B and the interjacent side c be given.

To find the sides a and b.

Omit C, and let c be the middle part; then A and B are the adjacent parts, and a and b the opposite parts; hence sin (A+B): sin (A~ B) = tan c: tan (a~ b), cos (A+B): cos(A~ B) = tanc: tan (a + b).

To find angle C.

Omit c, and make the supplement of C the middle part; then the sides a, b, are adjacent parts; and the angles A, B, are opposite parts; hence (354) since tan M = cot C,

sin (ab): sin(a+b) tan (A~ B): cot C.

EXAMPLE. The angles A and B are 130° 5' 22"-41 and 32° 26′ 6′′-41, and the side c is 51° 6′ 11′′-6; required the other parts.

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