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which gives

COS b=

To determine when there are two triangles, and also when there is but one, let us consider the second of equations (8.) R? cos B=sin A sin C cos 6-R cos A cos C,

R?cos B+R cos Accs C

sin A sin C Now, if cos B be greater than cos A we shall have

R2 cos B>R cos A cos C, and hence the sign of the second member of the equation will depend on that of cos B, and consequently cos b and cos B will have the same algebraic sign, or b and B will be of the same species. But when cos B>cos A the sin B<sin A : hence

If the sine of the angle opposite the required side be less than the sine of the other given angle, there will be but one solution.

If, however, sin B>sin A, the cos B will be less than cos A, and it is plain that such a value may then be given to cos C, as to render

Rocos B<R cos A cos C, or the sign of the second member of the equation may be made to depend on cos C. We can therefore give such values to C as to satisfy the two equations

R?cos B+R cos A cos C
+cos b=

and
sin A sin C
R2 cos B+R cos A cos C

sin A sin C Hence, if the sine of the angle opposite the required side be greater than the sine of the other given angle there will be two solutions.

Let us first suppose the side b to be less than 90°, or equal to 79° 12' 10".

If now, we let fall from the angle C a perpendicular on the base BA, the triangle will be divided into two right angled triangles, in each of which there will be two parts known besides the right angle. Calculating the parts by Napier's rules we find,

C=130° 54' 26"

c=119° 03' 26". If we take the side b=100° 47' 50", we shall find

C=156° 15' 04"

=152° 14' 18". Z *

cos b=

Ex. 2. In a spherical triangle ABC there are given A=103° 59' 57", B=46° 18' 7", and a=42° 8' 48" ; required the remaining parts. There will but one triangle, since sin B<sin A.

b=30° Ans. C=36° 7' 54"

c=24° 3' 56".

CASE III.

Having given the three sides of a spherical triangle to find the

angles.

For this case we use equations (3.).

sin is sin (1s—a) cos | A=RV

sin b sin c

sin

Ex. 1. In an oblique angled spherical triangle there are given a=56° 40', b=83° 13' and C=114° 30'; required the angles.

f(a+b+c)=4s =127° 11' 30"

(b+(-a)=(isa)=70° 31' 30". Log sin 1s 127° 11' 30"

9.901250 log sin (s—a) 70° 31' 30"

9.974413 -log sin b 830 13

ar.-comp.

0.003051 -log sin € 114° 30'

ar.-comp.

0.040977 Sum

19.919691 Half sum =log cos 1 A 24° 15', 39''

9.959845 Hence, angle A=48° 31' 18".

The addition of twice the logarithm of radius, or 20, to the numerator of the quantity under the radical just cancels the 20 which is to be subtracted on account of the arithmetical complements, to that the 20, in both cases, may be omitted. Applying the same formulas to the angles B and C, we find,

B= 62° 55' 46"

C=125° 19' 02". Ex. 2. In a spherical triangle there are given a 40° 18' 29“, b=67° 14' 28", and c=89° 47' 6" : required the three angles.

A= 34° 22' 18"
Ans.

B= 53° 35' 16"
C=119° 13' 32".

CASE IV.

Having given the three angles of a spherical triangle, to find the

three sides.

For this case we employ equations (7.)

cos(is—B)cos(IS-C) cos ļa=RV

sin B sin C

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Ex. 1. In a spherical triangle ABC there are given A=48° 30', B=125° 20', and C=62° 54'; required the sides.

(A+B+C)=1S= 118° 22'
(SMA)

690 52'
(IS—B)

6° 58' (isC)

550 28' Log cos (Is–B) -6° 58'

9.996782 log cos (įs—C) 55° 28°

9.753495 -log sin B 125° 20' ar.-comp. 0.088415 -log sin С 62° 54'

ar.-comp.

0.050506 Sum

19.889198 Half sum=log cos A=28° 19' 48"

9.944599

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Hence,

side a=56° 39' 36". In a similar manner we find,

b=114° 29' 58" c= 83° 12' 06".

Ex. 2. In a spherical triangle ABC, there are given A=109° 55' 42", B=116° 38' 33", and C=120° 43' 37" ; required the three sides.

a= 98° 21' 40" Ans. b=109° 50' 22"

c=115° 13' 26".

CASE V.

Having given in a spherical triangle, two sides and their in.

cluded angle, to find the remaining parts.

N

For this case we employ the two first of Napier's Analogies

cos į(a+b) : cos }(ab) : : cot;C : tang }(A+B)

sin }(a+b) : sin }(ab) : : cot ;C : tang :(A—B). Having found the half sum and the half difference of the angles A and B, the angles themselves become known; for, the greater angle is equal to the half sum plus the half difference, and the lesser is equal to the half sum minus the half diffe. rence.

The greater angle is then to be placed opposite the greater side. The remaining side of the triangle can then be found by Case II.

Ex. 1. In a spherical triangle ABC, there are given a=68° 46' 2", b=37° 10', and C=39° 23'; to find the remaining parts 3(a+b)=52° 58' 1", }(ab)=15° 48' 1", C=19° 41' 30".

i(a+b) 52° 58' 1" log. ar.-comp. 0.220210 Is to cos žab) 15° 48' 1"

9.983271 So is cot C 19° 41' 30"

10.446254 To tang }(A+B) 770 22 25

10.649735

As cos

As sin {(a+b) 52° 58' 1" log. ar.-comp. 0.097840 Is to sin ja6) 15° 48' 1"

9.435016 So is cot C 19° 41' 30"

10.446254 Totang }(A—B) 43° 37' 21"

9.979110 Hence, A=77° 22' 25" +43° 37' 21"=120° 59' 46" B=77° 22' 25"-43° 37' 21"= 33° 45' 04"

• = 43° 37' 37".

side c

Ex. 2. In a spherical triangle ABC, there are given b=83" 19' 42", c=23° 27' 46', the contained angle A=20° 39' 48" ; to find the remaining parts.

B=156° 30' 16" Ans. C= 9° 11' 48"

a = 61° 32' 12".

CASE VI.

In a spherical triangle, having given two angles and the included

side to find the remaining parts.

For this case we employ the second of Napier's Analogies. cos }(A+B) : cos }(A–B) : : tang ļc : tang i(a+b)

sin 3(A+B) : sin } (A–B) : : tang dc : tang }(ab). From which a and b are found as in the last case. The remaining angle can then be found by Case I.

Ex. 1. In a spherical triangle ABC, there are given A=81° 38' 20", B=70°°9' 38", c=59° 16' 23"; to find the remaining parts. }(A+B)=75° 53' 59", 1(A–B)=5° 44' 21", {c=29° 38' 11"

}(A+B) 75° 53' 59" log. ar.-comp. 0.613287 (A,B) 5° 44' 21"

9.997818 ic 29° 38' 11"

9.755051 i(a+b) 66° 42' 52"

10.366156

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Às cos
To cos
So is tang
To tang

As sin
To sin

So is tang To tang

3(A+B) 75° 53' 59" log. ar.-comp. 0.013286 FA—B) 5° 14' 21"

9.000000 с 29° 38' 11"

9.755051 }(ab) 3° 21' 25"

8.768337

angle C

Hence a=66° 42' 52'' +3° 21' 25'=70° 04' 17" b=66° 42' 52"-3° 21' 25"=63° 21' 27"

=64° 46' 33". Ex. 2. In a spherical triangle ABC, there are given A=34 15' 3", B=42° 15' 13", and c=76° 35' 36''; to find the remaining parts.

a=40° 0' 10" Ans.

b -50° 10' 30" C=58° 23' 41".

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