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angled triangles ADC and ADB.

* Though the projection generally determines such facts as this, the species of parts, and

the number of solutions in ambiguous cases, the student should not rely upon it, but determine

each such fact upon purely trigonometrical considerations, merely using the projection to give

clearness to the conception, and as a rough check against errors.

94 SPHERICAL TRIGONOMETRY.

Or, having computed C from the triangle ACD, we may find A and B more

expcditiously by using the proportions,

sin c : sin b : : sin C : sin B,

and sin c : sin a : : sin C : sin A.

The angles are C = 34 15' 03", A - 121 36' 12", and B = 42 15' 13".

4. Given A = 128 45', C = 30 35', and a = 68 50', to solve the

triangle. What values of A give two solutions ? What none ?

c = 37 28' 20", b = 40 09' 04", and B = 32 37' 58".

5. Given A = 129 05' 28", B = 142 12' 42", and C = 105 08' 10",

to solve the triangle.

a = 135 49' 20", b = 146 37' 15", and c = 60 04' 54".

6. Given a = 68 46' 02", b = 37 10', and C = 43 37' 38", to

project and solve the triangle.

A = 116 22' 22", B = 35 29' 54", and c. = 45 52' 34".

7. Given a = 40 16', b = 47 44', and A = 52 34', to project

and solve the triangle. What values of a give but one solution?

What none ?

There are tivo triangles. In the 1st, c = 53 12' 22", B = 65 23'

16", and C = 79 40' 26". In the 2d, c = 14 20' 32", B = 114 36'

44", and C = 17 43' 06".

8. Given a = 62 38', b = 10 13' 19", and C = 150 24' 12", to

project and solve the triangle.

A = 27 31' 44", B = 5 17' 58", and c = 71 37' 06".

9. Given a = 56 40', b = 83 13', and c = 114 30', to project

and solve the triangle.

A = 48 31' 18", B = 62 55' 44", and C = 125 18" 56".

10. Given A = 50 12', B = 58 08', and a = 62 42', to solve the

triangle. What values of A give but one solution ? What none ?

There are two solutions. 1st, b = 79 12' 10", c = 119 03' 26",

and C = 130 54' 28". 2d, b = 100 47' 50", c = 152 14' 18", and

C== 156 15' 06".

11. Given A = 36 25', B = 42 17' 10", and C = 95 10' 05", to

project and solve the triangle.

OBLIQUE ANGLED TRIANGLES SOLVED BY NAPIEll's ItULES. 95

12. Given a = 124 53', I = 31 19', and c = 171 48' 22", to

solve the triangle.

13. Given a = 150 17' 23", I = 43 12', and c= 82 50' 12," to

solve the triangle.

14. Given a = 115 20' 10", b = 57 30' 06", and A = 126 37' 3(T,_

to solve the triangle.

15. Given A = 109 55' 42", B = 116 38' 33", and C = 120 43'

37", to solve the triangle.

16. Given A = 50, I 60, and a 40, to solve the triangle.

17. Given a = 50 45' 20", I = 69 12' 40", and A = 44 22' 10",

to project and solve the triangle.

There are two solutions. 1st, B = 57 34' 51". C = 115 57' 51",

and c = 95 18' 16". 2d, B = 122 25' 09", C = 25 44' 32", and c

= 28 45' 05".

18. Given I = 99 40' 48", c = 100 49' 30", and A = 65 33' 10",

to project and solve the triangle.

a = 64 23' 15'', B = 95 38' 04", and c = 97 26' 29".

19. Given A = 48 30', B = 125 20', and c = 62 54', to solve the

triangle.

a = 56 39' 30", I = 114 29' 58", and c = 83 12' 06".

20. Given C = 54 15' 03", B = 40 18' 13", and a = 70 30' 30",

to solve the triangle.

21. Given A = 47 54' 21", c = 61 04' 56", and a = 40 31' 20",

to project and solve the triangle.

22. Given A = 50 10' 10", I = 69 34' 35", and a = 120 30' 30",

to project and solve the triangle.

96 SPHElilCAL TlUGOlNOMETIlY.

SECTION III.

GENERAL FORMULA.

[NOTE. This section is designed for such as

make mathematics a specialty. The preceding

sections are thought sufficient for the general

student]

143. Prop. In a Spherical Tri-

angle the cosine of any side is equal to

the product of the cosines of the other two

sides, plus the product of the sines of those

sides into the cosine of their included

angle ; that is,

(1) cos a = cos b cos c + sin b sin c cos A ;

(2) cos b = cos a cos c + sin a sin c cos B ;

(3) cos c = cos a cos b + sin a sin b cos C.

DEM. From Fig. 67, we have,

cos a = cos (c x) cosp

cos (c x} cos b r . cos b ~]

since cos p =

L cos#J

[expanding cos (c

cos x

cos b cos c cos x + cos b sin c sin x

cos x

= cos b cos c + cos b sin c tan x since - = tan x

L COS J

= cos b cos c + sin b sin c cos A since cos A = cot b tan x = i

L sin b J

In a similar manner (2) and (3) may be produced.

144. COR. 1. From set A, by passing to the polar triangle, we

have,

(1) cos A = cos B cos C + sin B sin C cos a ; \

(2) cos B = cos A cos C + sin A Bin C cos b ; /- B.

(3) cos C = cos A cos B + sin A sin B cos c. J

DEM. Letting a', 6', cf, A', B', and C' represent the parts of the polar triangle,

GENERAL FORMULA.

97

we have a - 180 - A', b = 180 B', c = 180 - C', A = 180 - a',__B =

180 &', and C 180 - c'. Whence, substituting in (1) A, we have,

cos (180 - A') = cos (180 - B') cos (180 - C')

+ sin (180 - B') sin (180 C') cos (180 - '),

or, cos A' = cos B' cos C' + sin B' sin C' cos a',

since cos (180 A') = cos A', etc. ; and sin (180 B') = sin B', etc.

Finally, dropping the accents, since the results are general, and treating (2)

and (3) of set A in the same way, we have set B.

145. COR. 2. From A and B we readily find the angles in terms

of the sides, and the sides in terms of the angles. Thus, from A,

,., v cos a cos b cos c ^

(1) cos A = . . -;

sin b sin c

(2) cos B =

(3) cosc =

cos b cos a cose

sin a sin c

cos c cos a cos b

sin a sin &

^ A'.

From B,

,., x COS A + COS B COS C 1

(1) cos a=. . = ;

sin B sm C

/ rt \ 2. cos B + cos A cose .

(2) cos b = ; -

(3) cos c =

smA sine

cos C + cos A cos B

sin A sin B

B'.

146. ~Prop. Formula A', and B', adapted to logarithmic com-

putation, become,

/-I N 1 j / sm (i s &) sm (is c )

(1) sm 4 A = A / -

v ' I/ sm i sm c

(2) sin f B =

sin (Js a) sin (J-s c)

sin a sin c

l*\ > <(,. A /

(3) smjc-j/-

sin ($s ~ a) sin (%s b)

7

. A".

SPHERICAL TRIGONOMETRY.

, ,, x t / cos is cos (is A )

And (1) sm \ci = A/ - s i n B 8inc~

4 / cos is cos (is B )

(2) smi&=4/ sinAsm c

cos JS cos (JS - C )

sinAsinB

. Subtracting each member of (1) A' from 1, we have,

cos a cos 6 cos c cos 5 cos c + sin b sin cos a

sin b sin c sin 6 sin c

1 _ cos A = 1 -

... 2 sin 2 iA = T=L, since 1 - cos A = 2 sin 2 *A (62, 5),

sin c

and cos 5 cos c + sin & sin c = cos (6 c) (55, D).

Now letting y b c, and x = a, we see from (59, D') that cos (b c)

cos a = 2 sin -J(a + 6 ) sin |(a + c 5).

Hence, 2 sin' JA = tott * - 4 Bto + ~ )

sm 6 sin c

or, sin A =

sin 6 sin c

Finally, putting s = a + b + c, whence %(a + b c)= s

we have,

sin

In like manner, (2) and (3) of set A' reduce to (2) and (3) of set A".

Again, subtracting each member of (1) set B', from 1, we have, 98

cos A + cos B cos C sin B sin C cos B cos C cos A

1 cos a = 1

sin B sin C sin B sin C

. o 2 i _ ~~ cos (B + C) cos A _ _ cos (B + C) + cos A

"~15nBsinC " sin B~s!n~C~~

_ 2 cos j(A + B + C) cos j(B + C + A)

sin B sin C

GENERAL FORMULA. 99

Now putting S = A + B + C, whence ^(B + C A) = S A, we have,

. / cos S cos (|S A)

= r ' -anras-c

In the same manner, (2) and (3) of B" are deduced from (2) and (3) ot Ls .

147* COR. 1. Passing to the polar triangle, A" and B" become

(1) cc

_ A /cos (-|-S - B) cos (|s C).

sin B sin C

(2)

-v-

_ A /cos(s - A)cos(j

sin A sin C

(3) cos 40 = ,Aos(iS-A)co S as-B).

I/ sin A sm B

B'

(1)

COS

* A ~~ y " sin <

*y

b sine

A /sm | sin (%s

sin a sin c

-A'

. /sin &s sin (As c)

W*i C =y - *maLk

Sen. Formulae A'" and B'" can be obtained directly from A and

B, in a manner altogether similar to that in which A" and B" were deduced, by

adding each member of the equations in sets A and B to 1, instead of subtract-

ing, and observing that 1 + cos x 2 cos 2 \x.

149. COR. 2. Dividing the formula of set A" by the correspond-

ing ones of set A"' ; and, in a similar manner, those of B'" by

/ -o// ? jj. - A / sm (~k s ~ a ) sm (i 5 ~ ^) sm

those o/B , and putting \ - ^ lg '

and

sm

- cos

=

A /sin (^s b) sin (%s c) _ k 9

sin %s sin (^,s- a) sin (-J-s a) '

sin -|5 sin (-J5 lj)

(3) tan JC =

sn

sm (-|x-

) sin (%s b) _ k

sm

sn

c)

sn

c)'

A 1

100

SPHERICAL TRIGONOMETRY.

a)

(2)

(3)

Y

COS

- B) cos (^S C) __

/cos (is A) cos

/

v

cos is cos (is A) cos (is A) '

~ C) K

- cos is cos (is B) cos (is - B) '

_ /cos (JS - A) cos (JS - B) K_

~~ 'I/ cos is cos (is C) ~ cos (is

- c).

Sen. In these formulae, k is the tangent of the arc with which the inscribed

circle is described, and K is the cotangent of the arc with which the circum-

FIG. 68.

FIG. 69.

scribed circle is described. Thus, using the common notation, we have in

Fig. 68, AD = AD' = %s a, and angle PAD = A; whence

tan PD

sin AD = cot PAD x tan PD =

tan PAD 1

or tan \k =

tan PD

k

, [(1) A"]. .-. k - lanPD.

sin ($s a) sin (^ )'

From Fig. 69, we have, AD = \c, and angle PAD = |S C.

Hence, cos (S C) = cot AP x tan^c, or tan \c - ~ ^^

cot AP K

or cot \c =

cos(iS- C)~cos(iS C)

, [(3) B IV ]. .-. K = cot A P.

GAUSS'S EQUATIONS.

. frob. To deduce Gauss's Equations, which are

sin |(A + B) _ cos(fl - b) .

(1)

cos c cos %c

sin -J(A B-) _ sin %(a b) <

cos Jc sin ^c

GAUSS'S EQUATIONS. _ , . 191

cosj(A + B) _

' sin^c

- B) _ sin \(a

sn C sn c

SOLUTION. From A, page 25, we have,

sin (iA + B), or sin K A + B ) = sin i A cos i B + cos i A sin B -

Substituting in the second member the values of sin A, cos 6, cos |A, and

sin ^B , from A" and A'", there results,

_ sin (\s - b) A /sin is sin (is -c) sin (ja - a) /sin -s sin(js

~~1E^ T ' sina"sln 6 ~ sin c V sin a sin

c;

a sin 6

_ sin (\B - V) + sin (fo a) ^ sin jg sin ^ - c) gee A//

sin c f sin sin 6

__ sin (js - 6) + sin Qg - a) cQg Q

shi c

But sin (is 6) + sin (i* a) = sin (%a + $ + \c b) + sin (}a + ib + |c a)

= sin [\c + i(a b)] + sin [ic |(a 6)]

= 2 sin |c cos i(a - b). (59, A').

Also, sin c = 2 sin |c cos |c.

Substitutmg these values, the preceding becomes,

2 sin -ic cos 4( b)

sin |(A + B) 1 - - J- cos ^C .

2 sin -Jc cos ic

. sin j(A + B) _ cos j(^ - &)

cos ^C cos i<;

In like manner starting with

sin (^A |B ), or sin |(A B) = sin -A cos |B cos |A sm ^B,

sin i(A - B) sin i(a - b) ....

there results, ____ ^ -^ (2 )

Starting with cos (A + |B), or cos |(A + B) =r cos^A cos |B - sin |A sin iB,

cos |(A + B) cos $(a + b) ...

there results, ^-^ - = - =r-s - (3)

sin |C cos ic

Starting with cos (|A B), or cos|(A B) = cos ^A cos ^B

cos KA B) sin \(a + b)

there r e8 ults, __J ^-.-J (4 )

102 SPHERICAL TltlGONOMETItY.

NAPIER'S ANALOGIES.

. Prob. To deduce Napier 's Analogies, -which an

\ tan |(A + B) _ cos(a - b) m

( tanj(A- B)

cot |C sin %(a + b)

(*\ ^ an i(fl 4- b) _ cos |(A B).

tan Jc ~ cos (A + B)'

... tan(a I) _ sin(A B)

tan Jc sin^(A 4- B)*

SOLUTION. To deduce (1), divide the 1st of Gauss's Equations by the 3d.

To deduce (2), divide the 2d of Gauss's by the 4th. To deduce (3\ divide the 4th

of Gauss's by the 3d. To deduce (4), divide the 2d of Gauss's by the 1st.

152. SCH. In using these formulsB the species must be carefully attended

to. Thus in (1), cot C and cos \(a b) are necessarily + ; hence tan i(A + B)

and cos-J(a + b) are of the same sign with each other. In (2), cot ^C and

sin $(a + b) are both + ; hence, tan |(A B) and sin %(a b) are of the same

sign with each other. And similar inspections may be made upon (3) and (4).

EXERCISES.

153. The proposition that " The sines of the angles are to each

other as the sines of their opposite sides" (135), Napier's Analogies

(151), and formulae A IV , B IV (149) are sufficient, in themselves, to

effect the solution of all cases of oblique spherical triangles; and

for practical purposes they generally require less labor than Napier's

Rules. We give a few solutions and refer the student to the pre-

ceding Exercises for further practice.

1. Given a = 100; c = 5 and I = 10, to solve the triangle.

(Prob. 1, Case 1st, 13V.)

1st. To find A and B we have,

cos i(a + b) : cos |(a b) : : cot C : tan i(A + B) ;

and sin $(a + b) : sin i(ti - b) : : cot C : tan i(A - B) [15O (1) (2)]

NAPIER'S ANALOGIES. 103

Computing by logarithms, we have,

ar. co. log cos [$(a + b) = 55] = 0.241409

+ log cos [i(a - b) = 45*1 = 9-849485

-f log cot [| C = 2 30'] = 11.359907

Rejecting 10 = 11.450801 = log tan K* + B )-

.-. KA + B) = 87 58' 18".

ar. co. log sin [K + b) = 55] = 0.086635

+ log sin [K - ft) = 451 = 9.849485

+ log cot [iC = 2 30'] = 11.359907

Rejecting 10 = 11.296027 = log tan (A - B).

... i(A _ B) = 87 06' 16"

The signs of all the terms being + , KA, + B) and K A B) are both lesa

than 90.

KA + B) + KA - B) = A = 87 58' 18" + 87 06' 16" = 175 04' 34"

KA + B) - KA - B) = B = 87 58' 18" - 87 06' 16" = 52' 02".

2d. To find c. This may be found from the proportion,

sin A : sin C : : sin a : sin c,

or from the 3d or 4th of Napier's Analogies. We use the last, though the first

is equally expeditious.

sin K A - B) : sin $(A + B) : : tan K ft) : tan \c.

ar. co. log sin [KA - B) = 87 06' 16"] = 0.000555

+ log sin [KA + B) zr 87 68' 18"] = 9.999728

+ log tan [Ka - 6) = 45] = 10.000000

Rejecting 10 = 10.000283 = log tan \c.

.'. c = 90 02' 14".

2. Given A = 135 05' 28".6, c = 50 30' 08".6, and I = 69 34

5 6 ".2, to solve the triangle.

1st. To find a and c. The 3d and 4th of Napier's Analogies give,

cos i(A + C) : cos KA C) : : tan $b : tan K + c ) ;

and sin K A + C) : sin KA - C) : : tan 6 : tan \(a c\

Computing by logarithms, we have

ar. co. log cos [KA + C) = 92 47' 48". 8] = 1.3116286 *

+ log cos [KA - C) = 42 17' 40"] = 9.8690535

+ log tan [& = 34 47' 28".l] : 98418527

Rejecting 10 = 11.0225348 = log tan K + c).

.-. K + c) = 95 25' 25".

(a + c) > 90, since cos i(A + C) is , cos KA C) is +, and tan ^b is +,

making tan (a + c) .

* These logarithms are taken from 7-place tables, in order to obtain the tenths of eeconds

accurately.

104

SPHERICAL TRIGONOMETRY.

ar. co. log sin |>(A + C) = 92 47' 48".6] = 0.0005176

+ log sin [(A - C) = 42 17' 40" ] = 9.8279768

+ log tan [p = 34 47' 28". 1] = 9.8418527

Rejecting 10 = 9lJ70347T = log tan \(a - e).

.: i(a -c) = 25 05' 05".

\(a c) < 90, since the signs of the terms are all + .

K + c) + \(a - c) = a = 120 30' 30", and #a + c) $(a c) = c = 70 20' 20".

3d. To find B. Either of the 1st two of Napier's Analogies will give B.

Thus (1) becomes,

cos $(a c) : cos $(a + c) : : tan |(A + C) : co

and (2) sin $(a c) : sin i(a + c) : : tan |(A C) : co

But as i(a + c) is so near 90, it will be better to use the second of these

than the first. Or we may with equal accuracy use,

sin c : sin b : : sin C : sin B.

ar. co. log sin (c = 70 20' 20" ) = 0.0260878

+ log sin (b = 69 34' 56".2) = 9.9718202

+ log sin(C = 50 30' 08". 6) = 9.8874210

Rejecting 10 = 9.8853290= log sin B. .-. B = 50 10' 10".

3. Given a = 50 45' 20", b = 69 12' 40", and A = 44 22' 10",

to solve the triangle.

1st. To find B.

ski a t sin b i t sin A : sin B.

ar. co. log sin (a = 50 45' 20'') = 0.1110044

+ log sin (b = 69 12' 40") = 9.9707626

+ log sin(A = 44 22' 10") = 9.8446525

Rejecting 10 = 9.9264195 = log sin B. .-. B = 57 34' 51".4,

and 122 25' 08". 6. There are two solutions, since a is intermediate in value

between p and both b and 180 b*

* The determination of the species of B, or what ia

the same thing, the number of solutions, can usually be

effected by a simple inspection without any computa-

tion whatever. Thus, sin p = sin b sin A, the loga-

rithms of which are given above, as ie log sin a. Now,

as both a and p are < 90, and log sinp < log sin a,

p <a. But o<6, and aleo less than 180 &. All

this can be seen at a glance,

NAPIER'S ANALOGIES. 105

To find C and c of the larger triangle in which B = 57 34' 51" A

Napier's 1st gives

ar. co. log cos [K& - a) = 9 13' 40"] = 0.0056570

+ log cos [K& + 1) = 59 59'] = 9.6991887

+ log tan [KB + A) = 50 58' 30".7] = 10.0912464

Rejecting 10 = 9.7960921 = log cot C.

.-. C = 115 57' 50".77

Napier's 3d gives

ar. co. log cos [KB - A) = 6 36' 20".7] = 0.0028928

+ log cos [KB + A) = 50 58' 30".7] = 9.7991039

+ log tan [K& + a) - 59 59'J = 10.2882689

Rejecting 10 = 10.0402656 - log tan fa

.-. c = 95 18' 16".4,

3d. To find C and c of the smaller triangle in which B = 122 25' 08".6.

Using the same formula as before.

ar. co. log cos [|(6 - a) = 9 13' 40"] = 0.0056570

+ log cos [K& + a) = 59 59'] = 9.6991887

+ log tan [|(B + A) = 83 23' 39".3] = 10.9362703

Rejecting 10 = 10.6411160 = log cot C.

/. C = 25 44' 31".6.

ar. co. log cos [KB - A) = 39 01' 29".3] = 0.1096506

+ log cos [KB + A) = 83 23' 39".3] = 9.0608369

+ log tan [i(b + a) = 59 59'] = 10.2382689

Rejecting 10 = 9.4087564 = log tan fa

.'. c = 28 45' 05".2.

154:. Sen. When Napier's Analogies are used for solving Prob.2nd(139], the

most expeditious and elegant method of resolving the ambiguity, is by means

of the analogies themselves. Thus, in the above example, after having found

that B = 57 34' 51".4, or 122 25' 08". 6, or both, a simple inspection of the anal-

ogy next used will determine the number of solutions.

Napier's 1st may be written

Now |C < 90, hence cot |C is +. If, therefore, neither of the values of

B renders cot ^C , there are two solutions. If one value renders cot -C + ,

and the other , there is one solution and it corresponds to the value of B

which makes cot |C +. If both values of B render cot |C , there is no solu-

tion. In the last example, we see that cos [K& + o) = 59 59'], and cos [K& )

= 9 13' 40"] are both + . Also tan [KB + A) = 50 58' 30". 7, or 83 23' 39".3,

or both] is + for both values of B. Therefore there are two solutions.

106 SPHERICAL TRIGONOMETRY.

4. Given A = 95 16', B = 80 42' 10", and a = 57 38', to solve

the triangle.

1st. To find 5, sin A : sin B : : sin a : sin 5.

ar. co. loo; sin (A = 95 16') = 0.001837

+ log siii (B = 80 42' 10") - 9.994257

+ log sin (a - 57 38') = 9.926671

Rejecting 10 = 97922765 - log sin b.

: b = 56 49' 57", or 123 10' 03", or both.

2d. To find <, tan $c = ^~^ ^_ |^ tan |( + b). Now for 6 = 56 49' 57",

tan \G is + ; but for b = 123 10' 03" tan $c is ; hence there is but one solu-

tion, and that corresponds to the smaller value of b.

ar. co. log cos [i(A - B) = 7 16' 55"] = 0.003517

+ log cos [|(A + B) = 87 59' 05"] = 8.546124

+ log tan [i(<j + b) - 57 13' 58"] = 10.191352

Rejecting 10= 8.740993 = log tan \c.

.'. Cr= 6 18' 19".

3d. To find C, we may use (1) or (2) of Napier's Analogies, or

sin a : sin c : : sin A : sin C,

the last of whicn is the most expeditious.

ar. co. log sin (a = 57 38') = 0.073329

+ log sin (c = 6 18' 19") = 9.040705

+ log sin (A = 95 16') = 9.998163

Rejecting 10 = 9.112197 = log sin C- .-. C = 7 26' 22'

This value is taken for C instead of its supplement, since C is opposite the

smallest side c.

5. Given a = 70 14' 20", I = 49 24' 10", and c = 38 46' 10"; to

solve the triangle.

COMPUTATION-.

a = 70 14' 20"

b = 49 24' 10"

c - 38 46' 10"

s = 158 24' 40"

is = 79 12' 20" ar. co. log sin = 0.007753

a = 8 58' 00" " " = 9.192734

b = 29 48' 10" " " = 9.696370

c = 40 26' 10" " " = 9.811977

2)18.708834

.-. log k = 9.354417

NAPIER'S ANALOGIES. 107

log tan |A = log k - log sin (is - a) + 10 = 10.161683. .-. A = 110 51 16".

log tan |B = log k - log sin (it - b) + 10 = 9.658047. .-. B = 48* 56' 04".

log tan |C = log k - log sin (fr - c) + 10 = 9.542440. .-. C = 38 26' 48".

6. Given A = 109 55' 42", B = 116 38' 33", andc = 12043' 37",

to solve the triangle.

COMPUTATION.

A = 109 55' 42"

B = 116 38' 33"

C = 120 43' 37"

S = 347 17'"52"

|S = 173 38' 56" ar. co. log cos = 0.002683

S - A - 63 43' 14" " " = 9.646158

|S - B = 57 00' 23" " " = 9.736035

iS - C = 52 55' 19" " " = 9.780247

2)19.165123

.-. log K = 9.582561

log cot $a = log K log cos (|S A) + 10 = 9.936403. /. a = 98 21' 38".

log cot # = log K - log cos (|S - B) + 10 = 9.846526. .'. b = 109 50' 20".

log cot $c - log K - log cos (iS - C) + 10 - 9.802314. .-. c = 115 13' 28".

Sen. 1. The student can use the exercises in the preceding section to famil-

iarize the methods here given. In doing so, it will be well for him to seek the

most expeditious solutions. He will find that

Examples under PROS. 1 require 11 logarithms by Napier's Analogies and

(135], and 12 logarithms by Napier's Rules and (135).

Examples under PROB. 2, when there is but one solution, require 10 loga-

rithms by Napier's Analogies and (135), and 12 logarithms by Napier's Rules

and (135). When there are two solutions, 15 logarithms are required by

Napier's Analogies and (135), and only 14 by Napier's Rules alone, or by these

rules and (135).

Examples under PROB. 3 require but 7 logarithms by the method given in this

section and 13 by the previous method.

Sen. 2. In cases in which the angles or sides are near the limits 0, 90, or

180, so that the functions used in the particular solution change very rapidly

in proportion to the arc, it is often possible to select one among the several

methods which will give more accurate results than the others. There are

also other methods which are better adapted to such cases than those here

given. For these, as well as for much other interesting matter, and especially

for the discussion of the General Spherical Triangle, American students have an

excellent resource in the treatise of Professor Chauvenet of Washington Univer-

sity, St. Louis.

108 SPHEBICAL TKIGONOMETKY.

s

SECTION IV.

AREA OF SPHERICAL TRIANGLES.

155 I*rob. Having the angles of a spherical triangle given, to

find the area.

SOLUTION. [The solution is given in PART II. (613), and we simply re-

produce the result in order to give completeness to this section.] The area is

equal to the ratio of the spherical excess to 90, or ITT, into the trirectangular tri-

angle. That is, letting the sum of the angles be S, the area K, and the radius of

the sphere 1, whence the area of the trirectangular triangle is #, we have

K = ^fl5 x \it = S - it.

i*

In the latter expression S is the sum of the angles in terms of the radius, i. e.,

S = 57^578' r a PP roximate ^ S = 5^3 W

EXERCISES.

1. What is the area of a spherical triangle whose angles are 100,

58, and 62, on a sphere whose diameter is 6 feet ?

220

SOLUTION. K = S it - - - 3.14159 = .698, the area of a similar tri-

57 .o

angle on a sphere whose radius is 1. Hence, the area of the required triangle

is .698 x 3 2 = 6.282. [The method given in PART II. (613) is more expedi-

tious, but it is our purpose to illustrate the form here given.]

2. What is the area of a spherical triangle whose angles are 170,

1 35, and 115, on a sphere whose radius is 10 feet?

Ans. 418.875 square feet.

3. What is the area of a spherical triangle whose angles are 150,

110, and 60, on a sphere whose radius is 3 feet ?

. Prob. Having the sides of a spherical triangle given, to

find the area.

SOLUTION. The angles mav be found by (148), and then the area by (155).

AREA OF SPHERICAL TRIANGLES. 103

But a more direct method is to find the spherical excess by means of Lh-utt*

tier's formula, which we will now produce.

*K = *(A 4- B + C - n)

Whence tan $K = tan }[A + B + C - it] = tan [4(A + B) - \(it - C)]

_ sin j(A + B) - sin #it - C)

~ cos i( A" + B) + cos fat - C) (7, page 31}

_ sin j(A + B) cos jC

~ cos-KA + B) + sin C

[cos i(a + b) + cos ^c] sin

lst and 3d)

cos ifrt - b) - cos jc / sin fo sin (fo - c) (146, 147)

cos $(a + b) + cos |c V sin (|s a) sin (|s 6)

_sin^i(a + c - b) sin j(6 + c - a) / sin jg sin (jg - c) c/

cos \(a + b + c) cos \(a + b c) V sin ($s a) sin ($s - b)

. /sin* i(is - ft) sin 2 -Kjs - ^"sin ^ sin (jT^) , g = a + ft + ,

r cos 2 is cos 2 i(|.s c) sin (|s a) sin ($8 b)

/ sin 2 j(js b) sin 2 j(k* a ) sm i g cos ^ gsm i(^ s ~ c ) cus ^ ~"0

11 T cos 2 |s cos 2 Kl*-c) sin |(is-a)"cos~i(|-a) sin |(i*-6) cos i(is-b)

.: Tan iK =* ^/tan i* tan i(|s - a) tan i(|s - 6) tan ^ - c). (A)

Having found K, the spherical excess, or what is the same thing, the area of

a similar triangle on a sphere whose radius is 1, we have but to multiply K by

the square of the radius in any given case.

EXERCISES.

1. Given a = 98, b = 110, and c = 115, to find the area of a

spherical triangle, on a sphere whose radius is 4000 miles.

COMPUTATION.

* Though the projection generally determines such facts as this, the species of parts, and

the number of solutions in ambiguous cases, the student should not rely upon it, but determine

each such fact upon purely trigonometrical considerations, merely using the projection to give

clearness to the conception, and as a rough check against errors.

94 SPHERICAL TRIGONOMETRY.

Or, having computed C from the triangle ACD, we may find A and B more

expcditiously by using the proportions,

sin c : sin b : : sin C : sin B,

and sin c : sin a : : sin C : sin A.

The angles are C = 34 15' 03", A - 121 36' 12", and B = 42 15' 13".

4. Given A = 128 45', C = 30 35', and a = 68 50', to solve the

triangle. What values of A give two solutions ? What none ?

c = 37 28' 20", b = 40 09' 04", and B = 32 37' 58".

5. Given A = 129 05' 28", B = 142 12' 42", and C = 105 08' 10",

to solve the triangle.

a = 135 49' 20", b = 146 37' 15", and c = 60 04' 54".

6. Given a = 68 46' 02", b = 37 10', and C = 43 37' 38", to

project and solve the triangle.

A = 116 22' 22", B = 35 29' 54", and c. = 45 52' 34".

7. Given a = 40 16', b = 47 44', and A = 52 34', to project

and solve the triangle. What values of a give but one solution?

What none ?

There are tivo triangles. In the 1st, c = 53 12' 22", B = 65 23'

16", and C = 79 40' 26". In the 2d, c = 14 20' 32", B = 114 36'

44", and C = 17 43' 06".

8. Given a = 62 38', b = 10 13' 19", and C = 150 24' 12", to

project and solve the triangle.

A = 27 31' 44", B = 5 17' 58", and c = 71 37' 06".

9. Given a = 56 40', b = 83 13', and c = 114 30', to project

and solve the triangle.

A = 48 31' 18", B = 62 55' 44", and C = 125 18" 56".

10. Given A = 50 12', B = 58 08', and a = 62 42', to solve the

triangle. What values of A give but one solution ? What none ?

There are two solutions. 1st, b = 79 12' 10", c = 119 03' 26",

and C = 130 54' 28". 2d, b = 100 47' 50", c = 152 14' 18", and

C== 156 15' 06".

11. Given A = 36 25', B = 42 17' 10", and C = 95 10' 05", to

project and solve the triangle.

OBLIQUE ANGLED TRIANGLES SOLVED BY NAPIEll's ItULES. 95

12. Given a = 124 53', I = 31 19', and c = 171 48' 22", to

solve the triangle.

13. Given a = 150 17' 23", I = 43 12', and c= 82 50' 12," to

solve the triangle.

14. Given a = 115 20' 10", b = 57 30' 06", and A = 126 37' 3(T,_

to solve the triangle.

15. Given A = 109 55' 42", B = 116 38' 33", and C = 120 43'

37", to solve the triangle.

16. Given A = 50, I 60, and a 40, to solve the triangle.

17. Given a = 50 45' 20", I = 69 12' 40", and A = 44 22' 10",

to project and solve the triangle.

There are two solutions. 1st, B = 57 34' 51". C = 115 57' 51",

and c = 95 18' 16". 2d, B = 122 25' 09", C = 25 44' 32", and c

= 28 45' 05".

18. Given I = 99 40' 48", c = 100 49' 30", and A = 65 33' 10",

to project and solve the triangle.

a = 64 23' 15'', B = 95 38' 04", and c = 97 26' 29".

19. Given A = 48 30', B = 125 20', and c = 62 54', to solve the

triangle.

a = 56 39' 30", I = 114 29' 58", and c = 83 12' 06".

20. Given C = 54 15' 03", B = 40 18' 13", and a = 70 30' 30",

to solve the triangle.

21. Given A = 47 54' 21", c = 61 04' 56", and a = 40 31' 20",

to project and solve the triangle.

22. Given A = 50 10' 10", I = 69 34' 35", and a = 120 30' 30",

to project and solve the triangle.

96 SPHElilCAL TlUGOlNOMETIlY.

SECTION III.

GENERAL FORMULA.

[NOTE. This section is designed for such as

make mathematics a specialty. The preceding

sections are thought sufficient for the general

student]

143. Prop. In a Spherical Tri-

angle the cosine of any side is equal to

the product of the cosines of the other two

sides, plus the product of the sines of those

sides into the cosine of their included

angle ; that is,

(1) cos a = cos b cos c + sin b sin c cos A ;

(2) cos b = cos a cos c + sin a sin c cos B ;

(3) cos c = cos a cos b + sin a sin b cos C.

DEM. From Fig. 67, we have,

cos a = cos (c x) cosp

cos (c x} cos b r . cos b ~]

since cos p =

L cos#J

[expanding cos (c

cos x

cos b cos c cos x + cos b sin c sin x

cos x

= cos b cos c + cos b sin c tan x since - = tan x

L COS J

= cos b cos c + sin b sin c cos A since cos A = cot b tan x = i

L sin b J

In a similar manner (2) and (3) may be produced.

144. COR. 1. From set A, by passing to the polar triangle, we

have,

(1) cos A = cos B cos C + sin B sin C cos a ; \

(2) cos B = cos A cos C + sin A Bin C cos b ; /- B.

(3) cos C = cos A cos B + sin A sin B cos c. J

DEM. Letting a', 6', cf, A', B', and C' represent the parts of the polar triangle,

GENERAL FORMULA.

97

we have a - 180 - A', b = 180 B', c = 180 - C', A = 180 - a',__B =

180 &', and C 180 - c'. Whence, substituting in (1) A, we have,

cos (180 - A') = cos (180 - B') cos (180 - C')

+ sin (180 - B') sin (180 C') cos (180 - '),

or, cos A' = cos B' cos C' + sin B' sin C' cos a',

since cos (180 A') = cos A', etc. ; and sin (180 B') = sin B', etc.

Finally, dropping the accents, since the results are general, and treating (2)

and (3) of set A in the same way, we have set B.

145. COR. 2. From A and B we readily find the angles in terms

of the sides, and the sides in terms of the angles. Thus, from A,

,., v cos a cos b cos c ^

(1) cos A = . . -;

sin b sin c

(2) cos B =

(3) cosc =

cos b cos a cose

sin a sin c

cos c cos a cos b

sin a sin &

^ A'.

From B,

,., x COS A + COS B COS C 1

(1) cos a=. . = ;

sin B sm C

/ rt \ 2. cos B + cos A cose .

(2) cos b = ; -

(3) cos c =

smA sine

cos C + cos A cos B

sin A sin B

B'.

146. ~Prop. Formula A', and B', adapted to logarithmic com-

putation, become,

/-I N 1 j / sm (i s &) sm (is c )

(1) sm 4 A = A / -

v ' I/ sm i sm c

(2) sin f B =

sin (Js a) sin (J-s c)

sin a sin c

l*\ > <(,. A /

(3) smjc-j/-

sin ($s ~ a) sin (%s b)

7

. A".

SPHERICAL TRIGONOMETRY.

, ,, x t / cos is cos (is A )

And (1) sm \ci = A/ - s i n B 8inc~

4 / cos is cos (is B )

(2) smi&=4/ sinAsm c

cos JS cos (JS - C )

sinAsinB

. Subtracting each member of (1) A' from 1, we have,

cos a cos 6 cos c cos 5 cos c + sin b sin cos a

sin b sin c sin 6 sin c

1 _ cos A = 1 -

... 2 sin 2 iA = T=L, since 1 - cos A = 2 sin 2 *A (62, 5),

sin c

and cos 5 cos c + sin & sin c = cos (6 c) (55, D).

Now letting y b c, and x = a, we see from (59, D') that cos (b c)

cos a = 2 sin -J(a + 6 ) sin |(a + c 5).

Hence, 2 sin' JA = tott * - 4 Bto + ~ )

sm 6 sin c

or, sin A =

sin 6 sin c

Finally, putting s = a + b + c, whence %(a + b c)= s

we have,

sin

In like manner, (2) and (3) of set A' reduce to (2) and (3) of set A".

Again, subtracting each member of (1) set B', from 1, we have, 98

cos A + cos B cos C sin B sin C cos B cos C cos A

1 cos a = 1

sin B sin C sin B sin C

. o 2 i _ ~~ cos (B + C) cos A _ _ cos (B + C) + cos A

"~15nBsinC " sin B~s!n~C~~

_ 2 cos j(A + B + C) cos j(B + C + A)

sin B sin C

GENERAL FORMULA. 99

Now putting S = A + B + C, whence ^(B + C A) = S A, we have,

. / cos S cos (|S A)

= r ' -anras-c

In the same manner, (2) and (3) of B" are deduced from (2) and (3) ot Ls .

147* COR. 1. Passing to the polar triangle, A" and B" become

(1) cc

_ A /cos (-|-S - B) cos (|s C).

sin B sin C

(2)

-v-

_ A /cos(s - A)cos(j

sin A sin C

(3) cos 40 = ,Aos(iS-A)co S as-B).

I/ sin A sm B

B'

(1)

COS

* A ~~ y " sin <

*y

b sine

A /sm | sin (%s

sin a sin c

-A'

. /sin &s sin (As c)

W*i C =y - *maLk

Sen. Formulae A'" and B'" can be obtained directly from A and

B, in a manner altogether similar to that in which A" and B" were deduced, by

adding each member of the equations in sets A and B to 1, instead of subtract-

ing, and observing that 1 + cos x 2 cos 2 \x.

149. COR. 2. Dividing the formula of set A" by the correspond-

ing ones of set A"' ; and, in a similar manner, those of B'" by

/ -o// ? jj. - A / sm (~k s ~ a ) sm (i 5 ~ ^) sm

those o/B , and putting \ - ^ lg '

and

sm

- cos

=

A /sin (^s b) sin (%s c) _ k 9

sin %s sin (^,s- a) sin (-J-s a) '

sin -|5 sin (-J5 lj)

(3) tan JC =

sn

sm (-|x-

) sin (%s b) _ k

sm

sn

c)

sn

c)'

A 1

100

SPHERICAL TRIGONOMETRY.

a)

(2)

(3)

Y

COS

- B) cos (^S C) __

/cos (is A) cos

/

v

cos is cos (is A) cos (is A) '

~ C) K

- cos is cos (is B) cos (is - B) '

_ /cos (JS - A) cos (JS - B) K_

~~ 'I/ cos is cos (is C) ~ cos (is

- c).

Sen. In these formulae, k is the tangent of the arc with which the inscribed

circle is described, and K is the cotangent of the arc with which the circum-

FIG. 68.

FIG. 69.

scribed circle is described. Thus, using the common notation, we have in

Fig. 68, AD = AD' = %s a, and angle PAD = A; whence

tan PD

sin AD = cot PAD x tan PD =

tan PAD 1

or tan \k =

tan PD

k

, [(1) A"]. .-. k - lanPD.

sin ($s a) sin (^ )'

From Fig. 69, we have, AD = \c, and angle PAD = |S C.

Hence, cos (S C) = cot AP x tan^c, or tan \c - ~ ^^

cot AP K

or cot \c =

cos(iS- C)~cos(iS C)

, [(3) B IV ]. .-. K = cot A P.

GAUSS'S EQUATIONS.

. frob. To deduce Gauss's Equations, which are

sin |(A + B) _ cos(fl - b) .

(1)

cos c cos %c

sin -J(A B-) _ sin %(a b) <

cos Jc sin ^c

GAUSS'S EQUATIONS. _ , . 191

cosj(A + B) _

' sin^c

- B) _ sin \(a

sn C sn c

SOLUTION. From A, page 25, we have,

sin (iA + B), or sin K A + B ) = sin i A cos i B + cos i A sin B -

Substituting in the second member the values of sin A, cos 6, cos |A, and

sin ^B , from A" and A'", there results,

_ sin (\s - b) A /sin is sin (is -c) sin (ja - a) /sin -s sin(js

~~1E^ T ' sina"sln 6 ~ sin c V sin a sin

c;

a sin 6

_ sin (\B - V) + sin (fo a) ^ sin jg sin ^ - c) gee A//

sin c f sin sin 6

__ sin (js - 6) + sin Qg - a) cQg Q

shi c

But sin (is 6) + sin (i* a) = sin (%a + $ + \c b) + sin (}a + ib + |c a)

= sin [\c + i(a b)] + sin [ic |(a 6)]

= 2 sin |c cos i(a - b). (59, A').

Also, sin c = 2 sin |c cos |c.

Substitutmg these values, the preceding becomes,

2 sin -ic cos 4( b)

sin |(A + B) 1 - - J- cos ^C .

2 sin -Jc cos ic

. sin j(A + B) _ cos j(^ - &)

cos ^C cos i<;

In like manner starting with

sin (^A |B ), or sin |(A B) = sin -A cos |B cos |A sm ^B,

sin i(A - B) sin i(a - b) ....

there results, ____ ^ -^ (2 )

Starting with cos (A + |B), or cos |(A + B) =r cos^A cos |B - sin |A sin iB,

cos |(A + B) cos $(a + b) ...

there results, ^-^ - = - =r-s - (3)

sin |C cos ic

Starting with cos (|A B), or cos|(A B) = cos ^A cos ^B

cos KA B) sin \(a + b)

there r e8 ults, __J ^-.-J (4 )

102 SPHERICAL TltlGONOMETItY.

NAPIER'S ANALOGIES.

. Prob. To deduce Napier 's Analogies, -which an

\ tan |(A + B) _ cos(a - b) m

( tanj(A- B)

cot |C sin %(a + b)

(*\ ^ an i(fl 4- b) _ cos |(A B).

tan Jc ~ cos (A + B)'

... tan(a I) _ sin(A B)

tan Jc sin^(A 4- B)*

SOLUTION. To deduce (1), divide the 1st of Gauss's Equations by the 3d.

To deduce (2), divide the 2d of Gauss's by the 4th. To deduce (3\ divide the 4th

of Gauss's by the 3d. To deduce (4), divide the 2d of Gauss's by the 1st.

152. SCH. In using these formulsB the species must be carefully attended

to. Thus in (1), cot C and cos \(a b) are necessarily + ; hence tan i(A + B)

and cos-J(a + b) are of the same sign with each other. In (2), cot ^C and

sin $(a + b) are both + ; hence, tan |(A B) and sin %(a b) are of the same

sign with each other. And similar inspections may be made upon (3) and (4).

EXERCISES.

153. The proposition that " The sines of the angles are to each

other as the sines of their opposite sides" (135), Napier's Analogies

(151), and formulae A IV , B IV (149) are sufficient, in themselves, to

effect the solution of all cases of oblique spherical triangles; and

for practical purposes they generally require less labor than Napier's

Rules. We give a few solutions and refer the student to the pre-

ceding Exercises for further practice.

1. Given a = 100; c = 5 and I = 10, to solve the triangle.

(Prob. 1, Case 1st, 13V.)

1st. To find A and B we have,

cos i(a + b) : cos |(a b) : : cot C : tan i(A + B) ;

and sin $(a + b) : sin i(ti - b) : : cot C : tan i(A - B) [15O (1) (2)]

NAPIER'S ANALOGIES. 103

Computing by logarithms, we have,

ar. co. log cos [$(a + b) = 55] = 0.241409

+ log cos [i(a - b) = 45*1 = 9-849485

-f log cot [| C = 2 30'] = 11.359907

Rejecting 10 = 11.450801 = log tan K* + B )-

.-. KA + B) = 87 58' 18".

ar. co. log sin [K + b) = 55] = 0.086635

+ log sin [K - ft) = 451 = 9.849485

+ log cot [iC = 2 30'] = 11.359907

Rejecting 10 = 11.296027 = log tan (A - B).

... i(A _ B) = 87 06' 16"

The signs of all the terms being + , KA, + B) and K A B) are both lesa

than 90.

KA + B) + KA - B) = A = 87 58' 18" + 87 06' 16" = 175 04' 34"

KA + B) - KA - B) = B = 87 58' 18" - 87 06' 16" = 52' 02".

2d. To find c. This may be found from the proportion,

sin A : sin C : : sin a : sin c,

or from the 3d or 4th of Napier's Analogies. We use the last, though the first

is equally expeditious.

sin K A - B) : sin $(A + B) : : tan K ft) : tan \c.

ar. co. log sin [KA - B) = 87 06' 16"] = 0.000555

+ log sin [KA + B) zr 87 68' 18"] = 9.999728

+ log tan [Ka - 6) = 45] = 10.000000

Rejecting 10 = 10.000283 = log tan \c.

.'. c = 90 02' 14".

2. Given A = 135 05' 28".6, c = 50 30' 08".6, and I = 69 34

5 6 ".2, to solve the triangle.

1st. To find a and c. The 3d and 4th of Napier's Analogies give,

cos i(A + C) : cos KA C) : : tan $b : tan K + c ) ;

and sin K A + C) : sin KA - C) : : tan 6 : tan \(a c\

Computing by logarithms, we have

ar. co. log cos [KA + C) = 92 47' 48". 8] = 1.3116286 *

+ log cos [KA - C) = 42 17' 40"] = 9.8690535

+ log tan [& = 34 47' 28".l] : 98418527

Rejecting 10 = 11.0225348 = log tan K + c).

.-. K + c) = 95 25' 25".

(a + c) > 90, since cos i(A + C) is , cos KA C) is +, and tan ^b is +,

making tan (a + c) .

* These logarithms are taken from 7-place tables, in order to obtain the tenths of eeconds

accurately.

104

SPHERICAL TRIGONOMETRY.

ar. co. log sin |>(A + C) = 92 47' 48".6] = 0.0005176

+ log sin [(A - C) = 42 17' 40" ] = 9.8279768

+ log tan [p = 34 47' 28". 1] = 9.8418527

Rejecting 10 = 9lJ70347T = log tan \(a - e).

.: i(a -c) = 25 05' 05".

\(a c) < 90, since the signs of the terms are all + .

K + c) + \(a - c) = a = 120 30' 30", and #a + c) $(a c) = c = 70 20' 20".

3d. To find B. Either of the 1st two of Napier's Analogies will give B.

Thus (1) becomes,

cos $(a c) : cos $(a + c) : : tan |(A + C) : co

and (2) sin $(a c) : sin i(a + c) : : tan |(A C) : co

But as i(a + c) is so near 90, it will be better to use the second of these

than the first. Or we may with equal accuracy use,

sin c : sin b : : sin C : sin B.

ar. co. log sin (c = 70 20' 20" ) = 0.0260878

+ log sin (b = 69 34' 56".2) = 9.9718202

+ log sin(C = 50 30' 08". 6) = 9.8874210

Rejecting 10 = 9.8853290= log sin B. .-. B = 50 10' 10".

3. Given a = 50 45' 20", b = 69 12' 40", and A = 44 22' 10",

to solve the triangle.

1st. To find B.

ski a t sin b i t sin A : sin B.

ar. co. log sin (a = 50 45' 20'') = 0.1110044

+ log sin (b = 69 12' 40") = 9.9707626

+ log sin(A = 44 22' 10") = 9.8446525

Rejecting 10 = 9.9264195 = log sin B. .-. B = 57 34' 51".4,

and 122 25' 08". 6. There are two solutions, since a is intermediate in value

between p and both b and 180 b*

* The determination of the species of B, or what ia

the same thing, the number of solutions, can usually be

effected by a simple inspection without any computa-

tion whatever. Thus, sin p = sin b sin A, the loga-

rithms of which are given above, as ie log sin a. Now,

as both a and p are < 90, and log sinp < log sin a,

p <a. But o<6, and aleo less than 180 &. All

this can be seen at a glance,

NAPIER'S ANALOGIES. 105

To find C and c of the larger triangle in which B = 57 34' 51" A

Napier's 1st gives

ar. co. log cos [K& - a) = 9 13' 40"] = 0.0056570

+ log cos [K& + 1) = 59 59'] = 9.6991887

+ log tan [KB + A) = 50 58' 30".7] = 10.0912464

Rejecting 10 = 9.7960921 = log cot C.

.-. C = 115 57' 50".77

Napier's 3d gives

ar. co. log cos [KB - A) = 6 36' 20".7] = 0.0028928

+ log cos [KB + A) = 50 58' 30".7] = 9.7991039

+ log tan [K& + a) - 59 59'J = 10.2882689

Rejecting 10 = 10.0402656 - log tan fa

.-. c = 95 18' 16".4,

3d. To find C and c of the smaller triangle in which B = 122 25' 08".6.

Using the same formula as before.

ar. co. log cos [|(6 - a) = 9 13' 40"] = 0.0056570

+ log cos [K& + a) = 59 59'] = 9.6991887

+ log tan [|(B + A) = 83 23' 39".3] = 10.9362703

Rejecting 10 = 10.6411160 = log cot C.

/. C = 25 44' 31".6.

ar. co. log cos [KB - A) = 39 01' 29".3] = 0.1096506

+ log cos [KB + A) = 83 23' 39".3] = 9.0608369

+ log tan [i(b + a) = 59 59'] = 10.2382689

Rejecting 10 = 9.4087564 = log tan fa

.'. c = 28 45' 05".2.

154:. Sen. When Napier's Analogies are used for solving Prob.2nd(139], the

most expeditious and elegant method of resolving the ambiguity, is by means

of the analogies themselves. Thus, in the above example, after having found

that B = 57 34' 51".4, or 122 25' 08". 6, or both, a simple inspection of the anal-

ogy next used will determine the number of solutions.

Napier's 1st may be written

Now |C < 90, hence cot |C is +. If, therefore, neither of the values of

B renders cot ^C , there are two solutions. If one value renders cot -C + ,

and the other , there is one solution and it corresponds to the value of B

which makes cot |C +. If both values of B render cot |C , there is no solu-

tion. In the last example, we see that cos [K& + o) = 59 59'], and cos [K& )

= 9 13' 40"] are both + . Also tan [KB + A) = 50 58' 30". 7, or 83 23' 39".3,

or both] is + for both values of B. Therefore there are two solutions.

106 SPHERICAL TRIGONOMETRY.

4. Given A = 95 16', B = 80 42' 10", and a = 57 38', to solve

the triangle.

1st. To find 5, sin A : sin B : : sin a : sin 5.

ar. co. loo; sin (A = 95 16') = 0.001837

+ log siii (B = 80 42' 10") - 9.994257

+ log sin (a - 57 38') = 9.926671

Rejecting 10 = 97922765 - log sin b.

: b = 56 49' 57", or 123 10' 03", or both.

2d. To find <, tan $c = ^~^ ^_ |^ tan |( + b). Now for 6 = 56 49' 57",

tan \G is + ; but for b = 123 10' 03" tan $c is ; hence there is but one solu-

tion, and that corresponds to the smaller value of b.

ar. co. log cos [i(A - B) = 7 16' 55"] = 0.003517

+ log cos [|(A + B) = 87 59' 05"] = 8.546124

+ log tan [i(<j + b) - 57 13' 58"] = 10.191352

Rejecting 10= 8.740993 = log tan \c.

.'. Cr= 6 18' 19".

3d. To find C, we may use (1) or (2) of Napier's Analogies, or

sin a : sin c : : sin A : sin C,

the last of whicn is the most expeditious.

ar. co. log sin (a = 57 38') = 0.073329

+ log sin (c = 6 18' 19") = 9.040705

+ log sin (A = 95 16') = 9.998163

Rejecting 10 = 9.112197 = log sin C- .-. C = 7 26' 22'

This value is taken for C instead of its supplement, since C is opposite the

smallest side c.

5. Given a = 70 14' 20", I = 49 24' 10", and c = 38 46' 10"; to

solve the triangle.

COMPUTATION-.

a = 70 14' 20"

b = 49 24' 10"

c - 38 46' 10"

s = 158 24' 40"

is = 79 12' 20" ar. co. log sin = 0.007753

a = 8 58' 00" " " = 9.192734

b = 29 48' 10" " " = 9.696370

c = 40 26' 10" " " = 9.811977

2)18.708834

.-. log k = 9.354417

NAPIER'S ANALOGIES. 107

log tan |A = log k - log sin (is - a) + 10 = 10.161683. .-. A = 110 51 16".

log tan |B = log k - log sin (it - b) + 10 = 9.658047. .-. B = 48* 56' 04".

log tan |C = log k - log sin (fr - c) + 10 = 9.542440. .-. C = 38 26' 48".

6. Given A = 109 55' 42", B = 116 38' 33", andc = 12043' 37",

to solve the triangle.

COMPUTATION.

A = 109 55' 42"

B = 116 38' 33"

C = 120 43' 37"

S = 347 17'"52"

|S = 173 38' 56" ar. co. log cos = 0.002683

S - A - 63 43' 14" " " = 9.646158

|S - B = 57 00' 23" " " = 9.736035

iS - C = 52 55' 19" " " = 9.780247

2)19.165123

.-. log K = 9.582561

log cot $a = log K log cos (|S A) + 10 = 9.936403. /. a = 98 21' 38".

log cot # = log K - log cos (|S - B) + 10 = 9.846526. .'. b = 109 50' 20".

log cot $c - log K - log cos (iS - C) + 10 - 9.802314. .-. c = 115 13' 28".

Sen. 1. The student can use the exercises in the preceding section to famil-

iarize the methods here given. In doing so, it will be well for him to seek the

most expeditious solutions. He will find that

Examples under PROS. 1 require 11 logarithms by Napier's Analogies and

(135], and 12 logarithms by Napier's Rules and (135).

Examples under PROB. 2, when there is but one solution, require 10 loga-

rithms by Napier's Analogies and (135), and 12 logarithms by Napier's Rules

and (135). When there are two solutions, 15 logarithms are required by

Napier's Analogies and (135), and only 14 by Napier's Rules alone, or by these

rules and (135).

Examples under PROB. 3 require but 7 logarithms by the method given in this

section and 13 by the previous method.

Sen. 2. In cases in which the angles or sides are near the limits 0, 90, or

180, so that the functions used in the particular solution change very rapidly

in proportion to the arc, it is often possible to select one among the several

methods which will give more accurate results than the others. There are

also other methods which are better adapted to such cases than those here

given. For these, as well as for much other interesting matter, and especially

for the discussion of the General Spherical Triangle, American students have an

excellent resource in the treatise of Professor Chauvenet of Washington Univer-

sity, St. Louis.

108 SPHEBICAL TKIGONOMETKY.

s

SECTION IV.

AREA OF SPHERICAL TRIANGLES.

155 I*rob. Having the angles of a spherical triangle given, to

find the area.

SOLUTION. [The solution is given in PART II. (613), and we simply re-

produce the result in order to give completeness to this section.] The area is

equal to the ratio of the spherical excess to 90, or ITT, into the trirectangular tri-

angle. That is, letting the sum of the angles be S, the area K, and the radius of

the sphere 1, whence the area of the trirectangular triangle is #, we have

K = ^fl5 x \it = S - it.

i*

In the latter expression S is the sum of the angles in terms of the radius, i. e.,

S = 57^578' r a PP roximate ^ S = 5^3 W

EXERCISES.

1. What is the area of a spherical triangle whose angles are 100,

58, and 62, on a sphere whose diameter is 6 feet ?

220

SOLUTION. K = S it - - - 3.14159 = .698, the area of a similar tri-

57 .o

angle on a sphere whose radius is 1. Hence, the area of the required triangle

is .698 x 3 2 = 6.282. [The method given in PART II. (613) is more expedi-

tious, but it is our purpose to illustrate the form here given.]

2. What is the area of a spherical triangle whose angles are 170,

1 35, and 115, on a sphere whose radius is 10 feet?

Ans. 418.875 square feet.

3. What is the area of a spherical triangle whose angles are 150,

110, and 60, on a sphere whose radius is 3 feet ?

. Prob. Having the sides of a spherical triangle given, to

find the area.

SOLUTION. The angles mav be found by (148), and then the area by (155).

AREA OF SPHERICAL TRIANGLES. 103

But a more direct method is to find the spherical excess by means of Lh-utt*

tier's formula, which we will now produce.

*K = *(A 4- B + C - n)

Whence tan $K = tan }[A + B + C - it] = tan [4(A + B) - \(it - C)]

_ sin j(A + B) - sin #it - C)

~ cos i( A" + B) + cos fat - C) (7, page 31}

_ sin j(A + B) cos jC

~ cos-KA + B) + sin C

[cos i(a + b) + cos ^c] sin

lst and 3d)

cos ifrt - b) - cos jc / sin fo sin (fo - c) (146, 147)

cos $(a + b) + cos |c V sin (|s a) sin (|s 6)

_sin^i(a + c - b) sin j(6 + c - a) / sin jg sin (jg - c) c/

cos \(a + b + c) cos \(a + b c) V sin ($s a) sin ($s - b)

. /sin* i(is - ft) sin 2 -Kjs - ^"sin ^ sin (jT^) , g = a + ft + ,

r cos 2 is cos 2 i(|.s c) sin (|s a) sin ($8 b)

/ sin 2 j(js b) sin 2 j(k* a ) sm i g cos ^ gsm i(^ s ~ c ) cus ^ ~"0

11 T cos 2 |s cos 2 Kl*-c) sin |(is-a)"cos~i(|-a) sin |(i*-6) cos i(is-b)

.: Tan iK =* ^/tan i* tan i(|s - a) tan i(|s - 6) tan ^ - c). (A)

Having found K, the spherical excess, or what is the same thing, the area of

a similar triangle on a sphere whose radius is 1, we have but to multiply K by

the square of the radius in any given case.

EXERCISES.

1. Given a = 98, b = 110, and c = 115, to find the area of a

spherical triangle, on a sphere whose radius is 4000 miles.

COMPUTATION.

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