Edward Olney.

# Elements of trigonometry, plane and spherical online

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* 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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