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greater arc has the less sine. If sin b = sin B,

b B, and BA BC = 90, and the side b can oc-

cupy but one position in the lune, thus giving rise

to but one triangle BAG, which satisfies the conditions

(or two equal triangles BAG and B'AC). If b > B,

which implies that sin b < sin B, the side b can

occupy two positions in the lune, as b' and b",

thus giving rise to two triangles BA'C', and BA"C", both of which satisfy the

conditions.

EXERCISES.

1. In a right angled spherical triangle BAG, A being the right angle,

B = 80 40', and a = 105 34', to project the triangle and compute

the other parts.

PROJECTION.* Projecting the triangle upon the plane of the side c (1O2\

we have, BCA, Fig. 51. [The student should give .

the process.]

SOLUTION. It is immaterial which of the re-

quired parts we seek first We will seek c. Now

the three circular parts under consideration are c,

compa, and comp B. Comp B is middle part,

and the extremes are adjacent ; hence, by Napier's

second rule we have,

cos B = tan c cot.

cos B cos 80 40'

or

tan c =

cot a cot 105 34'*

FIG. 51.

Now cos 80 40' is + , and cot 105 34' is . Therefore tan c is , and c > 90.

Computing by logarithms,

log cos 80 40' = 9.209993

- log cot 105 34' = 9-444947

= log tan c = 1.765045

Add 10 for tab. tan 10.

9.765045. .'. c = 149 47' 37"

* It is recommended that the projection be given always before the trigonometrical solution.

It is an excellent exercise, and gives clearness of perception.

82

SPHERICAL TRIGONOMETRY.

To find b, sin b = sin a sin B = sin 105 34' x sin 80 40'. Tiiis makes b

known by means of its sine, whence the signs of the formula do not determine

the species of b. But b is of the same species as B (124), and therefore less

than 90.

Computing by logarithms,

log sin 105 34' = 9.983770

+ log sin 80 40' = 9.994212

Deducting 10 = 9.977982 = log smb. /. b = 71 54' 33".

To find C, cos a = cot B cot C, or cotC = -^ = C ^5 f5. Whence

COt D COt OO 40

cot C is , and C > 90.

Computing by logarithms,

log cos 105 34' = 9.428717

- log cot 80 40' = 9.215780

Adding 10 = 10.212937 = log cot C. .-. C = 148 30' 54".

SCH. It is expedient to find each part directly from the parts given in the

example, in order that an error in finding one may not extend itself through

the whole solution.

2. Given a = 86 51', and B = 18 03' 32", to project the triangle

and compute the other parts.

c = 86 41' 14", I = 18 01' 50", C = 88 58' 25".

3. Given b = 155 27' 54", and c = 29 46' 08", to project the

triangle and compute the other parts. See

Fig. 52.

a = 142 09' 13", c = 54 01' 16",

B = 137 24' 21".

4. Given c = 73 41' 35", and B =

99 17' 33", to project the triangle and

compute the other parts.

C = 73 54' 46", I = 99 40' 30",

Fia. 52. = 9 2 42' 17".

5. Given B = 47 13' 43", and c = 126 40' 24", to project the

triangle and compute the other parts.

I = 32 g Q8' 56", a = 133 32' 26", c = 144 27' 03".

EXERCISES IN SOLVING EIGHT ANGLED TBIANGLES.

PROJECTION. In order to project this case, i. e.,

when the two oblique angles are given (I O,5), it is

most convenient to compute the base before pro-

jecting. It is also expedient, when two angles are

given, to project the larger at a point in the cir-

cumference of the primitive circle, as at C, espe-

cially if the smaller be quite small. In this case,

projecting the angle C at C, Fig. 53, conceive BA as

drawn through P (or, if desired, sketch it hypo-

thetically), and then compute b, from the relation

83

cos B = sin C cos 5, or cos b

Having

found b = 32 08' 56", take CA = 6, and draw AB through P.

6. Given B = 100, and I = 112, to project the triangle and com-

pute the other parts.

PROJECTION. See Fig. 54

NUMERICAL SOLUTION. To find c, we have

sin c tan b cot B = tan 112 cot 100.

Computing by logarithms,

log tan 112 = 10.393590

+ log cot 100 - 9.246319

Rejecting 10 = 9.639909 = log sin c.

.-. c = 25 52' 33".4, or 154 07' 26".6, i. e., BA, or BA'.

Fio. 54.

To find a, we have

sin b sin 112 C

sin B sin 100

sin b = sin a sin B ; whence sin a =

Computing by logarithms,

log sin 112 = 9.967166

- log sin 100 = 9.993351

Adding 10 = 9.973815 = log sin a. /. a = 7018'10".7,orl0941'49".3,

f. e., BC', or BC.

cos B cos 100

cos b cos 112'

To find C, we have

cos B = cos b sin C ; whence sin C =

Computing by logarithms,

log cos 100 = 9.239670

- log cos 112 = 9.573575

Adding 10 = 9.666095 = log sin C. .*. C = 27 36' 58" .8, or 152 23' 01".

i.e. y BCA, orBC'A'.

84 SPHERICAL TKIGONOMETIIY.

Thus we see that each of the two triangles BCA and BC'A' fulfills the con-

ditions of the problem.

7. Given one side of a right angled spherical triangle 160, and the

opposite angle 150, to project the triangle and compute the other

parts.

Results. There are two triangles. The other sides of the first

are 136 50' 23", and 39 04' 51"; and the angle opposite the latter

side is 67 09' 43". The corresponding parts of the other triangle

are 43 09' 37", 140 55' 09", and 112 50' 17".

8. In the spherical triangle DBF, right angled at E, given an oblique

angle 58, and the side opposite 64, to project the triangle and com-

pute the other parts.

9. In a right angled spherical triangle given an oblique angle

165, and the opposite side 112, to project the triangle and com-

pute the other parts.

10. In a right angled spherical triangle given one side 65 23' 12",

and the opposite angle 65 23' 12", to project the triangle and com-

pute the other parts.

11. Given c = 60 47' 24".3, B = 57 16' 20".2, and A = 90, to

project and compute.

a = 68 56' 28".9, c = 54 32' 32".l, and I = 51 43' 36".l.

12. Given c = 116, b = 16, and the included angle 90, to pro-

ject and compute.

QUADRANTAL TRIANGLES.

129. A Quadrantal Triangle is a spherical triangle one

of whose sides is a quadrant, or 90. Such a triangle is readily

solved by passing to its polar, solving it, and then passing back.

The polar triangle to a quadrantal triangle, being right angled, is

solved by Napier's rules.

Ex. 1.- Given a = 90, B = 75 42', and c = 18 37', to compute

the other parts.

SUG'S. Representing the supplemental parts of the polar triangle by A', B',

C', a', ft', and c', we have A' = 180 - a = 90, V = 180 - B = 104 18', and

OF OBLIQUE ANGLED SPHERICAL TRIANGLES.

85

C' = 180 - c = 101 23', from which to find B', a', and c'. This being rijrht

angled, we find, by applying Napier's rules, B' = 94 31' 21", a' = 76 25' 11",

and c' = 161 55' 20". Hence in the primitive triangle we have b = 180 B'

- 85 28' 39", A = 180 - a' = 103 34' 49", and C = 180 - c' = 18 04' 40".

Ex. 2. Given a = 90, C = 42 10', and A = 115 20', to find the

other parts.

B = 54 44' 24", I = 64 36' 40", c = 47 57' 47".

SECTION II.

OF OBLIQUE ANGLED SPHERICAL TRIANGLES.

130. All cases of oblique angled spherical triangles can be solved

by Napier's rules and the following proposition.

131. J?vop. In any spherical triangle, if a perpendicular be

let fall from either vertex upon the opposite side or side produced,

the tangent of half the sum of the segments* of that side is to the

tangent of half the sum of the other two sides, as the tangent of half

the difference of those sides is to the tangent of half the difference

of the segments.

DEM. In the triangle BAG let fall the perpen-

dicular p, from C upon the opposite side. Let

BD = s, and DA = s'. By Napier's first rule,

cos a = cos p cos s, and cos b = cos p cos s'.

COS (t COS 8

Dividing the former by the latter, r = , ;

' cos b cos s' '

whence, by composition and division,

cos b cos a cos s' cos s

cos a

But by (61),

and

+ cos b cos s

cos b cos a

+ COS

S 1 ' FIG. 55.

cos a + cos b

cos s' cos s

tan -1-

f? I i r \ tnn 1 fa *'\

COS 8 + COS '

* When the perpendicular falls without the base, as in Fig. 5<>, this term is to be understood

as meaning the distances from the foot of the perpendicular to each extremity of the base, as

BD and AD- This, in fact, is the general statement applying as well to the case when the

perpendicular falls on the base.

86

SPHERICAL TRIGONOMETRY.

tan

or, tan (* +

'a + ft) tan i(a ft) = tan i (s + s') tan (s s')

tan (a + ft) : : tan (a ft) : tan (s s'). Q. E. D.

j?3. Sen. 1. Since from a point in the surface

of a hemisphere two perpendiculars can always

be drawn to the circumference of the great circle

which forms its base, and since the feet of these

perpendiculars are 180 apart, and no side of a

spherical triangle can equal 180, the foot of one

perpendicular will always fall within the base or

upon one extremity of it, and the other without

the base; or both will fall without the base. If

we take the foot of the perpendicular which falls

within the base, or the nearer one when both fall

without, the sum of the distances from the foot of

the perpendicular to the extremities of the base is always less than 180,

i. e., s + s' < 180. When the perpendicular falls wilhin, * + *' makes up one

side of the triangle, and hence is less than 180. If both perpendiculars fall

without, let D, Fig. 56, be the foot of the nearer one. Now DB + BD' = 180 ;

but by hypothesis DA < BD', .% DB + DA < 180. When DA = BD', DB + DA

= 180.

133. Sen. 2. As in spherical triangles the greater segment is not always

adjacent to the greater side, it becomes necessary to determine the position of

the segments. This can be done by the signs of the proportion

tan (s + s') : tan \ (a + ft) : : tan (a ft) : tan (s s').

1st. Tan ^ (.<? + s') is always + , since, if D falls in the base, s + s' < 180 ;

and if D falls without, by taking the nearer perpendicular, s + s' is still < 180

(132). .'. H* + s') < 90, and tan i (* + *') is +

2d. When a + ft < 180, tan i (a + ft) is + ; and when a + ft > 180.

tan i (a + ft) is .

3d. When a > ft, a ft is a positive arc less than 180, hence tan (a ft)

is + ; and when a < ft, (a ft) is a negative arc and less than 180, hence

tan i (a, ft) is .

4th. The signs of these terms being determined, that of tan (s s') becomes

known. Now, as ^(s s') cannot be numerically greater than 90, tan i(* s')

is + when s > s', and when s < '.

5th. When s + s' = 180, tan $(s + s')= oo. Now as a ft < 180, tan \ (a -ft)

cannot be oo, nor can tan \ (s s 1 ) = when the perpendicular falls without.

Hence to make the proportion possible, tan (a + ft) must be oc, or a + ft = 180

In this case we project on the plane of a or ft. If a + b = 180, aud a + c, = 180,

we project on the plane of a. If a + b = 180, a + c = 180, and b + c = 180, the

triangle is trirectangular.

134. Sen. 3. If either segment is greater than the whole base, the perpen-

dicular falls without the triangle. In this case the shorter segment lies in an

opposite direction from its angle to that considered in the demonstration, and

heiice is to be considered ; and the algebraic sum of the segments is still

equal to the side upon which the perpendicular is let fall.

OBLIQUE ANGLED TRIANGLES SOLVED BY NAPIER'S RULES. 87

o. Prop. In a spherical triangle, the sines of the sides are

to each other as the sines of their opposite angles.

DEM. By Napier's first rule we have from either Fig. 55 or Fig. 56,

sin p = sin a sin B, and sin p sin b sin A.

.-. sin a sin B = sin b sin A, or sin a : sin b : : sin A : sin B. Q. E. D.

Sen. This proposition is not introduced here because it is necessary for the

solution, of spherical triangles, but because of its essential importance. It is

often convenient to use it in the solution of a triangle, but never necessary, as

will appear hereafter. It affords a ready method of determining a part OPPOSITE

a given part, provided the species of the part be determined by other considerations.

SOLUTION OF OBLIQUE ANGLED SPHERICAL TRIANGLES BY

NAPIER'S RULES FOR RIGHT ANGLED SPHERICAL TRIANGLES.

136. The examples which arise in the solution of oblique angled

spherical triangles are all comprised under the three following

problems, each of which consists of two cases :

1. When the given parts are all adjacent to each other.

2. When two of the given parts are adjacent and one separate.

3. When the given parts are all separate from each other.

137 I*rob 1. Given three adjacent parts of an oblique angled

spherical triangle, to solve the triangle.

CASE 1st. Given two sides and the included angle.

SOLUTION. Project the triangle qn the plane of one oftJie given sides (106\

and let fall a perpendicular from the angle opposite upon this side or upon

D ""

FIG. 57. FlG. 58.

this side produced, as the case may be. There are thus formed two right

angled triangles, as BDC and DCA, each of which can be solved by Napier's

88 SPHERICAL TRIGONOMETRY.

rules, by first solving the one containing the given angle. Thus, in the triangle

BDC right angled at D, a and B are supposed known ; whence CD, BD, and the

angle BCD, can be computed. As BA = c is known, the segment DA can

be found, it being the difference between c and the arc BD. When the solu-

tion of this triangle gives BD > r, it is evident that the perpendicular falls

without the triangle, which will agree with the projection. Passing to the

triangle ADC, right angled at D, we now know CD and DA ; whence the

other parts can be found. Finally, the angle BCA of the required triangle =:

BCD + DCA when the perpendicular falls within the triangle, and BCD DCA

when the perpendicular falls without.

CASE 2cL Given two angles and the included side.

SOLUTION. The solution of this case is effected by passing to the polar

triangle, projecting and solving it by Case 1st, and then passing back.

138. Sen. A slight saving of labor is effected by using (135} in the solu-

tion. Thus, in the triangle BCD, compute CD and BD as before, and (not com-

puting angle BCD) then passing to the triangle DCA, compute b and A. Finally,

compute C (the entire angle) from the proportion

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

139. Prob. 2. In an oblique angled spherical triangle, given

two parts adjacent to each other and one separated from both of them,

to solve the triangle.

CASE 1st. Given two sides and an angle opposite one of them.

SOLUTION. Project the triangle on the plane of the unknown side, with the

given angle at B ; and let fall the perpendicular from the angle C opposite the

unknown side. Compute the triangle BDC. Having computed this triangle,

FIG. 59. FIG. 60.

compare the side opposite the given angle, as b, with the perpendicular and

the arcs BC and CB', i. e., with p, a, and 180 a. If b = p there is but

OBLIQUE ANGLED TRIANGLES SOLVED BY NAPIER* S RULES. 89

one solution and the triangle is right angled, A falling at D. If b is inter-

mediate in value between p and both a and 180 , it can occupy two positions

as in Fig. 59, and there are two solutions. If b is intermediate in value between

p and only one of the arcs a or 180 a, there is but one solution. When B < 90

the perpendicular is less than any oblique arc; hence in this case, if b <p, there

is no solution. But if B > 90, the perpendicular is greater than the oblique

airs; hence in this case, if b > p, there is no solution. [These results should be

obtained independently of the results given by the projection, and one be made

a check upon the other.] The solution is now completed by computing the

parts of DCA, and adding or subtracting the segments BD and AD, and the angles

BCD and A CD, as the case may require.

CASE 3d. Given two angles and a side opposite one of them.

SOLUTION. Pass to the polar triangle ; solve it, and then pass back.*

140. Sen. The relation established in (135} may also be used in the

solution of this problem. Thus, having projected the triangle, computed

p, and determined whether there are one or two solutions, to find A, when

, 5, and B are given, we have, sin b : sin a : : sin B : sin A. Then computing

the third side c (or sides), by means of the right angled triangles BCD and DCA

as before, we may use the proportion (135) to find the angle BCA and BCA'.

But the use of this proportion gives no advantage except in cases in which

there is only one solution.

141. Prob. 3. In an oblique angled spherical triangle, given

three parts all separated from each other, to solve the triangle.

CASE 1st. Given the sides to find the angles.

SOLUTION. Project the triangle on the plane of one of its sides, as c. From

the proportion,

tan -JO -f- s') : tan \(a + b) :: tan|(a b) : tan l(ss') y

* This case can be projected and solved in a manner

altogether similar to the first, without passing to the

polar triangle. Thus, let Bi #1 and A be the given parts.

Project the triangle on the plane of c, as in the figure.

Project B in the usual way, and make BC the projec-

tion of a. Through C draw DD', and. make BDO = the

projection of A- Drawing the small circle with radius

PC, draw diameters through the intersections Q and O'-

Then will A and A' be the vertices of the triangle re-

quired. The student may prove that the figures POD

and PCA are equal, and also PO'D and PCA', and

hence that angle BDO = A A'

Fia. 61.

90

SPHERICAL TRIGONOMETRY.

FIG. 62.

find |(* *'). ' Then half the sum of the segments

+ half the difference gives the greater segment, and

half the sum half the difference gives the less.

Determine from the signs of the terms whether * is

greater or less than s' : and also determine whether

the perpendicular lies within or without the tri-

angle (134.). Observe that these results correspond

to those given by the projection. Finally, in each

of the two right angled triangles BCD and DC A,

there are two sides given ; whence the angles can

be found by Napier's rules. If the perpendicular

falls within, C = BCD + DCA, and A, of the re-

quired triangle DAC. If the perpendicular falls

without, C = BCD DCA, and A of the triangle

= 180 - DAC.

CASE 2d. Given the angles to find the

SOLUTION. Pass to the polar triangle ; solve it,

and then pass back.

142, Sen. Here, again, (135) affords a slightly

more expeditious solution. Having projected the

triangle, found and located the segments, and com-

puted one angle, as B, by the methods given above, the other angles may be

found from the proportions,

sin b : sin a : : sin B : sin A,

and sin 5 : sin c : : sin B : sin C.

EXERCISES.

1. Given I = 120 30' 30", c = 70 20' 20", and A = 50 10' 10",

to project and solve the triangle.

PROJECTION. See Fig. 64.

TRIGONOMETRICAL SOLUTION. 1st. To solve the

triangle ABD, in which the two known parts are

situated.

(a) To find p, sin p = sin c sin A.

log sin 70 20' 20" = 9.973912

4- log sin 50 10' 10" = 9.885329

Rejecting 10 = 9.859241 = log- sin p

.'. p = 46 19' 01", the species being determined

by the opposite angle (125). [Observe that the re-

sult corresponds with the projection].

)BLIQUE ANGLED TKIANGLES BOLTED BY SAPIENS EULES. 91

(ft) To find AD, cos A = cot c tan AD, or tan AD = ^-^ .

log cos 50 10' 10" = 9.806532

- log cot 70 20' 20" = 9.553016

Adding 10 = 10.253516 = log tan AD. /. AD = 60 50' 49",

the species being determined by the signs of the formula.

(c) To find angle ABD, cos c = cot A cot ABD, or cot ABD = CC

log cos 70 20' 20" = 9.526929

log cot 50 10' 10" = 9.921204

Adding 10 = 9.605725 = log cot ABD. .-. ABD = 68 01' 53"

the species being determined by the signs of the formula.

3d. To solve the triangle DBC.

(a) To find DC. Since AD < ft, the foot of the perpendicular falls in the base,

and DC = AC - AD = b - AD = 120 30' 30" - 60 50' 49" = 59 39' 41".

(b) To find a, cos a cosp cos DC

log cos 46 19' 01" = 9.839270

+ log cos 59 39' 41" = 9.703386

Rejecting 10 = 9.542656 = log cos a. .: a = 69 34' 56", the

species being determined by the signs of the formula.

(c) To find C, sin p = sin a sin C, or sin C =

log sin p = 9.859241

- log sin 69 34' 56" = 9.971820

Adding 10 = 9.887421 = log sin C. .*. C = 50 30' 08", the

species being determined by the side opposite.

(d) To find angle DBC, sin p = tan DC cot DBC, or cot DBC = ^ ^ .

log sin p = 9.859241

- log tan 59 39' 41" = 10.232653

Adding 10 = ~9.626588 = log cot DBC. .'. DBC = 67 03' 36",

the species being determined by the signs of the formula.

Finally, B = ABD + DBC = 68 01' 53" + 67 03' 36" = 135 05' 29"

Sen. We might have omitted the computation of angle ABD in the first

part, and DBC in the second, and have found instead the entire angle B from

sin a : shift : : sin A : sin B. To compute this requires the looking out of but

two logarithms, since sin a is given in the second part (c), and sin A in the first

part (a).

2. Given a = 97 35', I = 27 08' 22", and A == 40 51' 18", to

project and compute the triangle. Between what limits must the

92 SPHERICAL TRIGONOMETRY.

value of a be assigned in order that there may be two solutions ? Be-

yond what limiting values of a is a solution impossible ?

PROJECTION. See Fig. 65.

TRIGONOMETRICAL SOLUTION. To find p, sin p

sin b sin A.

log sin 27 08' 22" = 9.659115

+ log sin 40 51' 18" = 9.815675

Rejecting 10 = 9.474790 = log sin p.

.'. p 17 21' 40". [Reason for the species.]

We now observe that there is one and only out

solution, since the arc a (97 35') cannot lie between

CD (17 21' 40") and 6(27 08' 22"), but can lie be-

tween CD and CA' (180 - b = 152 51' 38").

To find AD, cos A = cot b tan AD or tan AD =

log cos 40 51' 18" = 9.878733

log cot 27 08' 22" = 10.290226

Adding 10 = 9.588507 = log tan AD. .'. AD = 21 11' 30".6.

To find angle ACD, cos b cot A cot ACD, or cot ACD =

log cos 27 08' 22" = 9.949340

- log cot 40 51' 18" = 10.0630575

Adding 10 = 9.8862825 = log cot ACD. /. ACD = 52 25' 01".

[Reason for the species. ]

To find B, sin p = sin a sin B, or sin B = !!EJ?.

sines

log sin p (as above) = 9.474790

- log sin 97 35' = 9.996185

Adding 10 = 9.478605 = log sin B. .-. B = 17 31' 09".

[Reason tor the species. ]

To find DB, cos a = cos p cos DB, or cos DB = - .

cosp

log cos 97 35' = 9.120469

- log cos 17 21' 40" = 9.979750

Adding 10 = 9.140719 = log cos DB. .-. DB = 97 56' 51".8.

[Reason for the species.]

AB = AD + DB = 21 11' 30".6 + 97 56' 51".3 = 119 08' 21".9.

To find DCS, sin p = tan DB cot DCB, or cot DCB =

log sin p (as above) = 9.474790

- log tan 97 56' 51 ".3 = 10.855090

Adding 10 =" 8.619700 = log cot DCB. .-. DCB = 92 23' 7".7.

[Reason for the species.]

ACB = C = ACD -f DCB = 52 35' 01" + 92 23' 7".7 = 144 48' 8".7.

OBLIQUE ANGLED TKIANGLES SOLVED BY NAPIER'S RULE. OS

Finally, we observe that, if any value were assigned to a between b (27 08' 22")

and CD (17 21' 40") there would be two solutions ; since for such values the side

a could lie on both sides of CD. But, for any value of a less than CD (17 21' 40"),

there would be no solution ; since CD is the shortest distance from C to the arc

ABA'. Also, for any value of a greater than the arc CA' (152 51' 38"), there

would be no solution, as such an arc would fall between CA' and CD' (if not

> CD'), and consequently would make c > 180.

SCH. Such examples as this and the preceding can be more expeditiously

solved by using p in each equation in solving the triangles ACD and DCB. By

this means and using (135) to determine the side c, the solution can be effected

with only 12 logarithms. Thus in Ex. 2.

1st. To find p, sin^ = sin b sin A< requires 3 logarithms

2<1. To find ACD, cos ACD = cot 6 tan/?, requires 3 "

3d. To find DCB, cos DCB = cot a tan p, requires 2 " (log tan p being known).

4th. To find B, n# = sin a sin B> requires 2 (log sin P being known).

5th. To find c, sin A : sin C : ' sin a : sin c, requires 2_ " (log's of sin A and sin a

Total 12 logarithms being known.)

3. Given a = 76 35' 36", I = 50 10' 30", and c = 40 00' 10", to

project and solve the triangle.

PROJECTION. See Fig. 66.

TRIGONOMETRICAL SOLUTION. 1st. To find the

segments CD and DB, we have,

tan i(s + a'),

or

tan a : tan |(5 + c) : : tan i(b c) : tan i(s a').

Computing by logarithms.

a. c. log tan \a = log tan 38 17' 48" = 0.102561

+ log tan K& + c) = log tan 45 05' 20" = 10.001347

+ log tan i(5 c) = log tan 5 05' 10" = 8.949406

Rejecting 10 = 9.053314 = log tan i(a - a').

.-. i( - 8 ') = 6 27' 02".

In order to determine whether s or s' is the greater,* we observe the signs

of the proportion, and rinding tan %(s s') positive, know that * > s'.

Hence, s = K* + *') + fi - ') = 38 17' 48" + 6 27' 02" = 44 44' 50",

and s' = i(a + s') - i(a - s') = 38* 17' 48" - 6 27' 02" = 31 50' 46".

The angles sought are now readily found by computing the two right

b B, and BA BC = 90, and the side b can oc-

cupy but one position in the lune, thus giving rise

to but one triangle BAG, which satisfies the conditions

(or two equal triangles BAG and B'AC). If b > B,

which implies that sin b < sin B, the side b can

occupy two positions in the lune, as b' and b",

thus giving rise to two triangles BA'C', and BA"C", both of which satisfy the

conditions.

EXERCISES.

1. In a right angled spherical triangle BAG, A being the right angle,

B = 80 40', and a = 105 34', to project the triangle and compute

the other parts.

PROJECTION.* Projecting the triangle upon the plane of the side c (1O2\

we have, BCA, Fig. 51. [The student should give .

the process.]

SOLUTION. It is immaterial which of the re-

quired parts we seek first We will seek c. Now

the three circular parts under consideration are c,

compa, and comp B. Comp B is middle part,

and the extremes are adjacent ; hence, by Napier's

second rule we have,

cos B = tan c cot.

cos B cos 80 40'

or

tan c =

cot a cot 105 34'*

FIG. 51.

Now cos 80 40' is + , and cot 105 34' is . Therefore tan c is , and c > 90.

Computing by logarithms,

log cos 80 40' = 9.209993

- log cot 105 34' = 9-444947

= log tan c = 1.765045

Add 10 for tab. tan 10.

9.765045. .'. c = 149 47' 37"

* It is recommended that the projection be given always before the trigonometrical solution.

It is an excellent exercise, and gives clearness of perception.

82

SPHERICAL TRIGONOMETRY.

To find b, sin b = sin a sin B = sin 105 34' x sin 80 40'. Tiiis makes b

known by means of its sine, whence the signs of the formula do not determine

the species of b. But b is of the same species as B (124), and therefore less

than 90.

Computing by logarithms,

log sin 105 34' = 9.983770

+ log sin 80 40' = 9.994212

Deducting 10 = 9.977982 = log smb. /. b = 71 54' 33".

To find C, cos a = cot B cot C, or cotC = -^ = C ^5 f5. Whence

COt D COt OO 40

cot C is , and C > 90.

Computing by logarithms,

log cos 105 34' = 9.428717

- log cot 80 40' = 9.215780

Adding 10 = 10.212937 = log cot C. .-. C = 148 30' 54".

SCH. It is expedient to find each part directly from the parts given in the

example, in order that an error in finding one may not extend itself through

the whole solution.

2. Given a = 86 51', and B = 18 03' 32", to project the triangle

and compute the other parts.

c = 86 41' 14", I = 18 01' 50", C = 88 58' 25".

3. Given b = 155 27' 54", and c = 29 46' 08", to project the

triangle and compute the other parts. See

Fig. 52.

a = 142 09' 13", c = 54 01' 16",

B = 137 24' 21".

4. Given c = 73 41' 35", and B =

99 17' 33", to project the triangle and

compute the other parts.

C = 73 54' 46", I = 99 40' 30",

Fia. 52. = 9 2 42' 17".

5. Given B = 47 13' 43", and c = 126 40' 24", to project the

triangle and compute the other parts.

I = 32 g Q8' 56", a = 133 32' 26", c = 144 27' 03".

EXERCISES IN SOLVING EIGHT ANGLED TBIANGLES.

PROJECTION. In order to project this case, i. e.,

when the two oblique angles are given (I O,5), it is

most convenient to compute the base before pro-

jecting. It is also expedient, when two angles are

given, to project the larger at a point in the cir-

cumference of the primitive circle, as at C, espe-

cially if the smaller be quite small. In this case,

projecting the angle C at C, Fig. 53, conceive BA as

drawn through P (or, if desired, sketch it hypo-

thetically), and then compute b, from the relation

83

cos B = sin C cos 5, or cos b

Having

found b = 32 08' 56", take CA = 6, and draw AB through P.

6. Given B = 100, and I = 112, to project the triangle and com-

pute the other parts.

PROJECTION. See Fig. 54

NUMERICAL SOLUTION. To find c, we have

sin c tan b cot B = tan 112 cot 100.

Computing by logarithms,

log tan 112 = 10.393590

+ log cot 100 - 9.246319

Rejecting 10 = 9.639909 = log sin c.

.-. c = 25 52' 33".4, or 154 07' 26".6, i. e., BA, or BA'.

Fio. 54.

To find a, we have

sin b sin 112 C

sin B sin 100

sin b = sin a sin B ; whence sin a =

Computing by logarithms,

log sin 112 = 9.967166

- log sin 100 = 9.993351

Adding 10 = 9.973815 = log sin a. /. a = 7018'10".7,orl0941'49".3,

f. e., BC', or BC.

cos B cos 100

cos b cos 112'

To find C, we have

cos B = cos b sin C ; whence sin C =

Computing by logarithms,

log cos 100 = 9.239670

- log cos 112 = 9.573575

Adding 10 = 9.666095 = log sin C. .*. C = 27 36' 58" .8, or 152 23' 01".

i.e. y BCA, orBC'A'.

84 SPHERICAL TKIGONOMETIIY.

Thus we see that each of the two triangles BCA and BC'A' fulfills the con-

ditions of the problem.

7. Given one side of a right angled spherical triangle 160, and the

opposite angle 150, to project the triangle and compute the other

parts.

Results. There are two triangles. The other sides of the first

are 136 50' 23", and 39 04' 51"; and the angle opposite the latter

side is 67 09' 43". The corresponding parts of the other triangle

are 43 09' 37", 140 55' 09", and 112 50' 17".

8. In the spherical triangle DBF, right angled at E, given an oblique

angle 58, and the side opposite 64, to project the triangle and com-

pute the other parts.

9. In a right angled spherical triangle given an oblique angle

165, and the opposite side 112, to project the triangle and com-

pute the other parts.

10. In a right angled spherical triangle given one side 65 23' 12",

and the opposite angle 65 23' 12", to project the triangle and com-

pute the other parts.

11. Given c = 60 47' 24".3, B = 57 16' 20".2, and A = 90, to

project and compute.

a = 68 56' 28".9, c = 54 32' 32".l, and I = 51 43' 36".l.

12. Given c = 116, b = 16, and the included angle 90, to pro-

ject and compute.

QUADRANTAL TRIANGLES.

129. A Quadrantal Triangle is a spherical triangle one

of whose sides is a quadrant, or 90. Such a triangle is readily

solved by passing to its polar, solving it, and then passing back.

The polar triangle to a quadrantal triangle, being right angled, is

solved by Napier's rules.

Ex. 1.- Given a = 90, B = 75 42', and c = 18 37', to compute

the other parts.

SUG'S. Representing the supplemental parts of the polar triangle by A', B',

C', a', ft', and c', we have A' = 180 - a = 90, V = 180 - B = 104 18', and

OF OBLIQUE ANGLED SPHERICAL TRIANGLES.

85

C' = 180 - c = 101 23', from which to find B', a', and c'. This being rijrht

angled, we find, by applying Napier's rules, B' = 94 31' 21", a' = 76 25' 11",

and c' = 161 55' 20". Hence in the primitive triangle we have b = 180 B'

- 85 28' 39", A = 180 - a' = 103 34' 49", and C = 180 - c' = 18 04' 40".

Ex. 2. Given a = 90, C = 42 10', and A = 115 20', to find the

other parts.

B = 54 44' 24", I = 64 36' 40", c = 47 57' 47".

SECTION II.

OF OBLIQUE ANGLED SPHERICAL TRIANGLES.

130. All cases of oblique angled spherical triangles can be solved

by Napier's rules and the following proposition.

131. J?vop. In any spherical triangle, if a perpendicular be

let fall from either vertex upon the opposite side or side produced,

the tangent of half the sum of the segments* of that side is to the

tangent of half the sum of the other two sides, as the tangent of half

the difference of those sides is to the tangent of half the difference

of the segments.

DEM. In the triangle BAG let fall the perpen-

dicular p, from C upon the opposite side. Let

BD = s, and DA = s'. By Napier's first rule,

cos a = cos p cos s, and cos b = cos p cos s'.

COS (t COS 8

Dividing the former by the latter, r = , ;

' cos b cos s' '

whence, by composition and division,

cos b cos a cos s' cos s

cos a

But by (61),

and

+ cos b cos s

cos b cos a

+ COS

S 1 ' FIG. 55.

cos a + cos b

cos s' cos s

tan -1-

f? I i r \ tnn 1 fa *'\

COS 8 + COS '

* When the perpendicular falls without the base, as in Fig. 5<>, this term is to be understood

as meaning the distances from the foot of the perpendicular to each extremity of the base, as

BD and AD- This, in fact, is the general statement applying as well to the case when the

perpendicular falls on the base.

86

SPHERICAL TRIGONOMETRY.

tan

or, tan (* +

'a + ft) tan i(a ft) = tan i (s + s') tan (s s')

tan (a + ft) : : tan (a ft) : tan (s s'). Q. E. D.

j?3. Sen. 1. Since from a point in the surface

of a hemisphere two perpendiculars can always

be drawn to the circumference of the great circle

which forms its base, and since the feet of these

perpendiculars are 180 apart, and no side of a

spherical triangle can equal 180, the foot of one

perpendicular will always fall within the base or

upon one extremity of it, and the other without

the base; or both will fall without the base. If

we take the foot of the perpendicular which falls

within the base, or the nearer one when both fall

without, the sum of the distances from the foot of

the perpendicular to the extremities of the base is always less than 180,

i. e., s + s' < 180. When the perpendicular falls wilhin, * + *' makes up one

side of the triangle, and hence is less than 180. If both perpendiculars fall

without, let D, Fig. 56, be the foot of the nearer one. Now DB + BD' = 180 ;

but by hypothesis DA < BD', .% DB + DA < 180. When DA = BD', DB + DA

= 180.

133. Sen. 2. As in spherical triangles the greater segment is not always

adjacent to the greater side, it becomes necessary to determine the position of

the segments. This can be done by the signs of the proportion

tan (s + s') : tan \ (a + ft) : : tan (a ft) : tan (s s').

1st. Tan ^ (.<? + s') is always + , since, if D falls in the base, s + s' < 180 ;

and if D falls without, by taking the nearer perpendicular, s + s' is still < 180

(132). .'. H* + s') < 90, and tan i (* + *') is +

2d. When a + ft < 180, tan i (a + ft) is + ; and when a + ft > 180.

tan i (a + ft) is .

3d. When a > ft, a ft is a positive arc less than 180, hence tan (a ft)

is + ; and when a < ft, (a ft) is a negative arc and less than 180, hence

tan i (a, ft) is .

4th. The signs of these terms being determined, that of tan (s s') becomes

known. Now, as ^(s s') cannot be numerically greater than 90, tan i(* s')

is + when s > s', and when s < '.

5th. When s + s' = 180, tan $(s + s')= oo. Now as a ft < 180, tan \ (a -ft)

cannot be oo, nor can tan \ (s s 1 ) = when the perpendicular falls without.

Hence to make the proportion possible, tan (a + ft) must be oc, or a + ft = 180

In this case we project on the plane of a or ft. If a + b = 180, aud a + c, = 180,

we project on the plane of a. If a + b = 180, a + c = 180, and b + c = 180, the

triangle is trirectangular.

134. Sen. 3. If either segment is greater than the whole base, the perpen-

dicular falls without the triangle. In this case the shorter segment lies in an

opposite direction from its angle to that considered in the demonstration, and

heiice is to be considered ; and the algebraic sum of the segments is still

equal to the side upon which the perpendicular is let fall.

OBLIQUE ANGLED TRIANGLES SOLVED BY NAPIER'S RULES. 87

o. Prop. In a spherical triangle, the sines of the sides are

to each other as the sines of their opposite angles.

DEM. By Napier's first rule we have from either Fig. 55 or Fig. 56,

sin p = sin a sin B, and sin p sin b sin A.

.-. sin a sin B = sin b sin A, or sin a : sin b : : sin A : sin B. Q. E. D.

Sen. This proposition is not introduced here because it is necessary for the

solution, of spherical triangles, but because of its essential importance. It is

often convenient to use it in the solution of a triangle, but never necessary, as

will appear hereafter. It affords a ready method of determining a part OPPOSITE

a given part, provided the species of the part be determined by other considerations.

SOLUTION OF OBLIQUE ANGLED SPHERICAL TRIANGLES BY

NAPIER'S RULES FOR RIGHT ANGLED SPHERICAL TRIANGLES.

136. The examples which arise in the solution of oblique angled

spherical triangles are all comprised under the three following

problems, each of which consists of two cases :

1. When the given parts are all adjacent to each other.

2. When two of the given parts are adjacent and one separate.

3. When the given parts are all separate from each other.

137 I*rob 1. Given three adjacent parts of an oblique angled

spherical triangle, to solve the triangle.

CASE 1st. Given two sides and the included angle.

SOLUTION. Project the triangle qn the plane of one oftJie given sides (106\

and let fall a perpendicular from the angle opposite upon this side or upon

D ""

FIG. 57. FlG. 58.

this side produced, as the case may be. There are thus formed two right

angled triangles, as BDC and DCA, each of which can be solved by Napier's

88 SPHERICAL TRIGONOMETRY.

rules, by first solving the one containing the given angle. Thus, in the triangle

BDC right angled at D, a and B are supposed known ; whence CD, BD, and the

angle BCD, can be computed. As BA = c is known, the segment DA can

be found, it being the difference between c and the arc BD. When the solu-

tion of this triangle gives BD > r, it is evident that the perpendicular falls

without the triangle, which will agree with the projection. Passing to the

triangle ADC, right angled at D, we now know CD and DA ; whence the

other parts can be found. Finally, the angle BCA of the required triangle =:

BCD + DCA when the perpendicular falls within the triangle, and BCD DCA

when the perpendicular falls without.

CASE 2cL Given two angles and the included side.

SOLUTION. The solution of this case is effected by passing to the polar

triangle, projecting and solving it by Case 1st, and then passing back.

138. Sen. A slight saving of labor is effected by using (135} in the solu-

tion. Thus, in the triangle BCD, compute CD and BD as before, and (not com-

puting angle BCD) then passing to the triangle DCA, compute b and A. Finally,

compute C (the entire angle) from the proportion

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

139. Prob. 2. In an oblique angled spherical triangle, given

two parts adjacent to each other and one separated from both of them,

to solve the triangle.

CASE 1st. Given two sides and an angle opposite one of them.

SOLUTION. Project the triangle on the plane of the unknown side, with the

given angle at B ; and let fall the perpendicular from the angle C opposite the

unknown side. Compute the triangle BDC. Having computed this triangle,

FIG. 59. FIG. 60.

compare the side opposite the given angle, as b, with the perpendicular and

the arcs BC and CB', i. e., with p, a, and 180 a. If b = p there is but

OBLIQUE ANGLED TRIANGLES SOLVED BY NAPIER* S RULES. 89

one solution and the triangle is right angled, A falling at D. If b is inter-

mediate in value between p and both a and 180 , it can occupy two positions

as in Fig. 59, and there are two solutions. If b is intermediate in value between

p and only one of the arcs a or 180 a, there is but one solution. When B < 90

the perpendicular is less than any oblique arc; hence in this case, if b <p, there

is no solution. But if B > 90, the perpendicular is greater than the oblique

airs; hence in this case, if b > p, there is no solution. [These results should be

obtained independently of the results given by the projection, and one be made

a check upon the other.] The solution is now completed by computing the

parts of DCA, and adding or subtracting the segments BD and AD, and the angles

BCD and A CD, as the case may require.

CASE 3d. Given two angles and a side opposite one of them.

SOLUTION. Pass to the polar triangle ; solve it, and then pass back.*

140. Sen. The relation established in (135} may also be used in the

solution of this problem. Thus, having projected the triangle, computed

p, and determined whether there are one or two solutions, to find A, when

, 5, and B are given, we have, sin b : sin a : : sin B : sin A. Then computing

the third side c (or sides), by means of the right angled triangles BCD and DCA

as before, we may use the proportion (135) to find the angle BCA and BCA'.

But the use of this proportion gives no advantage except in cases in which

there is only one solution.

141. Prob. 3. In an oblique angled spherical triangle, given

three parts all separated from each other, to solve the triangle.

CASE 1st. Given the sides to find the angles.

SOLUTION. Project the triangle on the plane of one of its sides, as c. From

the proportion,

tan -JO -f- s') : tan \(a + b) :: tan|(a b) : tan l(ss') y

* This case can be projected and solved in a manner

altogether similar to the first, without passing to the

polar triangle. Thus, let Bi #1 and A be the given parts.

Project the triangle on the plane of c, as in the figure.

Project B in the usual way, and make BC the projec-

tion of a. Through C draw DD', and. make BDO = the

projection of A- Drawing the small circle with radius

PC, draw diameters through the intersections Q and O'-

Then will A and A' be the vertices of the triangle re-

quired. The student may prove that the figures POD

and PCA are equal, and also PO'D and PCA', and

hence that angle BDO = A A'

Fia. 61.

90

SPHERICAL TRIGONOMETRY.

FIG. 62.

find |(* *'). ' Then half the sum of the segments

+ half the difference gives the greater segment, and

half the sum half the difference gives the less.

Determine from the signs of the terms whether * is

greater or less than s' : and also determine whether

the perpendicular lies within or without the tri-

angle (134.). Observe that these results correspond

to those given by the projection. Finally, in each

of the two right angled triangles BCD and DC A,

there are two sides given ; whence the angles can

be found by Napier's rules. If the perpendicular

falls within, C = BCD + DCA, and A, of the re-

quired triangle DAC. If the perpendicular falls

without, C = BCD DCA, and A of the triangle

= 180 - DAC.

CASE 2d. Given the angles to find the

SOLUTION. Pass to the polar triangle ; solve it,

and then pass back.

142, Sen. Here, again, (135) affords a slightly

more expeditious solution. Having projected the

triangle, found and located the segments, and com-

puted one angle, as B, by the methods given above, the other angles may be

found from the proportions,

sin b : sin a : : sin B : sin A,

and sin 5 : sin c : : sin B : sin C.

EXERCISES.

1. Given I = 120 30' 30", c = 70 20' 20", and A = 50 10' 10",

to project and solve the triangle.

PROJECTION. See Fig. 64.

TRIGONOMETRICAL SOLUTION. 1st. To solve the

triangle ABD, in which the two known parts are

situated.

(a) To find p, sin p = sin c sin A.

log sin 70 20' 20" = 9.973912

4- log sin 50 10' 10" = 9.885329

Rejecting 10 = 9.859241 = log- sin p

.'. p = 46 19' 01", the species being determined

by the opposite angle (125). [Observe that the re-

sult corresponds with the projection].

)BLIQUE ANGLED TKIANGLES BOLTED BY SAPIENS EULES. 91

(ft) To find AD, cos A = cot c tan AD, or tan AD = ^-^ .

log cos 50 10' 10" = 9.806532

- log cot 70 20' 20" = 9.553016

Adding 10 = 10.253516 = log tan AD. /. AD = 60 50' 49",

the species being determined by the signs of the formula.

(c) To find angle ABD, cos c = cot A cot ABD, or cot ABD = CC

log cos 70 20' 20" = 9.526929

log cot 50 10' 10" = 9.921204

Adding 10 = 9.605725 = log cot ABD. .-. ABD = 68 01' 53"

the species being determined by the signs of the formula.

3d. To solve the triangle DBC.

(a) To find DC. Since AD < ft, the foot of the perpendicular falls in the base,

and DC = AC - AD = b - AD = 120 30' 30" - 60 50' 49" = 59 39' 41".

(b) To find a, cos a cosp cos DC

log cos 46 19' 01" = 9.839270

+ log cos 59 39' 41" = 9.703386

Rejecting 10 = 9.542656 = log cos a. .: a = 69 34' 56", the

species being determined by the signs of the formula.

(c) To find C, sin p = sin a sin C, or sin C =

log sin p = 9.859241

- log sin 69 34' 56" = 9.971820

Adding 10 = 9.887421 = log sin C. .*. C = 50 30' 08", the

species being determined by the side opposite.

(d) To find angle DBC, sin p = tan DC cot DBC, or cot DBC = ^ ^ .

log sin p = 9.859241

- log tan 59 39' 41" = 10.232653

Adding 10 = ~9.626588 = log cot DBC. .'. DBC = 67 03' 36",

the species being determined by the signs of the formula.

Finally, B = ABD + DBC = 68 01' 53" + 67 03' 36" = 135 05' 29"

Sen. We might have omitted the computation of angle ABD in the first

part, and DBC in the second, and have found instead the entire angle B from

sin a : shift : : sin A : sin B. To compute this requires the looking out of but

two logarithms, since sin a is given in the second part (c), and sin A in the first

part (a).

2. Given a = 97 35', I = 27 08' 22", and A == 40 51' 18", to

project and compute the triangle. Between what limits must the

92 SPHERICAL TRIGONOMETRY.

value of a be assigned in order that there may be two solutions ? Be-

yond what limiting values of a is a solution impossible ?

PROJECTION. See Fig. 65.

TRIGONOMETRICAL SOLUTION. To find p, sin p

sin b sin A.

log sin 27 08' 22" = 9.659115

+ log sin 40 51' 18" = 9.815675

Rejecting 10 = 9.474790 = log sin p.

.'. p 17 21' 40". [Reason for the species.]

We now observe that there is one and only out

solution, since the arc a (97 35') cannot lie between

CD (17 21' 40") and 6(27 08' 22"), but can lie be-

tween CD and CA' (180 - b = 152 51' 38").

To find AD, cos A = cot b tan AD or tan AD =

log cos 40 51' 18" = 9.878733

log cot 27 08' 22" = 10.290226

Adding 10 = 9.588507 = log tan AD. .'. AD = 21 11' 30".6.

To find angle ACD, cos b cot A cot ACD, or cot ACD =

log cos 27 08' 22" = 9.949340

- log cot 40 51' 18" = 10.0630575

Adding 10 = 9.8862825 = log cot ACD. /. ACD = 52 25' 01".

[Reason for the species. ]

To find B, sin p = sin a sin B, or sin B = !!EJ?.

sines

log sin p (as above) = 9.474790

- log sin 97 35' = 9.996185

Adding 10 = 9.478605 = log sin B. .-. B = 17 31' 09".

[Reason tor the species. ]

To find DB, cos a = cos p cos DB, or cos DB = - .

cosp

log cos 97 35' = 9.120469

- log cos 17 21' 40" = 9.979750

Adding 10 = 9.140719 = log cos DB. .-. DB = 97 56' 51".8.

[Reason for the species.]

AB = AD + DB = 21 11' 30".6 + 97 56' 51".3 = 119 08' 21".9.

To find DCS, sin p = tan DB cot DCB, or cot DCB =

log sin p (as above) = 9.474790

- log tan 97 56' 51 ".3 = 10.855090

Adding 10 =" 8.619700 = log cot DCB. .-. DCB = 92 23' 7".7.

[Reason for the species.]

ACB = C = ACD -f DCB = 52 35' 01" + 92 23' 7".7 = 144 48' 8".7.

OBLIQUE ANGLED TKIANGLES SOLVED BY NAPIER'S RULE. OS

Finally, we observe that, if any value were assigned to a between b (27 08' 22")

and CD (17 21' 40") there would be two solutions ; since for such values the side

a could lie on both sides of CD. But, for any value of a less than CD (17 21' 40"),

there would be no solution ; since CD is the shortest distance from C to the arc

ABA'. Also, for any value of a greater than the arc CA' (152 51' 38"), there

would be no solution, as such an arc would fall between CA' and CD' (if not

> CD'), and consequently would make c > 180.

SCH. Such examples as this and the preceding can be more expeditiously

solved by using p in each equation in solving the triangles ACD and DCB. By

this means and using (135) to determine the side c, the solution can be effected

with only 12 logarithms. Thus in Ex. 2.

1st. To find p, sin^ = sin b sin A< requires 3 logarithms

2<1. To find ACD, cos ACD = cot 6 tan/?, requires 3 "

3d. To find DCB, cos DCB = cot a tan p, requires 2 " (log tan p being known).

4th. To find B, n# = sin a sin B> requires 2 (log sin P being known).

5th. To find c, sin A : sin C : ' sin a : sin c, requires 2_ " (log's of sin A and sin a

Total 12 logarithms being known.)

3. Given a = 76 35' 36", I = 50 10' 30", and c = 40 00' 10", to

project and solve the triangle.

PROJECTION. See Fig. 66.

TRIGONOMETRICAL SOLUTION. 1st. To find the

segments CD and DB, we have,

tan i(s + a'),

or

tan a : tan |(5 + c) : : tan i(b c) : tan i(s a').

Computing by logarithms.

a. c. log tan \a = log tan 38 17' 48" = 0.102561

+ log tan K& + c) = log tan 45 05' 20" = 10.001347

+ log tan i(5 c) = log tan 5 05' 10" = 8.949406

Rejecting 10 = 9.053314 = log tan i(a - a').

.-. i( - 8 ') = 6 27' 02".

In order to determine whether s or s' is the greater,* we observe the signs

of the proportion, and rinding tan %(s s') positive, know that * > s'.

Hence, s = K* + *') + fi - ') = 38 17' 48" + 6 27' 02" = 44 44' 50",

and s' = i(a + s') - i(a - s') = 38* 17' 48" - 6 27' 02" = 31 50' 46".

The angles sought are now readily found by computing the two right

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