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PROJECTION OF RIGHT ANGLED SPHERICAL TRIANGLES. 6&

from P (towards N if B < 90, and towards N' if B > 90) a distance equal to

the natural cosine of B.

QUERY. Is the solution of this problem possible for all values of the hypot-

enuse or adjacent side, and the angle?

EXAMPLES.

1. Having given B = 65, and the hypotenuse a = 120, to proiect

the triangle. See Fig. 30.

2. Having given C = 45, and the adjacent

side b 50, to project the triangle.

SUG'S. Project the angle C as the angle B of

the preceding, and lay off b = 50 from its vertex

on the circumference of the primitive circle.

3. Having given C = 170, and hypote-

nuse a = 160, to project the triangle.

FIG. 30.

4. Having given B = 150, and c = 40, to project the triangle.

1O4. Prob. 4. To project a right angled spherical triangle

when an oblique angle and side opposite are given.

SOLUTION. Project the given angle at B, Figs. 31, 32, as in the last problem.

Then, from any point in the circumference of the

primitive circle, as N, take NO', in the diameter

passing through that point, equal to the projection

of the given side. (This is done by taking Ntf =

Nd' = b, as in Prob. 1, and drawing dd'}. Now,

with P as a centre, and radius PO', describe a cir-

cumference cutting BOB'. One extremity of the

given side b will be projected in this circumference,

since this circumference contains the projections of

all the points in the surface of the hemisphere

which are at a distance b from the circumference

BNB'N'. But the vertex C is also projected in the

FIG. 31.

enuse, belong to a treatise on Conic Sections. In this case, BB' and 20P are * ne axes of the

ellipse, and the curve can be constructed by taking gp as a radius, and striking arcs from Q

as a centre, cutting BB'- These intersections are the foci of the required ellipse. Then take

a string equal in length to BB'i and, fastening its ends to the foci, place a pencil against the

string, and keeping the string tight, carry the pencil around the carve.

70

SPHERICAL TRIGONOMETRY.

FIG. 32.

arc BOB'; hence, it must be at the intersections

C, C'. Drawing the projections of the perpendic-

ulars CA, and C'A', the projection is complete.

QUERIES. When will there be two triangles

fulfilling the given conditions ? When but one ?

When none ? When there is but one triangle what

kind of a triangle is it? If B > 90, must b be

greater or less than B in order to have two solu-

tions ? If B < 90, how is it ? If B > 90, can b be

If B < 90, can b > 90 ?

EXAMPLES.

1. Given C = 120, c 150, to project the triangle. See Fig. 33.

2 Given c = 80, c = 60, to project the triangle. See Fig. 34.

FIG. 38.

FIG. 34.

3. Given B = 70, b = 70, to project the triangle. See Fig. 35.

4. Given B = 64, I = 75, to project the triangle. See Fig. 36.

FIG. 35. FIG. 36.

5. Given B = 160, I = 110 g , to project the triangle.

PROJECTION OF OBLIQUE ANGLED SPHERICAL TRIANGLES. 71

1O5. SCH. When the given parts are the two oblique angles, the projection

is most readily effected by first computing one of the sides. The projection in

this case will be considered in connection with the numerical solution of the

case, in the next section.

PROJECTION OF OBLIQUE ANGLED SPHERICAL TRIANGLES.

.106. frob. 1. To project a spherical triangle when two sides

and the included angle are given.

SOLUTION. Let a, c, and B denote the given parts.

B at some point in the circumference of the primi-

tive circle, as B. Lay off one of the given sides, as

c, from B on BNB'. Let BA = c. Determine the

extremity Cof the projection of the other given side

a, as in Prob. 2, etc., and drawing the diameter

AA', pass the arc through ACA' ; BCA is the pro-

jection sought.

QUERY. Is this projection always possible, what-

ever the relative magnitude of the given parts ?

Project the given angle

FIG. 37.

EXAMPLES.

1. Given A = 130, c = 85, and b = 100, to project the triangle.

2. Given c = 40, a = 37, and b = 80, to project the triangle.

A'"

107. Prob. 2. To project a spherical triangle when two sides

and an angle opposite one of them are given.

SOLUTION. Let the given parts be a, &, and B. Make the projection upon

the plane of the unknown side c. Thus, drawing the primitive circle and the

conjugate diameters BB', NNT, conceive c as projected from B on the arc BNB',

and project the given angle B as in preceding

problems. On the arc BB', take BC = the pro-

jection of the given adjacent side a. To deter-

mine the projection of the opposite side 5, describe

ta circle about P, as a centre, with a radius PC.

Through C draw PD, and taking Dd = Dd 1 b

draw dd'. Through the intersections 0, o' of dd'

with the circumference of the small circle, draw

the radii PA, PA'. Finally, passing arcs through

the points A, C, A", and A', C, A'", BAG, and

BA'C are the projections of triangles which fulfill

the given conditions. The projections of B and

72

SPHERICAL TRIGONOMETRY.

a were made upon principles previously established ; and it only remains to

show that AC, and A'C are projections of b. Since by construction DL is the

projection of an arc equal to , the projections of arcs of great circles connect-

ing D with o and o' are projections of arcs equal to b. But the figure DoP

ACP, and Do'P = A'CP; therefore AC = A'C = Do = the projection of b.

108. Sen. It is evident that this problem has one solution, two solutions, or

no solution, according to the value of b as compared with a and B. Thus, if the

projection of b DC, o and o' coincide, there is but one solution, and the tri-

angle is right angled at A (which in this case falls at D). Also, if the projection

of b is intermediate in value between DC and only one of the arcs BC, B'C, there

is only one solution. If, however, as in the figure given, the projection of b is

intermediate in value between DC and both BC and B'C, there are Resolutions.

Finally, if b is given of such value that the chord dd' does not touch the small

circle, there is no solution. The latter case occurs when B < 90, if the projec-

tion of b < DC ; and when B > 90, if the projection of b > DC, as will appear

from Figs. 88, 39.

We may observe, also, that there are two solutions when o and o' both fall on

the same side of BB' as c ; one solution when o and o' coincide, and when they

fall on opposite sides of BB' ; and no solution when o and o' are imaginary, i. c.,

when dd' does not touch the small circle, or when both fall on the opposite side

of BB' from c.

FIG. 39.

EXAMPLES.

1. Given B = 110, a - 120, and I = 83,

to project the triangle. See Fig. 39.

2. Given B = 110, a = 120, and b = 130,

to project the triangle.

3. Given c = 64, a = 120, and c = 75,

to project the triangle.

4. Given C = 80, b = 60, and c = 40,

to project the triangle.

5. Given c = 112, b = 75, and c = 150,

to project the triangle.

109. Prob. 3. To project a spherical triangle when the three

sides are given.

SOLUTION OF RIGHT ANGLED SPHERICAL TRIANGLES.

73

SOLUTION. Drawing the primitive circle and the conjugate diameters, as

nsual, take BA = c, the side on the plane of which it is proposed to project the

triangle. Take Be = Be' = a and draw ed ; then

as before shown, the projection of the vertex C

lies in ee'. In like manner taking Ad = Aef = b,

and drawing dd' the projection of C lies in dd'.

The intersection of the chords ee' and dd' is there-

fore the projection of the vertex C. Finally,

passing arcs through ACA' and BCB', the projec-

tion is complete.

QUERIES. How does the projection show the

impossibility when the sum of the three sides is

greater than 360 ? How when the sum of two

sides is less than the third side ?

EXAMPLES.

1. Given a = 100, I = 80, and c = 68,

to project the triangle.

2. Given a = 108, I = 120, and c = 25,

to project the triangle. See Fig. 41.

3. Given a = 120, I = 65, and c = 40,

to project the triangle.

4. Given a = 150, I = 140, and c =

170, to project the triangle.

FIG. 41.

SECTION L

SOLUTION OF RIGHT ANGLED SPHERICAL TRIANGLES.

110. Spherical Trigonometry treats of the relations be-

tween the trigonometrical functions of the sides and angles of

spherical triangles, and of the solution of such triangles by means of

these relations.

111. Sen. In the present treatise we shall confine our attention to triangles

none of whose sides or angles exceed 180.

112. The Six farts of a spherical triangle are the three

sides and the three angles : any three of these being given, the others

may be found.

74 SPHEKICAL TRIGONOMETRY.

113. Sen. The last statement involves the assertion that the three angles

of a spherical triangle determine the sides, whereas we are accustomed to say

in Plane Trigonometry, that, at least one given part of a plane triangle must be

a side, in order that the triangle may he determined. There is really no such

difference as these two statements imply. For example, it' we have the angles

of a plane triangle given, we know the ratios of the sides to each other, since

the sides are to each other as the sines of the angles opposite ; but we cannot

determine the absolute values of the sides. This is in accordance with the state-

ment that mutually equiangular plane triangles are similar figures (not neces-

sarily equal). Now these are exactly the facts in the case of spherical triangles,

if we do not limit tJiem to the same or equal spJieres. Thus, the angles of a spheri-

cal triangle being given as 48 30', 125 20', and 62 54', we solve the triangle by

the rules of spherical trigonometry and find that the sides are 56 39' 30",

114 29' 58", and 83 12' 6". But, so long as the radius of the sphere is unknown,

these results are merely the relative values of the sides, not their absolute lengths.

Moreover, consider two concentric spheres whose radii are m and n. Now, any

triangle being constructed on the one, if planes are passed through its sides

intersecting at the common centre, their intersection with the surface of the

other sphere will form a triangle mutually equiangular with the first, and any

one side of the one triangle is to the corresponding side of the other, as the radii'

of the spheres ; hence the homologous sides are proportional. We see, therefore,

that to determine absolutely a spherical triangle, it is necessary to know one of

the sides in linear extent as well as angular measure, or, what is equivalent,

the radius of the sphere must be known.

114. The Species of a part of a spherical triangle is deter-

mined by its relation to 90. Two parts, both greater or both less

than 90, are of the same species ; two parts, one of which is greater

and the other less than 90, are of different species. Thus, two parts

which are 58, and 63, respectively, are of the same species ; two

which are 160, and 115, are of the same species; two which are

120, and 48, are of different species.

Napier's Five Circular farts are the two sides of a

right angled spherical triangle about the right angle, and the comple-

ments of the hypotenuse and of the oblique angles. These terms are

merely conventional, and are applied exclusively to right angled

triangles.

ILL. In a spherical triangle ABC, right angled at A, the

sides b and c, the complement of hypotenuse , and the com-

plements of the angles B and C, are the circular parts. We

may designate them 5, c, comp a, comp B, and comp C.

[It is essential that this nomenclature and the statements

of the two following paragraphs be clearly understood, and

firmly fixed in the mind, in order that the phraseology of

the fundamental rules may be intelligible.!

SOLUTION OF BIGHT ANGLED SPHERICAL TRIANGLES. 75

116. When five things occur in succession, as it were in a circle,

like the circular parts b, c, comp B, comp a, and comp C, in the

preceding figure (no account being made of A), it will be observed,

that, taking any three at pleasure, one of the three may always be

selected which lies adjacent to each of the others, or separated from

each of them. Of the three parts thus considered, the Middle l?ait

is the one which has the other two adjacent to or separated from it ;

while the latter are called the ExtTewies, adjacent or opposite, as

the case may be.

ILL'S. Let the three parts under consideration be comp a, b, and c; comp a

is the middle part, and b and c are the opposite extremes. If 5, e, and comp C

are under consideration, b is the middle part, and c and comp C are the adjacent

extremes.

117. Sen. In solving a right angled spherical triangle, there are always

three parts under consideration at once, viz., the two given parts and the part

sought, no mention being made of the right angle. Now, the first question to be

settled in order to a solution is, Which of the three parts under consideration is the

middle part, and are the extremes opposite or adjacent? Beginners are very liable

to make mistakes by failing to use the complements of the proper parts ; or by

not correctly distinguishing the middle part, and the character of the extremes,

as opposite or adjacent. The student should practise upon such simple exer-

cises as the following until the questions can be answered instantly, and with-

out mistake.

EXERCISES.

1. In Fig. 42, given a and c, to find C. What are the circular parts

under consideration ? Which is the middle part ? Are the extremes

adjacent or opposite?

Answers. The circular parts are comp #, c, and comp C. cis the

middle part, and the extremes are opposite.

2. Having c, and a given, to find I. What are the circular parts ?

Which is the middle part ? Are the extremes adjacent or opposite ?

3. Having c, and b given, to find B. What are the circular parts ?

Which is the middle part ? Are the extremes adjacent or opposite ?

4. Having a, and b given, to find C. What are the circular parts ?

Which is the middle part? Are the extremes adjacent or opposite?

5. Having B and C given, to find b. What are the circular parts ?

Which is the middle part ? Are the extremes adjacent or opposite ?

76

SPHERICAL TRIGONOMETRY.

G. What are the opposite extremes when b is the middle part?

What the adjacent extremes ? Which are the opposite and which the

adjacent extremes when c is the middle part? When comp B is the

middle part? When comp C is the middle part? When comp a is

the mi Idle part ?

7. What part is middle part to comp c and c as adjacent ex-

tremes? As opposite extremes?

Ans., I, and none.

8. In Fig. 43, M "being the right angle,

what are the circular parts ? Given and m t

to find o. What are the circular parts under

consideration ? Are the extremes adjacent or

opposite ?

Fie. 43.

9. What are the opposite extremes to comp

O ? What the adjacent ? To comp m ? To o

To n?

NAPIER'S RULES.

118. Hule I. Prop. In any right angled spherical triangle,

the sine of the middle part equals the pro-

duct of the cosines of the opposite ex-

tremes.

DEM. In the spherical triangle BAG, right

angled at A, taking b, c, comp B, comp C, and

comp a in succession as middle parts, we are to

prove that

sin b = cos (comp a) x cos (comp B), or sin b = sin a sin B ; (1)

sin c cos (comp a} x cos (comp C), or sin c sin a sin C ; (2)

sin (comp B) = cos b x cos (comp C), or cos B = cos b sin C ; (3)

sin (comp C) = cos c x cos (comp B), or cos C = cos c sin B (4)

sin (comp a) = cos b x cos c, or cos a = cos b cos c. (5)

To demonstrate these relations, let O be the centre of the sphere, and draw

the radii OA, OB, and OC. The angles BOC, AOC, and AOB, are measured

respectively by , b, and c, the sides of the triangle ; hence these angles at the

crntre and their measuring arcs may be used interchangeably. From one of

,

NAPIER S RULES.

77

;he oblique angles, as C, let fall a perpendicular upon the radius OA. From

the foot of this perpendicular draw DE perpendicular to OB, and join C and E.

Now CDE is a right angle (PAKT II., 426), CE is perpendicular to OB (PAKT II.,

399), and DEC is equal to angle B of the triangle (PAKT II., 558).

CD = sin b, OD = cos b, CE = sin a, and OE = cos a.

From the triangle CDE, right angled at D, we have

CD = CE x sinCED, or sin b = sin sin B. (1)

Generalized, (1) becomes, The sine of either side about the right angle = the

sine of the hypotenuse into the sine of the angle opposite the side. Hence, from

analogy to (1), we may write

sin c = sin a sin C. (2)

Or (2) may be proved in the same manner as (1),

by letting fall from B a perpendicular upon OA,

from its foot drawing DE perpendicular to OC,

and joining E and B. Then BD sin c, OD

ccs c, BE = sin , OE = cos a, and angle BED =

angle C. From the triangle BDE, we have

BD = BE x sin BED, or sin c = sin a sin C. (2)

To prove (3), we have from the triangle CDE,

Fig. 44,

cos CED =

ED

CE

or cos B =

ED

sin a'

But from triangle OED, right angled at E, ED = OD x sin DOE = cos b sin e =

[from (2)J, cos b sin a sin C. Substituting this value of ED, we have

cos b sin a sin C

cos B = r = cos b sm C. (3)

sin a

We may write (4) from (3) by analogy, as (2) was from (1) ; or, better, let the

student produce it from Fig. 45, as (3) was produced from Fig. 44.

Finally, to produce (5), consider the triangle ODE, in either figure, right

angled at E. This gives

OE=OD x cos DOE, or cos a = cos b cos c. (5)

119. Rule 2. Prop. In any right angled spher-

ical triangle, the sine of the middle part equals the pro-

duct of the tangents of the adjacent extremes.

DEM. In the spherical triangle BAG, right angled at A, taking

b, c, comp B, comp C, and comp a, in succession as middle parts,

we are to prove that,

FIG. 46.

78 SPHERICAL TRIGONOMETRY.

sin b tan c x tan (comp C), or sin b = tan c cot C ; (1)

sin c = tan b x tan (comp B), or sin c = tan b cot B ; (2)

sin (comp B) = tan c x tan (comp a), or cos B = cot a tan c ; (3)

sin (comp C) = tan b x tan (comp a), or cos C = cot a tan b \ (4)

sin (comp a) = tan (comp B) x tan (comp C), or cos a = cot B cot C (5)

Taking the formula of RULE 1st, and in the second member of each substitu-

ting the value of each factor as found in some other of the set, we readily

write the following

sin c cos C sin e cos C

sm 6 = sm a sin B = ^ = x -; = = tan c cot C ; (1)

sin C cose cose sm C

sin b cos B sin b cos B

sin c = sin a sin C = -= ? = 7 x -r = = tan b cot B ; (2)

sin B cos b cos b sin B

cos a sin c cos a sin c

cos B = cos b sin C = : = x = cot a tan c ; (3)

cos c sin a sm a cos c

cos a sin b cos a sin b

cos C = cos c sin B = r : = - x r = cot tan b ; (4)

cos 6 sin a sin a cos b

cos B cos C cos B cos C

cos a = cos b cos c = . = ^ =. x -r ^ = cot B cot C. (5)

sin C sin B sin B sin C

Q. E. D.

120. Sen. 1. It will be a good exercise for the student to demonstrate

RULE 2d from the an-

E. nexed figures, as RULE 1st

was from Figs. 44 and 45.

The 5th is not as readily

obtained from the figure

as the others. The stu-

dent may trace the fol-

lowing relations, some in

one figure, and some in

FIG. 47. FIG. 48. the other.

-. , -,

OE sec e AE tanc AE

sin c sin b sin c *.*/-

= . ; ; whence, cot B cot C = - = cos b cos c. .'. cos a = cot B cot C.

tan 6' tan 6 tanc

121. Sen. 2. It is of much importance, especially for the purposes of

Spherical Astronomy, that the student observe that the relations expressed in

the above formula, and in fact all the relations between the sides and angles of

spherical triangles, are also the relations between the facial and diedral angles

of triedrals. Thus, if a, b, and c represent the facial angles, and A, B, and C the

opposite diedrals, all these relations can be established, and in exactly the same

manner as above, without any allusion to the spforical triangle. [The student

should do it.J

DETERMINATION OF SPECIES. 79

DETERMINATION OF SPECIES.

In the solution of spherical triangles the determination of

the species of a part sought becomes of essential importance, since

any part of such a triangle may have any value between and 180.

Hence, when we have learned the numerical value of any function of

a part, we have yet to determine whether the part itself is less or

greater than 90, i. e., the species of the part. This may always be

effected by some one of the following propositions.

123. Prop. If the part sought is found in terms of its cos,

tan, or cot, its species can be determined by the algebraic signs of the

functions in the formula used.

DEM. In each of the formula arising from the application of Napier's rules,

there are three functions, the arcs corresponding to two of which are always

known, hence the algebraic signs of their functions are known, and the signs

of these two determine the sign of the third or unknown function. Now, when

a cos, tan, or cot is + , the corresponding angle is less than 90 (if less than

180) ; and when one of these functions is , the corresponding angle is greater

than 90 ; i. e., in a triangle, it is between 90 and 180.

124. When the part sought is found in terms of its sine, the

species cannot be determined by the signs of the formula, since the

part being less than 180 its sine is always +. The three following

propositions dispose of such cases.

125. Prop. An oblique angle of a right angled spherical tri-

angle and its opposite side are always of the same species.

DEM. From Napier's first rule we have, cos B = cos b sin C. But sin C is

necessarily + ; therefore, cos B and cos b always have the same sign, and B and

b are of the same species. In like manner, we see from cos C = cos c sin B, that

C and c are of the same species. Q. E. D.

126. Prop. When the hypotenuse of a right angled spherical

triangle is less than 90, the other two sides (and consequently thff

80 SPHERICAL TRIGONOMETRY.

oblique angles) are of the same species with each other. But when the

hypotenuse is greater than 90, the other two sides (and consequently

the oblique angles) are of different species from each other.

DEM. From Napier's first rule we have, cos a = cos b cos c. Now, if

a < 90, cos a is + ; hence cos b and cos c must have like signs, and b and c be

both less or both greater than 90 But if a > 90 (and less than 180, as it is),

cos a is ; hence, cos b and cos c must have different signs, and b and c be one

greater and the other less than 90. Finally, since the oblique angles are of the

same species as their opposite sides, they are of the same species with each other

when a < 90, and of different species from each other when a > 90.

. l?rop When a side and its opposite angle are given in a

right angled spherical triangle, there is NO solution if the sine of the

side is greater than the sine of its opposite angle ; there is ONE

solution and the triangle is bi-rectangular if the sine of the side

equals the sine of its opposite angle ; and there are TWO solutions if

the sine of the side is less than the sine of its opposite angle.

DEM. We have sin b = sin a sin B, or sin a = . = . Now, sin b > sin B

sm B

makes sin a > 1, which is manifestly impossible. Sin b sin B makes b = B,

since they are of the same species. But when the arc included by the sides of

an oblique angle of a right angled spherical triangle is equal to the angle, the

vertex of the angle is the pole of the arc. Hence, in this case the other sides of

the triangle are each 90. Finally, if sin b < sin B, a has two values, one

greater and the other less than 90. Hence there are two triangles.

128. Sen. These relations between an angle and its opposite side may be

observed directly from a figure. When B < 90, the measure of it, that is

b = B, is the greatest included perpendicular which

can be drawn to one side of the lune BAB'C. Hence,

in this case, b cannot exceed B, which implies that

sin b cannot be > sin B, as when the arcs are less

than 90, the greater arc has the greater sine. If

sin b = sin B, b = B, and BA = BC = 90, and the

side b can occupy but one position in the lime, thus

giving rise to but one triangle BAG, which satisfies

the conditions (or two equal triangles BAG and

B'AC). If b < B, which, when B is less than 90?

implies that sin b < sin B, the side can occupy two positions in the lune, b' and

b" giving rise to two triangles, BA'C' and BA"C", both of which fulfill the

conditions.

EXERCISES IN SOLVING RIGHT ANGLED TRIANGLES.

81

Again, when B > 90, the measure of it, i. e., b = B, is the least included

perpendicular that can be drawn to one side of the

lune. Hence, in this case we cannot have b < B,

which implies that sin b cannot be > sin B, since the

PROJECTION OF RIGHT ANGLED SPHERICAL TRIANGLES. 6&

from P (towards N if B < 90, and towards N' if B > 90) a distance equal to

the natural cosine of B.

QUERY. Is the solution of this problem possible for all values of the hypot-

enuse or adjacent side, and the angle?

EXAMPLES.

1. Having given B = 65, and the hypotenuse a = 120, to proiect

the triangle. See Fig. 30.

2. Having given C = 45, and the adjacent

side b 50, to project the triangle.

SUG'S. Project the angle C as the angle B of

the preceding, and lay off b = 50 from its vertex

on the circumference of the primitive circle.

3. Having given C = 170, and hypote-

nuse a = 160, to project the triangle.

FIG. 30.

4. Having given B = 150, and c = 40, to project the triangle.

1O4. Prob. 4. To project a right angled spherical triangle

when an oblique angle and side opposite are given.

SOLUTION. Project the given angle at B, Figs. 31, 32, as in the last problem.

Then, from any point in the circumference of the

primitive circle, as N, take NO', in the diameter

passing through that point, equal to the projection

of the given side. (This is done by taking Ntf =

Nd' = b, as in Prob. 1, and drawing dd'}. Now,

with P as a centre, and radius PO', describe a cir-

cumference cutting BOB'. One extremity of the

given side b will be projected in this circumference,

since this circumference contains the projections of

all the points in the surface of the hemisphere

which are at a distance b from the circumference

BNB'N'. But the vertex C is also projected in the

FIG. 31.

enuse, belong to a treatise on Conic Sections. In this case, BB' and 20P are * ne axes of the

ellipse, and the curve can be constructed by taking gp as a radius, and striking arcs from Q

as a centre, cutting BB'- These intersections are the foci of the required ellipse. Then take

a string equal in length to BB'i and, fastening its ends to the foci, place a pencil against the

string, and keeping the string tight, carry the pencil around the carve.

70

SPHERICAL TRIGONOMETRY.

FIG. 32.

arc BOB'; hence, it must be at the intersections

C, C'. Drawing the projections of the perpendic-

ulars CA, and C'A', the projection is complete.

QUERIES. When will there be two triangles

fulfilling the given conditions ? When but one ?

When none ? When there is but one triangle what

kind of a triangle is it? If B > 90, must b be

greater or less than B in order to have two solu-

tions ? If B < 90, how is it ? If B > 90, can b be

If B < 90, can b > 90 ?

EXAMPLES.

1. Given C = 120, c 150, to project the triangle. See Fig. 33.

2 Given c = 80, c = 60, to project the triangle. See Fig. 34.

FIG. 38.

FIG. 34.

3. Given B = 70, b = 70, to project the triangle. See Fig. 35.

4. Given B = 64, I = 75, to project the triangle. See Fig. 36.

FIG. 35. FIG. 36.

5. Given B = 160, I = 110 g , to project the triangle.

PROJECTION OF OBLIQUE ANGLED SPHERICAL TRIANGLES. 71

1O5. SCH. When the given parts are the two oblique angles, the projection

is most readily effected by first computing one of the sides. The projection in

this case will be considered in connection with the numerical solution of the

case, in the next section.

PROJECTION OF OBLIQUE ANGLED SPHERICAL TRIANGLES.

.106. frob. 1. To project a spherical triangle when two sides

and the included angle are given.

SOLUTION. Let a, c, and B denote the given parts.

B at some point in the circumference of the primi-

tive circle, as B. Lay off one of the given sides, as

c, from B on BNB'. Let BA = c. Determine the

extremity Cof the projection of the other given side

a, as in Prob. 2, etc., and drawing the diameter

AA', pass the arc through ACA' ; BCA is the pro-

jection sought.

QUERY. Is this projection always possible, what-

ever the relative magnitude of the given parts ?

Project the given angle

FIG. 37.

EXAMPLES.

1. Given A = 130, c = 85, and b = 100, to project the triangle.

2. Given c = 40, a = 37, and b = 80, to project the triangle.

A'"

107. Prob. 2. To project a spherical triangle when two sides

and an angle opposite one of them are given.

SOLUTION. Let the given parts be a, &, and B. Make the projection upon

the plane of the unknown side c. Thus, drawing the primitive circle and the

conjugate diameters BB', NNT, conceive c as projected from B on the arc BNB',

and project the given angle B as in preceding

problems. On the arc BB', take BC = the pro-

jection of the given adjacent side a. To deter-

mine the projection of the opposite side 5, describe

ta circle about P, as a centre, with a radius PC.

Through C draw PD, and taking Dd = Dd 1 b

draw dd'. Through the intersections 0, o' of dd'

with the circumference of the small circle, draw

the radii PA, PA'. Finally, passing arcs through

the points A, C, A", and A', C, A'", BAG, and

BA'C are the projections of triangles which fulfill

the given conditions. The projections of B and

72

SPHERICAL TRIGONOMETRY.

a were made upon principles previously established ; and it only remains to

show that AC, and A'C are projections of b. Since by construction DL is the

projection of an arc equal to , the projections of arcs of great circles connect-

ing D with o and o' are projections of arcs equal to b. But the figure DoP

ACP, and Do'P = A'CP; therefore AC = A'C = Do = the projection of b.

108. Sen. It is evident that this problem has one solution, two solutions, or

no solution, according to the value of b as compared with a and B. Thus, if the

projection of b DC, o and o' coincide, there is but one solution, and the tri-

angle is right angled at A (which in this case falls at D). Also, if the projection

of b is intermediate in value between DC and only one of the arcs BC, B'C, there

is only one solution. If, however, as in the figure given, the projection of b is

intermediate in value between DC and both BC and B'C, there are Resolutions.

Finally, if b is given of such value that the chord dd' does not touch the small

circle, there is no solution. The latter case occurs when B < 90, if the projec-

tion of b < DC ; and when B > 90, if the projection of b > DC, as will appear

from Figs. 88, 39.

We may observe, also, that there are two solutions when o and o' both fall on

the same side of BB' as c ; one solution when o and o' coincide, and when they

fall on opposite sides of BB' ; and no solution when o and o' are imaginary, i. c.,

when dd' does not touch the small circle, or when both fall on the opposite side

of BB' from c.

FIG. 39.

EXAMPLES.

1. Given B = 110, a - 120, and I = 83,

to project the triangle. See Fig. 39.

2. Given B = 110, a = 120, and b = 130,

to project the triangle.

3. Given c = 64, a = 120, and c = 75,

to project the triangle.

4. Given C = 80, b = 60, and c = 40,

to project the triangle.

5. Given c = 112, b = 75, and c = 150,

to project the triangle.

109. Prob. 3. To project a spherical triangle when the three

sides are given.

SOLUTION OF RIGHT ANGLED SPHERICAL TRIANGLES.

73

SOLUTION. Drawing the primitive circle and the conjugate diameters, as

nsual, take BA = c, the side on the plane of which it is proposed to project the

triangle. Take Be = Be' = a and draw ed ; then

as before shown, the projection of the vertex C

lies in ee'. In like manner taking Ad = Aef = b,

and drawing dd' the projection of C lies in dd'.

The intersection of the chords ee' and dd' is there-

fore the projection of the vertex C. Finally,

passing arcs through ACA' and BCB', the projec-

tion is complete.

QUERIES. How does the projection show the

impossibility when the sum of the three sides is

greater than 360 ? How when the sum of two

sides is less than the third side ?

EXAMPLES.

1. Given a = 100, I = 80, and c = 68,

to project the triangle.

2. Given a = 108, I = 120, and c = 25,

to project the triangle. See Fig. 41.

3. Given a = 120, I = 65, and c = 40,

to project the triangle.

4. Given a = 150, I = 140, and c =

170, to project the triangle.

FIG. 41.

SECTION L

SOLUTION OF RIGHT ANGLED SPHERICAL TRIANGLES.

110. Spherical Trigonometry treats of the relations be-

tween the trigonometrical functions of the sides and angles of

spherical triangles, and of the solution of such triangles by means of

these relations.

111. Sen. In the present treatise we shall confine our attention to triangles

none of whose sides or angles exceed 180.

112. The Six farts of a spherical triangle are the three

sides and the three angles : any three of these being given, the others

may be found.

74 SPHEKICAL TRIGONOMETRY.

113. Sen. The last statement involves the assertion that the three angles

of a spherical triangle determine the sides, whereas we are accustomed to say

in Plane Trigonometry, that, at least one given part of a plane triangle must be

a side, in order that the triangle may he determined. There is really no such

difference as these two statements imply. For example, it' we have the angles

of a plane triangle given, we know the ratios of the sides to each other, since

the sides are to each other as the sines of the angles opposite ; but we cannot

determine the absolute values of the sides. This is in accordance with the state-

ment that mutually equiangular plane triangles are similar figures (not neces-

sarily equal). Now these are exactly the facts in the case of spherical triangles,

if we do not limit tJiem to the same or equal spJieres. Thus, the angles of a spheri-

cal triangle being given as 48 30', 125 20', and 62 54', we solve the triangle by

the rules of spherical trigonometry and find that the sides are 56 39' 30",

114 29' 58", and 83 12' 6". But, so long as the radius of the sphere is unknown,

these results are merely the relative values of the sides, not their absolute lengths.

Moreover, consider two concentric spheres whose radii are m and n. Now, any

triangle being constructed on the one, if planes are passed through its sides

intersecting at the common centre, their intersection with the surface of the

other sphere will form a triangle mutually equiangular with the first, and any

one side of the one triangle is to the corresponding side of the other, as the radii'

of the spheres ; hence the homologous sides are proportional. We see, therefore,

that to determine absolutely a spherical triangle, it is necessary to know one of

the sides in linear extent as well as angular measure, or, what is equivalent,

the radius of the sphere must be known.

114. The Species of a part of a spherical triangle is deter-

mined by its relation to 90. Two parts, both greater or both less

than 90, are of the same species ; two parts, one of which is greater

and the other less than 90, are of different species. Thus, two parts

which are 58, and 63, respectively, are of the same species ; two

which are 160, and 115, are of the same species; two which are

120, and 48, are of different species.

Napier's Five Circular farts are the two sides of a

right angled spherical triangle about the right angle, and the comple-

ments of the hypotenuse and of the oblique angles. These terms are

merely conventional, and are applied exclusively to right angled

triangles.

ILL. In a spherical triangle ABC, right angled at A, the

sides b and c, the complement of hypotenuse , and the com-

plements of the angles B and C, are the circular parts. We

may designate them 5, c, comp a, comp B, and comp C.

[It is essential that this nomenclature and the statements

of the two following paragraphs be clearly understood, and

firmly fixed in the mind, in order that the phraseology of

the fundamental rules may be intelligible.!

SOLUTION OF BIGHT ANGLED SPHERICAL TRIANGLES. 75

116. When five things occur in succession, as it were in a circle,

like the circular parts b, c, comp B, comp a, and comp C, in the

preceding figure (no account being made of A), it will be observed,

that, taking any three at pleasure, one of the three may always be

selected which lies adjacent to each of the others, or separated from

each of them. Of the three parts thus considered, the Middle l?ait

is the one which has the other two adjacent to or separated from it ;

while the latter are called the ExtTewies, adjacent or opposite, as

the case may be.

ILL'S. Let the three parts under consideration be comp a, b, and c; comp a

is the middle part, and b and c are the opposite extremes. If 5, e, and comp C

are under consideration, b is the middle part, and c and comp C are the adjacent

extremes.

117. Sen. In solving a right angled spherical triangle, there are always

three parts under consideration at once, viz., the two given parts and the part

sought, no mention being made of the right angle. Now, the first question to be

settled in order to a solution is, Which of the three parts under consideration is the

middle part, and are the extremes opposite or adjacent? Beginners are very liable

to make mistakes by failing to use the complements of the proper parts ; or by

not correctly distinguishing the middle part, and the character of the extremes,

as opposite or adjacent. The student should practise upon such simple exer-

cises as the following until the questions can be answered instantly, and with-

out mistake.

EXERCISES.

1. In Fig. 42, given a and c, to find C. What are the circular parts

under consideration ? Which is the middle part ? Are the extremes

adjacent or opposite?

Answers. The circular parts are comp #, c, and comp C. cis the

middle part, and the extremes are opposite.

2. Having c, and a given, to find I. What are the circular parts ?

Which is the middle part ? Are the extremes adjacent or opposite ?

3. Having c, and b given, to find B. What are the circular parts ?

Which is the middle part ? Are the extremes adjacent or opposite ?

4. Having a, and b given, to find C. What are the circular parts ?

Which is the middle part? Are the extremes adjacent or opposite?

5. Having B and C given, to find b. What are the circular parts ?

Which is the middle part ? Are the extremes adjacent or opposite ?

76

SPHERICAL TRIGONOMETRY.

G. What are the opposite extremes when b is the middle part?

What the adjacent extremes ? Which are the opposite and which the

adjacent extremes when c is the middle part? When comp B is the

middle part? When comp C is the middle part? When comp a is

the mi Idle part ?

7. What part is middle part to comp c and c as adjacent ex-

tremes? As opposite extremes?

Ans., I, and none.

8. In Fig. 43, M "being the right angle,

what are the circular parts ? Given and m t

to find o. What are the circular parts under

consideration ? Are the extremes adjacent or

opposite ?

Fie. 43.

9. What are the opposite extremes to comp

O ? What the adjacent ? To comp m ? To o

To n?

NAPIER'S RULES.

118. Hule I. Prop. In any right angled spherical triangle,

the sine of the middle part equals the pro-

duct of the cosines of the opposite ex-

tremes.

DEM. In the spherical triangle BAG, right

angled at A, taking b, c, comp B, comp C, and

comp a in succession as middle parts, we are to

prove that

sin b = cos (comp a) x cos (comp B), or sin b = sin a sin B ; (1)

sin c cos (comp a} x cos (comp C), or sin c sin a sin C ; (2)

sin (comp B) = cos b x cos (comp C), or cos B = cos b sin C ; (3)

sin (comp C) = cos c x cos (comp B), or cos C = cos c sin B (4)

sin (comp a) = cos b x cos c, or cos a = cos b cos c. (5)

To demonstrate these relations, let O be the centre of the sphere, and draw

the radii OA, OB, and OC. The angles BOC, AOC, and AOB, are measured

respectively by , b, and c, the sides of the triangle ; hence these angles at the

crntre and their measuring arcs may be used interchangeably. From one of

,

NAPIER S RULES.

77

;he oblique angles, as C, let fall a perpendicular upon the radius OA. From

the foot of this perpendicular draw DE perpendicular to OB, and join C and E.

Now CDE is a right angle (PAKT II., 426), CE is perpendicular to OB (PAKT II.,

399), and DEC is equal to angle B of the triangle (PAKT II., 558).

CD = sin b, OD = cos b, CE = sin a, and OE = cos a.

From the triangle CDE, right angled at D, we have

CD = CE x sinCED, or sin b = sin sin B. (1)

Generalized, (1) becomes, The sine of either side about the right angle = the

sine of the hypotenuse into the sine of the angle opposite the side. Hence, from

analogy to (1), we may write

sin c = sin a sin C. (2)

Or (2) may be proved in the same manner as (1),

by letting fall from B a perpendicular upon OA,

from its foot drawing DE perpendicular to OC,

and joining E and B. Then BD sin c, OD

ccs c, BE = sin , OE = cos a, and angle BED =

angle C. From the triangle BDE, we have

BD = BE x sin BED, or sin c = sin a sin C. (2)

To prove (3), we have from the triangle CDE,

Fig. 44,

cos CED =

ED

CE

or cos B =

ED

sin a'

But from triangle OED, right angled at E, ED = OD x sin DOE = cos b sin e =

[from (2)J, cos b sin a sin C. Substituting this value of ED, we have

cos b sin a sin C

cos B = r = cos b sm C. (3)

sin a

We may write (4) from (3) by analogy, as (2) was from (1) ; or, better, let the

student produce it from Fig. 45, as (3) was produced from Fig. 44.

Finally, to produce (5), consider the triangle ODE, in either figure, right

angled at E. This gives

OE=OD x cos DOE, or cos a = cos b cos c. (5)

119. Rule 2. Prop. In any right angled spher-

ical triangle, the sine of the middle part equals the pro-

duct of the tangents of the adjacent extremes.

DEM. In the spherical triangle BAG, right angled at A, taking

b, c, comp B, comp C, and comp a, in succession as middle parts,

we are to prove that,

FIG. 46.

78 SPHERICAL TRIGONOMETRY.

sin b tan c x tan (comp C), or sin b = tan c cot C ; (1)

sin c = tan b x tan (comp B), or sin c = tan b cot B ; (2)

sin (comp B) = tan c x tan (comp a), or cos B = cot a tan c ; (3)

sin (comp C) = tan b x tan (comp a), or cos C = cot a tan b \ (4)

sin (comp a) = tan (comp B) x tan (comp C), or cos a = cot B cot C (5)

Taking the formula of RULE 1st, and in the second member of each substitu-

ting the value of each factor as found in some other of the set, we readily

write the following

sin c cos C sin e cos C

sm 6 = sm a sin B = ^ = x -; = = tan c cot C ; (1)

sin C cose cose sm C

sin b cos B sin b cos B

sin c = sin a sin C = -= ? = 7 x -r = = tan b cot B ; (2)

sin B cos b cos b sin B

cos a sin c cos a sin c

cos B = cos b sin C = : = x = cot a tan c ; (3)

cos c sin a sm a cos c

cos a sin b cos a sin b

cos C = cos c sin B = r : = - x r = cot tan b ; (4)

cos 6 sin a sin a cos b

cos B cos C cos B cos C

cos a = cos b cos c = . = ^ =. x -r ^ = cot B cot C. (5)

sin C sin B sin B sin C

Q. E. D.

120. Sen. 1. It will be a good exercise for the student to demonstrate

RULE 2d from the an-

E. nexed figures, as RULE 1st

was from Figs. 44 and 45.

The 5th is not as readily

obtained from the figure

as the others. The stu-

dent may trace the fol-

lowing relations, some in

one figure, and some in

FIG. 47. FIG. 48. the other.

-. , -,

OE sec e AE tanc AE

sin c sin b sin c *.*/-

= . ; ; whence, cot B cot C = - = cos b cos c. .'. cos a = cot B cot C.

tan 6' tan 6 tanc

121. Sen. 2. It is of much importance, especially for the purposes of

Spherical Astronomy, that the student observe that the relations expressed in

the above formula, and in fact all the relations between the sides and angles of

spherical triangles, are also the relations between the facial and diedral angles

of triedrals. Thus, if a, b, and c represent the facial angles, and A, B, and C the

opposite diedrals, all these relations can be established, and in exactly the same

manner as above, without any allusion to the spforical triangle. [The student

should do it.J

DETERMINATION OF SPECIES. 79

DETERMINATION OF SPECIES.

In the solution of spherical triangles the determination of

the species of a part sought becomes of essential importance, since

any part of such a triangle may have any value between and 180.

Hence, when we have learned the numerical value of any function of

a part, we have yet to determine whether the part itself is less or

greater than 90, i. e., the species of the part. This may always be

effected by some one of the following propositions.

123. Prop. If the part sought is found in terms of its cos,

tan, or cot, its species can be determined by the algebraic signs of the

functions in the formula used.

DEM. In each of the formula arising from the application of Napier's rules,

there are three functions, the arcs corresponding to two of which are always

known, hence the algebraic signs of their functions are known, and the signs

of these two determine the sign of the third or unknown function. Now, when

a cos, tan, or cot is + , the corresponding angle is less than 90 (if less than

180) ; and when one of these functions is , the corresponding angle is greater

than 90 ; i. e., in a triangle, it is between 90 and 180.

124. When the part sought is found in terms of its sine, the

species cannot be determined by the signs of the formula, since the

part being less than 180 its sine is always +. The three following

propositions dispose of such cases.

125. Prop. An oblique angle of a right angled spherical tri-

angle and its opposite side are always of the same species.

DEM. From Napier's first rule we have, cos B = cos b sin C. But sin C is

necessarily + ; therefore, cos B and cos b always have the same sign, and B and

b are of the same species. In like manner, we see from cos C = cos c sin B, that

C and c are of the same species. Q. E. D.

126. Prop. When the hypotenuse of a right angled spherical

triangle is less than 90, the other two sides (and consequently thff

80 SPHERICAL TRIGONOMETRY.

oblique angles) are of the same species with each other. But when the

hypotenuse is greater than 90, the other two sides (and consequently

the oblique angles) are of different species from each other.

DEM. From Napier's first rule we have, cos a = cos b cos c. Now, if

a < 90, cos a is + ; hence cos b and cos c must have like signs, and b and c be

both less or both greater than 90 But if a > 90 (and less than 180, as it is),

cos a is ; hence, cos b and cos c must have different signs, and b and c be one

greater and the other less than 90. Finally, since the oblique angles are of the

same species as their opposite sides, they are of the same species with each other

when a < 90, and of different species from each other when a > 90.

. l?rop When a side and its opposite angle are given in a

right angled spherical triangle, there is NO solution if the sine of the

side is greater than the sine of its opposite angle ; there is ONE

solution and the triangle is bi-rectangular if the sine of the side

equals the sine of its opposite angle ; and there are TWO solutions if

the sine of the side is less than the sine of its opposite angle.

DEM. We have sin b = sin a sin B, or sin a = . = . Now, sin b > sin B

sm B

makes sin a > 1, which is manifestly impossible. Sin b sin B makes b = B,

since they are of the same species. But when the arc included by the sides of

an oblique angle of a right angled spherical triangle is equal to the angle, the

vertex of the angle is the pole of the arc. Hence, in this case the other sides of

the triangle are each 90. Finally, if sin b < sin B, a has two values, one

greater and the other less than 90. Hence there are two triangles.

128. Sen. These relations between an angle and its opposite side may be

observed directly from a figure. When B < 90, the measure of it, that is

b = B, is the greatest included perpendicular which

can be drawn to one side of the lune BAB'C. Hence,

in this case, b cannot exceed B, which implies that

sin b cannot be > sin B, as when the arcs are less

than 90, the greater arc has the greater sine. If

sin b = sin B, b = B, and BA = BC = 90, and the

side b can occupy but one position in the lime, thus

giving rise to but one triangle BAG, which satisfies

the conditions (or two equal triangles BAG and

B'AC). If b < B, which, when B is less than 90?

implies that sin b < sin B, the side can occupy two positions in the lune, b' and

b" giving rise to two triangles, BA'C' and BA"C", both of which fulfill the

conditions.

EXERCISES IN SOLVING RIGHT ANGLED TRIANGLES.

81

Again, when B > 90, the measure of it, i. e., b = B, is the least included

perpendicular that can be drawn to one side of the

lune. Hence, in this case we cannot have b < B,

which implies that sin b cannot be > sin B, since the

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