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B



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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