In the right angled triangle ADC there are given
A and 6, whence AD, P, and angle ACD, can be
computed. Then passing to triangle CDB, we
know P, and DB (since we have c given and
have computed AD). Hence, we can compute B, FIG. 18.
, and DCB. Thus, the parts of ACB become known .... When the given
parts are two angles and the included side, find the third angle by taking the
supplement of the two given. Let fall a perpendicular from one extremity of
the given side upon the opposite side. The two right angled triangles thus
formed can then be computed. Thus, if A, 6, and C are given, having found B,
let fall CD. The triangle ACD has the angle A and side b known, whence its
parts can be computed. Having computed P we can pass to the triangle CDB,
ind knowing P and B, can compute it. Thus the parts of ACB become known.
66 PLANE TRIGONOMETRY.
2nd. When two of the given parts are adjacent and one separated ; i. e , when
two angles and a side opposite one are given ; or two sides and an angle oppo
site one are given. The first of these cases is virtually the same as the last given.
To solve the other, let fall a perpendicular from the angle between the given sides,
and two right angled triangles will be formed which can readily be computed.
Thus a, 6, and A being given, and CD let fall from C, the triangle ACD can
first be computed, and then CDB. This is the ambiguous case, but it is easily
determined. Having computed P, if the given side a is less than P there is no
solution ; if = to P, one solution (a right angled triangle) ; if a > P and < Z>, there
sue two solutions, i. e., it will go in between CD and AC, and also beyond CD ; if
a > P and also > b there is only one solution, as it will not go in between CD
and AC.
3rd. Wlien the three given parts are all separated from each other. This is the
case in which the three sides are given to find the angles. It is readily solved
by letting fall a perpendicular from the angle opposite the greatest side, upon
that side, as CD upon AB. Then compute the segments AD (which call m), and
DB (n), from the following relation (PART II, Ex. 12, page 162) :
m + n (or c) : b + a : : b a : m n.
Knowing m and n, the angles of the two right angled triangles ACD and
CDB can be computed, and these make known the angles of ACB. Q. E D
[NOTE. A few additional examples are here given which the pupil can use
to illustrate the theory presented in (91). If more are needed the preceding
can be used : these may also be used to apply the methods before given. Again,
a veiy great variety and number of examples may be made from these by as
signing different parts as known.]
EXERCISES.
1. In a plane triangle BDF, the side BF is 123.75, DF 500, and the
included angle F 120. Find the angles B and D, and the side BD.
B = 49 12' 4"; D = 10 47' 56"; BD = 572.006.
2. In a plane triangle ABC, the angle A is 70 21', the angle B
54 22', and the side BC 125. Find the angle C, and the sides AB
and AC.
C = 55 17' ; AB = 109.1 ; AC = 107.88.
3. In a plane triangle ABC, the side AB is 98, the side BC 95.12.
and the angle C 33 21'. Find the angles A and B, and the side AC.
A = 32 14' 55"; B = 114 24' 5"; AC = 162.33.
4. In a plane triangle DAC, given AD = 450, AC = 309, and D =
27 50', to find the other parts.
C = 137 9' 36", or 42 50' 24" ; A = 15 0'24", or 109 19' 36";
OC = 171.36, 01 624 5.
SOLUTION OF OBLIQUE ANGLED PLANE TRIANGLES.
COMPUTATION.
It will give definiteness to the stu
dent's thought, if he first sketch the
9gure geometrically. Thus, lay off
D . 27 50', and taking AD = 450,
let fall the perpendicular AP.
1st. To compute p.
v=c sin D = 450 sin 27 50',
log 450 = 2.653213
log sin 27 50' = 9.669225
log p = 2.322438. .'. p  210.106.
57
FIG. 19.
Knowing p, we see by inspection that AC can lie in both the positions AC
nd AC', and hence that there are two solutions.
2nd. To compute C, from the triangle ACP, hi which d and p are now
inown.
p 210.106
Sin C d 309
log 210.106 = 2.322438
log 309 = 2.489958
log sin C = 9.832480. .'. C' = 42 50' 24", and C = 137 9' 36"
3rd. To find the angle A. DAC = 180  (D + ACD) = 180  164 59' 36" =
15 0' 24". DAC' = 180  (D + AC'D) = 180  70 40' 24" = 109 19' 36".
4th. To find DC. Compute DP and CP from the triangles APD and APC.
DP  CP = DC, and DP + CP = DC'.
5. In a plane triangle ABC, the side AB is 460, BC is 340, and AC
280. Find the tingles A, B, and C.
A = 47 23' 16"; B = 37 18' 31"; C = 95 18' 13".
6. The sides of a plane triangle are 40, 34, and 25 feet respect
ively ; required the angles.
38 25' 30", 57 41' 24", 83 63' 16".
7> The sides of a plane triangle are 390, 350, and 270 feet respect
' vely ; required the angles.
42 22' 06", 60 52' 33", and 76 45' 21".
8. Given two sides of a plane triangle 450 and 540, and the in
cluded angle 80, to find the remaining parts.
Angles, 56 11', 43 49'; and the side, 640.08.
9. Given two sides of a plane triangle 76 and 109, and the in
cluded angle 101 30', to find the remaining parts.
Angles, 30 57' 30", 47 32' 30"; and the side, 144.8.
58
PLANE TTUGONOMETUY.
FUNCTIONS OF THE ANGLES OF A TRIANGLE IN TERMS OF THE
SIDES.
92. Prop. Any side of a plane triangle equals the sum of the
products of each of the other sides into the cosine of the angle which it
makes witli the first side.
DKM. In the first figure AB = AD + DB.
But AD = b cos A, and DB = a cos B. .'. c
b cos A + a cos B. In the second figure AB =
AD DB. But AD = b cos A, and DB =
a cos CBD = a ( cos CBA) = a cos B. ..
= b cos A (a cos B) = b cos A + a cos B.
In like manner, we have a = b cos C + c cos
B, and b a cos C + c cos A. Collecting and
P arranging.
(1) a = b cos C + c cos B ;
(2) & = a cos C + c cos A ;
(3) c = a cos B + b cos A. Q. E. D.
Fia. 20.
93. COR. The square of any side of a plane triangle equals the
sum of the squares of the other two, minus twice their rectangle into
the cosine of their included angle.
DEM. From (3) (92), we have, by transposing and squaring,
a* cos 2 B C* + b' 1 cos* A 2bc cos A ; and
from (85) a? sin 2 B = & 2 sin 2 A.
Adding,
a* = c* + b* 2bc cos A.
In like manner, & 2 = a 2 + c 2 2ac cos B ;
and c* = a 2 + & 2 2ab cos C. Q. E. D.
94. Sen. These formula affor J another means for finding the angles of a
plane triangle when the sides are given. Thus,
(li) cosA = "
(2i) cos B =
2bc
These formula give directly the natural cosines of the angles in terms of
the sides. To adapt them to logarithmic computation, we transform them as
follows :
* + c *  a 2
Subtracting each member of (li) from unity, 1  cos A = 1 
2bc
Zbc
SOLUTION OP OBLIQUE ANGLED PLANE TRIANGLES. 59
But 1 cos A = 2sin*iA (57, O) ; and letting p = a + b + c, \(a + b c)
%p Cy and i( + c b) = $p b. Whence, substituting, 2sm*iA =
(1.) sin iA = 4~ c ; ~. In like manner,
00
(2,) sin *B = .P'gl^L. and
In a manner altogether similar, by adding each member of (1) to unity, and
reducing, we get
(1.) cos A = A ~ ; and from (2.) and (3.),
oc
(2.)
,) by (1 3 ), (&) by (2 3 ), and (3 a ) by (3,), we have.
(1 4 ) tan A :
12 ^ tan IB 

EXERCISES.
[NOTE. In order to render these formula familiar, and to giye the student
exercise in apply ing formula, a few examples are appended. If necessary, any
which precede can be used.]
1. The sides of a plane triangle being 40, 34, and 25, finl the
angles.
SOLUTION. By natural functions.
Let the sides be represented by a, 6, and c in order, and the angles opposite
Dy A, B, and C ; then
Cos A = ?1 = 1156+6  1GO = .10647. ... A = 88* 53' IV.
60
PLANE TRIGONOMETRY.
There is no ambiguity in this case, since the cosine is +, and hence the angle
C90.
The same angle is found by logarithmic computation, thus :
a = 40
ft = 34 . .
p =99 . .
.log = 1.602060
.log = 1.531479
.log = 1.397940
. log = 1.995635
. log = 1.694605
log (ip  c) = 1.389166
log (kp  ft) = 1.190332
a. c. log 6 = 8.468521
a. c. log c = 8.602060
2)1.650079
%p a = 9.5 ..
IP _ &  15.5 . .
. log = 0.977724
. log  1.190332
1.825039
 c = 24.5 . . . log = 1.389166 log sin
= 9.825039.
.'. *A = 41 56' 38" and A = 83 53' 16".
In like manner the other angles may be found.
2. The sides of a plane triangle being 6, 5, and 4, find the
angles.
The angles are 82 49' 09", 55 46' 16", and 41 24' 35".
3. The sides of a plane triangle being 8601.5, 4082, and 7068, find
the angles.
The angles are 54 35' 12", 28 4' 44", and 97 20' 4".
4. The sides of a plane triangle being .5123864, .3538971, and
3090507, find the angles.
The angle opposite the last side is 36 18' 10".2.
AREA OF PLANE TRIANGLES.
95. Prop. The area of a plane triangle is equal to half the
product of any two sides into the sine of the included angle.
DEM. Let ABC be any plane triangle, and ft and
c any two sides with A as the included angle. From
the extremity of one of these two sides remote from
A, let fall a perpendicular p, upon the otbor side
Now,
Area ACB = %pc.
But, from ACD, p = b sin A. /. Area ACB =
IftcsinA. Q. B. D.
Pie. 21.
00. COR. Tli? area of a plane triangle is equal to the square root
of the continued product of half its perimeter into half its perinietei
minus each side separately.
AREA OF PLANE TRIANGLES. 61
DEM. From the proposition, and since sin A = 2sin A cos A. we have,
1A ' x
Q. B. D.
EXERCISES.
1. Given two sides of a plane triangle 125.81 and 57.65, and the
included angle 57 25'. Find the area.
Area = 3055.7.
2. Given the sides of a plane triangle 103.5 and 90, and the
included angle 100, to find the area.
Area = 4586.74.
3. How many square yards are there in a triangle whose sides aro
30, 40, and 50 feet?
4. Find the area of a triangle whose sides are 20, 30, and 40.
Area = 290.4737.
5. What is the area of a triangle whose sides are 30 and 40, and
their included angle 28 57' ?
Area = 290.427.
6. What is the number of square yards in a triangle, of which the
sides are 25 feet and 21.25 feet, and their included angle 45 ?
Area = 20.8694.
7. Find the area of a triangle in which two of the angles are 80
and 60 respectively, and the included side 32 feet.
Area = 679.33 square feet.
8. Find the area of a triangular field having one of its sides 45
poles in length, and the two adjacent angles, respectively, 70 and
69 40'.
Area = 1378.411 square poles.
9. Find the urea of a triangular piece of ground having two
angles respectively 73 10' and 90 50', and the side opposite the
latter 75.3 poles.
Area = 748.03 square poles.
PRACTICAL APPLICATIONS.
[NOTE. The following problems aro inserted, not as any part of a treatise
tipon the subject of trigonometry as pure science, but as affording the student
pood mental exercise, and valuable and interesting information.]
62 PLANE TRIGONOMETRY.
1. To find the length (in miles) of a degree of longitude at Ann
Arbor, Mich.
E SOLUTION. Let NESQ be a meridian section of
the earth, EQ the equatorial diameter, and EL the lati
tude of Ann Arbor, 42 16' 48".3. A degree of longi
tude at L is 3^5 of the circumference of the circle
J whose radius is LD. CL the radius of the earth at
this point = 3957.* Now in the right angled triangle
LCD, we have CLD = ECL  42 16' 48".3, and CL 
3957; whence, LD = CL x cos 42 16' 48".3, and LD
= 2927.6. .'. A degree = 51.1 miles. As a degree iu
longitude makes 4 minutes difference hi time, 51.1
Q
FIG. 22.
miles east or west on this parallel is equivalent to 4 minutes difference in time.
QUERY. How does it appear from the above solution that the length of a
degree of longitude varies as the cosine of the latitude?
2. To find the distance of a planet from the earth at any par
ticular time.
SOLUTION To render the problem as simple as possible, we will suppose two
observatories on the same meridian, at N,and
N'; and that when the planet P is on the same
meridian, the angles ZNP, and Z'N'P (the
zenith distances) are measured. With these
data and the radius of the earth, CN, CM',
known, the problem comes quite within the
scope of the present study. The process is
as follows: The arc NN' being known, the
angle NCN' is known. Then in the triangle
NCN', two sides and the included angle are
known, whence the other parts can be found.
Now, knowing the angles PNC, PN'C, and
CNN', CN'N, we can find the angles PNN',
PN'N. This affords sufficient parts of the triangle PNN' to determine the triangle.
and we find PN, or PN'. Finally, in the triangle PNC, we know PN, NC, and
the included angle ; whence the other parts can be computed. But PC is the
distance sought.
3. Suppose in case of the moon, the angles PNZ, and PN'Z', being
measured, are found to be respectively 44 54' 21", and 48 42' 57",
the distance between the points of observation N and N' is 92 14',
and the radius of the earth is 3956.2 miles; find the distance to the
moon.
Distance = 237,954.098 miles.
FIG. 23.
* The equatorial radius of the earth is 3962.8 miles ; but in consequence of the flattening in
the directi on of the polar diameter it is less here.
PRACTICAL PROBLEMS.
63
4. Required the height of a
hill D above a horizontal plane
AB, the distance between A and
B being equal to 975 yards, and
the angles of elevation at A and
B being respectively 15 36' and
27 29'.
B C
DC = 587.61 yards
5. Find the area of a regular hexagon, and also of a regular
octagon, whose sides are each 10 feet.
Areas, 259.8, and 482.84 square feet
6. Find the area of a regular pentagon, and also of a regular dec
agon, whose sides are each 12 feet.
Areas, 247.74, and 1107.96 square feet.
7 Wishing to know the length of a certain pond of water, 1
measured a line 100 yards in length, and at each of its extremities
observed the angles subtended by the other extremity and a couple
of trees at the extremities of the pond. These angles were, at one
end of the line, 32 and 98, and at the other, 37 and 118 ; what
was the length of the pond ?
Draw the horizontal line AB equal to 100;
make the angle BAD 32, BAG 98, ABC 37,
and ABD 118. The intersections of the lines
AC and BC, AD and BD, determine the extremi
ties of the pond; the straight line CD is the
length of the pond.
CD = 161.868 yards.
8. The distances AB, AC, and BC, between
the 'points A, B, and C, are known ; viz., AB =
800 yds., AC = 600 yds., and BC = 400 yds.
From a fourth point P, the angles ARC and BPC
are measured; viz., APC = 33 45'',
and BPC = 22 30'.
Required the distances AP, BP, and CP.
AP = 710.193 yds.
Distances, \ BP = 934.291 yds.
CP = 1042.522 yds.
64 PLANE TRIGONOMETRY.
BUG'S. Conceive the circumference passed through A, B, and P, and AD ;ind
DB drawn. In the triangle ADB, angle DAB the given angle DPB, and DBA
= APD. Hence, all the parts of triangle ADB can be found. Again, since UK
sides of the triangle ACB are given, its angles can be found. Then, since angK 1
CAB DAB = CAD, there are two sides and the included angle known in
triangle ACD ; whence angle ACD can be found. Thus we reach the triangle
ACP, in which there are now known AC and the angles.
9. From the top of a mountain, three miles high, the angle of
depression of a line tangent to the earth's surface is taken, and
found to be 2 13' 27". What is the diameter of the earth, considered
as a sphere ?
Ans. 7958.45 miles.
10. Taking the sun's mean apparent diameter as 32' 3".4, and his
distance from the earth 91,430,000 miles, show that, if his centre
were coincident with the earth's, his body would extend in all direc
tions nearly 200,000 miles beyond the moon. (See Ex. 3.)
Sun's diameter = 852,574 miles.
11. Assuming the height of the Great Pyramid to be 480 feet, how
far off may it be seen across the desert ?
Ans., 27 miles.
12. What was the perpendicular height of a balloon, when its
angles of elevation were 35 and 04, as taken by two observers on
the same level, at the same time, both on the same side of it, and
in the same vertical plane ; the distance between the two observers
being 880 yards ?
Ans., 935.757 yards.
13. Given two sides of a parallelogram 60 and 80, and a diagonal
100. Is this the longer or shorter diagonal ? What is the other ?
What are the angles of the parallelogram ?
14. A balloon being directly over one of two towns standing on
i v ,he same horizontal plane, at a distance of eight miles from each
ether, the angle of depression to the more remote town was observed
by the aeronaut to be 10. What was the height of the balloon ?
Ans., 1.41 miles.
15. The most recent observations make the sun's horizontal par
allax 8''.i)4, and the earth's equatorial radius 3962.8 miles. Show
that the distance of the sun from the earth is nearly as given in Ex.
10, instead of 95 millions of miles, as it has been heretofore con
sidered.
CHAPTER II
SPHERICAL TRIGONOMETRY.
INTRODUCTION.
PROJECTION OF SPHERICAL TRIANGLES.
97. To Project a Spherical Triangle on a plane surface
is to draw the triangle on that surface so that it will present the
same appearance to the eye, situated at a particular point, as when
drawn on the surface of a sphere.
98. The Simplest Method of projecting a spherical triangle
is to project it on the plane of one of its sides, the eye being supposed
situated in the axis of the sphere perpendicular to this plane, and at
an infinite distance from it. The plane is called the Plane of Pro
jection ; and its intersection with the sphere is called the Primitive
Circle, and is the base of the hemisphere on which the triangle is
conceived as situated.
99. Fundamental Propositions. 1st. WJien the parts of
a spherical triangle are each conceived as less than 180, any such
triangle can be represented on a hemisphere.
2d. The primitive Circle has its axis, and consequently its pole,
projected at its centre.
3d. The semicircumference of any circle of the sphere, perpendic
ular to the Primitive Circle, is projected in the chord representing
the intersection of the circles ; and, if the perpendicular circle he a
great circle, its semicircumference is projected in a diameter of the
Primitive Circle.
i
SPHERICAL TRIGONOMETRY.
ILL'S. These propositions are direct consequences of the fundamental con
ception. Thus, let ABA'B' represent the base of the hemisphere on which the
triangle is conceived as situated. This is the
Primitive Circle, and the ey& is supposed situated
at an infinite distance, and in a line perpendicular
to the plane of the paper at P. The pole of the
Primitive Circle being in this line is projected
(seen as) at P. As all great circles perpendicular
to the Primitive Circle pass through its pole and
include its axis, the eye is in all such planes, and
any lines of these planes, as the semicircumfer
ences of the great circles in which they intersect
the sphere, are projected (appear to the eye) as
diameters of the Primitive Circle. Moreover,
since the eye is at infinity, it is to be conceived as in
the plane of any small circle which is perpendicular to the primitive, and which
is therefore projected in a chord, as CC'.
PROJECTION OF BIGHT ANGLED SPHERICAL TRIANGLES.
100. JProb. 1. To project a right angled spherical triangle on
the plane of one of its sides, when the two sides about the right angle
are given*
SOLUTION. Let the angles of the triangle be represented by A, B, and C, A
being the right angle. Let the sides opposite these angles respectively be rep
resented by a, b, and c ; whence b and c are the
given sides. Draw the primitive circle and the
diameters BB', NN' at right angles to each other.
Prom B lay off BA c.f Let the right angle be
at A ; whence the side b is perpendicular to the
primitive circle, and projected in the diameter AA'.
To project the vertex C, conceive the semicircum
ference, of which A A' is the projection, to revolve
on AA' until it falls upon the semicircumference
A'BA, then will the point C fall at d. Hence make
A$ = b. In like manner revolving the semicir
cumference, of which AA' is "he projection, until
it falls upon A'B'A, the point C will foil at d'.
Hence make M f = b. The point C will de
scribe the semicircumference of a small circle
perpendicular to the primitive circle, and whose projection is dd'. Now, as the
* In this treatise the discussions embrace only such triangles as have each part less than
180.
t For the purposes for which we shall use these projections, an arc can be laid off with
sufficient accuracy by observing its relation to 90, 60, 30 a , or some aliquot part of the circum
ference, which is readily obtained on geometrical principles.
FIG. 25.
PROJECTION OF RIGHT ANGLED SPHERICAL TRIANGLES.
67
projection of the vertex of the triangle C is at the same time in AA' and dd', it
must be at their intersection. Finally, the hypotenuse a is projected in a curve
passing through B, C, and B', since two great circles intersect at the extremities
of a diameter. This curve is really an ellipse, but for our present purpose it
may be considered as the arc of a circle passing through B, C, and B'. There
fore, BAG is the projection required.
QUERIES. Will a solution of this problem be possible for all values of b and
c ? How does it appear from the projection ?
EXAMPLES.
1. Having given I = 110, and c = 60, to
project the triangle. See Fig. 26.
2. Having given I = 50, and c = 130, to
project the triangle.
3. Having given I = 90, and c = 30, to
project the triangle.
4. Having given b = 90, and c 90, to
project the triangle.
101. I*rob. 2. To project a right angled spherical triangle
when the hypotenuse and one side are given.
SOLUTION. Using the common notation, let c represent the known side.
Drawing the primitive circle and the conjugate* diameters BB', NN', lay off
BA = c, and draw the diameter AA'. The projec
tion of b will lie in AA', and the projection of the
vertex C will fall somewhere in this line. Now
the arc a lies in a semicircumference passing
through B and B'. Conceive this semicircumfer
ence to revolve on BB', as an axis, till it coincides,
first, with BNB', and then with BN'B'. The point
C will trace the semicircumference of a small
circle perpendicular to the primitive circle, and
whose projection is dd'. Hence, making Bd = Bd'
= a, and drawing dd' ', the projection of the vertex
C lies at the same time in AA' and dd', and is there
fore at their intersection. Passing an arc of a
circle (strictly an ellipse) through BCB', we have
ABC, the projection desired.
FIG. 27.
* Two diameters of a circle which are at right angles to each other are called Conjugate
Uiameters.
68
SPHERICAL TRIGONOMETRY.
QUERIES. Will a solution of this problem be possible for all values of a and
c ? If a had been greater than c in the above case, would the solution have
been possible ? Will dd' and AA' always intersect, whatever may be the relative
values of a and c ?
EXAMPLES.
1. Having given c = 75, and a = 64, to project the triangle.
2. Having given c = 45, and a = 136, to
project the triangle.
3. Having given b = 110, and a = 85, to
project the triangle. See Fig. 28.
4. Having given I = 110, and a = 120,
to project the triangle.
5. Having given c = 90, and a = 75, or
a = 120, to project the triangle.
Pio. 28.
QUERY. If one side of a right angled spherical
triangle is 90, what must the hypotenuse be ?
Why?
102. Prob. 3. To project a right angled spherical triangle
when an oblique angle and the hypotenuse, or the oblique angle and
the adjacent side are given.
SOLUTION. Draw the primitive circle and the conjugate diameters BB', NN'
as usual. To construct the given angle B, we observe that this angle is
measured by an arc of the great Circle which is
projected in NN'. Hence, lay off Nd = Nd' = B ,
and draw M ; then is NO the projection of the
arc which measures B, and the projection of
a lies in the arc passing through BO3'.* Having
the angle B projected, if the hypotenuse a is the
other given part, find the projection of C by
taking Be = Be' = a, and drawing ee'. Com
plete the projection by drawing AC through P.
When the adjacent side c is given, take BA = c,
and draw AC as before. [The student will be
able to give the reasons.]
Fle> 29. 103. Sen. As OP is the cosine of the angle
ABC, the point O may be found by measuring
* As has been remarked, this arc is really a semiellipse. This fact, together with the
method of constructing the semiellipse, and thus getting the correct projection of the hypot