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X \ii«.«+,'an
HARVARD
COLLEGE
LIBRARY
THE GIFT OF
Miss Ellen Lang Wentworth \
of Exeter J New Hampshire
3 2044 097 045 4^
-tL^^
PLANE AMD SPHEEICAL
TRIGONOMETRY.
BY
G. A. WENTWOKTH, A.M.,
rB0FE8S0& OP KATHBKATIC8 IH PBILLIFS EXETB& ACADEMY.
BOSTON:
PUBLISHED BY GINN, HEATH, * CO.
1884.
EduucT \C.^. S^4.^1"I
HARVARD COLLEGE LIBRARY
GIFT OF
MISS ELLEN L. WENTWORTH
MAY 8 1939
Entered, according to Act of Congress, in the year 1882, by
G. A. WENTWORTH,
in the office of the Librarian of Congress, at Washington.
GiMN, HsATH, & Co., Printers:
J. S. Gushing, Supt., i6 Hawley Street,
Boston.
PREFACE.
r preparing this work the aim has been to fomiBh jnst so much
of Trigonometry as is actaally taught in oar best schools and
colleges. Consequently, development of functions in series and all
other investigations that are important only for the special student
have been omitted. The principles have been unfolded with the
utmost brevity consistent with simplicity and clearness, and inter-
esting problems have been selected with a view to awaken a real
love for the study. Much time and labor have been spent in devis-
ing the simplest proo& for the propositions, and in exhibiting the
best methods of arranging the logarithmic work.
The author is under particular obligation for assistance to G. A. Hill,
A.M., of Cambridge, Mass., to whom is chiefly due whatever value
the Trigonometry possesses.
a A. WENTWORTH.
Phillips Exeteb Acadext,
September, 1882.
COKTEl^TS.
PLANE TBIOONOMETBY.
CHAPTER I. FxTFcnoFs op Acute Aholes:
Definitions, 1 ; representation of fdnctions by lines, 7; changes in
the functions as the angle changes, 9 ; fdnctions of complementary
angles, 10 ; relations of the fdnctions of an angle, 11 ; formnlas for
finding all the other fdnctions of an angle when one function of the
angle is given, 13 ; functions of iS**, 30*», 60®, 16.
CHAPTER II. Thb Right Tbiaitole :
Solution : Case I., when an acute angle and the hypotenuse are
given, 16 ; Case II., when an acute angle and the opposite leg are
given, 17; Case III., when an acute angle and the adjacent leg
are given, 17 ; Case IV., when the hypotenuse and a leg are given,
18 ; Case V., when the two legs are given, 18 ; general method of
solving the right triangle, 19; area of the right triangle, 20; the
isosceles triangle, 24 ; the regular polygon, 26.
CHAPTER III. GOKIOMETET:
Definition of Goniometry, 28; angles of any magnitude, 28; gen-
eral definitions of the functions of angles, 29 ; algebraic signs of the
functions, 31 ; functions of a variable angle, 32 ; Ainctions of angles
larger than 360®, 34 ; formulas for acute angles extended to all angles,
35 ; reduction of the functions of all angles to the functions of angles
in the first quadrant, 38 ; functions of angles that differ by 90®, 40 ;
functions of a negative angle, 41 ; functions of the sum of two angles,
43 ; functions of the difference of two angles, 45 ; functions of twice
an angle, 47 ; functions of half an angle, 47 ; sums and differences of
functions, 48.
VI TRIGONOMETRY.
CHAPTER IV. The Oblique Trianole :
Law of sines, 50 ; law of cosines, 52 ; law of tangents, 52 ; Solu-
tion : Case I., when one side and two angles are given, 54 ; Case II.,
when two sides and the angle opposite to one of them are given, 56 ;
Case III., when two sides and the included angle are given, 60 ; Case
IV., when the three sides are given, 64 ; area of a triangle, 68.
SPHERICAL TRIGONOMETRY.
CHAPTER V. The Right Spherical Triangle:
Introduction, 71 ; formulas relating to right spherical triangles,
73 ; Napier's rules, 76. Solution : Case I., when th^two legs are given,
78 ; Case II., when the hypotenuse and a leg are given, 78 ; Case III.,
when a leg and the opposite angle are given, 79 ; Case IV., when a
leg and an adjacent angle are given, 79 ; Case V., when the hypote-
nuse and an oblique angle are given, 79 ; Case Vl., when the two
oblique angles are given, 79 ; solution of the isosceles spherical tri-
angle, 84 ; solution of a regular spherical polygon, 84.
CHAPTER VI. The Oblique Spherical Triangle:
Fundamental formulas, 85 ; formulas for half angles and sides, 87 ;
Gauss's equations and Napier's analogies, 89. Solution ; Case I., when
two sides and the included angle are given, 91 ; Case II., when two
angles and the included side are given, 93 ; Case III., when two sides
and an angle opposite to one of them are given, 95 ; Case IV., when
two angles and a side opposite to one of them are given, 97; Case V.,
when the three sides are given, 98 ; Case VI., when the three angles
are given, 99 ; area of a spherical triangle, 101.
CHAPTER VII. Applications op Spherical Trigonometry:
Problem, to reduce an angle measured in space to the horizon, 104 ;
problem, to find the distance between two places on the earth's sur-
face when the latitudes of the places and the difference of their lonjgi-
tudes are known, 105; the celestial sphere, 105 ; spherical co-ordinates,
108; the astronomical triangle, 110; astronomical problems, 111-
114.
PLAlifE TRIGOIS"OMETRT.
CHAPTER I.
TBIOONOMETRIC FUNCTIONS OF ACUTE ANGLES.
§ 1. Definitions.
The Bides and angles of a plane triangle are so related that
any three given parts, provided at least one of them is a side,
determine the shape and the size of the triangle.
Geometry shows how, from three such parts, to construct the
triangle and find the values of the unknown parts.
Trigonometrj shows how to compute the unknown parts of a
triangle from the numerical values of the given parts.
Geometry shows in a general way that the sides and angles
of a triangle are mutually dependent. Trigonometry begins
by showing the exact nature of this dependence in the right
triangle^ and for this purpose employs the ratios of its sides.
Let MAI^(Fig. 1) be an acute angle. If from any points
J5, JD, Fy in one of its sides
perpendiculars BC^ DE, FO^
are let fall to the other side, then
the right triangles ABC, ADE^
AFO, thus formed have the
angle A common, and are there-
fore mutually equiangular and
similar. Hence, the ratios of
their corresponding sides, pair by
pair, are equal. That is.
AC _AE_^AO,
AB AD AF'
AO^AE^AO, etc
BC DE FO' '
TEIGONOMETRY.
Hence, for every value of an acute angle A there are certain
numhers that express the values of the ratios of the sides in
all right triangles thai have this acute angle A,
There are altogether six different ratios :
I. The ratio of the opposite leg to the hypotenuse is called
the Sine of -4, and is written sin A,
II. The ratio of the adjacent leg to the hypotenuse is called
the Cosine of -4, and written cosJ..
III. The ratio of the opposite leg to the adjacent leg is
called the Tangent of -4, and written tan -4.
IV. The ratio of the adjacent leg to the opposite leg is
called the Cotangent of A, and written cot -4.
V. The ratio of the hypotenuse to the adjacent leg is called
t^e Secant of A, and written sec -4.
VI. The ratio of the hypotenuse to the opposite leg is called
the Cosecant of -4, and written esc A.
In the right triangle ABC
(Fig. 2) let a, 5, c denote the
lengths of the sides opposite to
the acute angles -4, JB, and the
right angle (7, respectively, these
lengths being all expressed in
terms of a common unit. Then,
sm
j__a_ opposite leg
c hypotenuse
tan^ = ^ = ^E25?^^^i?S.
b adjacent leg
j_c _ hypotenuse
b adjacent leg
ji_i __ adjacent leg
c hypotenuse
J __b adjacent leg
CSC -4 =
a opposite leg
c _ hypotenuse
a opposite leg
These six ratios are called the Trigonometrio Fimctioiifl of the
angle A,
TRIGONOMETBIO FUNCTIONS.
Exercise I.
1. What are the functions of the other acute angle -B of the
triangle ^^a (Fig. 2)?
2. Prove that if two angles, A and -B, are complements of
each other (i.e., iiA + ]3 = 90^), then,
8in-4 = cos5, tan-4 = cot5, sec^ = C8c^;
cos A = sin Bf cot A = tan -B, esc A = sec J3.
3. Find the values of the functions of A, if a, 5, c respec-
tively have the following values :
(i.) 3, 4, 5. (iii.) 8, 15, 17. (v.) 3.9, 8, 8.9.
(ii.) 5, 12, 13. (iv.) 9, 40, 41. (vi.) 1.19, 1.20, 1.69.
4. What condition must be fulfilled by the lengths of the
three lines a, 5, c (Fig. 2) in order to make them the sides of
a right triangle ? Is this condition fulfilled in Example 3 ?
5. Find the values of the functions of ^, if a, J, c respec-
tively have the following values :
(i.) 2mn, Tnf — rf, m^-^-n^ (iii.) pqr^ qrs, rsp,
(ii.) -2^, x+y, =^±t. (iv.) ran, r^, vl.
x — y x — y pq sq pa
6. Prove that the values of a, J, c, in (i.) and (ii.), Example
5, satisfy the condition necessary to make them the sides of
a right triangle.
7. What equations of condition must be satisfied by the
values of a, i, c, in (iii.) and (iv.). Example 5, in order that the
values may represent the sides of a right triangle?
Compute the functions of A and J3 when,
8. a = 24, i = 143. 11. a = VfT?, J = V2^
9. a = 0.264, c = 0.265. 12. a = Vf+pq, c = p + q.
10. J = 9.5, . = 19.3. 18. J = 2y^, c=p + q.
TRIGONOMETRY.
Compute the functions of A when,
14. a = 25. 16. a-\-l = \c,
15. a = f^. 17. a-i = ~.
18. Find a if sin ^ = f and c = 20.5.
19. Find h if cos -4 = 0.44 and c = 3.5.
20. Find aiftan^ = V- and J = 2^.
21. Find h if cot JL = 4 and a = 17.
22. Find c if sec ^ = 2 and h = 20.
23. Find c if csc^ = 6.45 and a = 35.6.
Construct a right triangle ; given,
24. c = 6, tan^=f. 26. J = 2, sin ^ = 0.6.
25. a=3.5, cos-4=|. 27. 5 = 4, esc ^ = 4.
28. In a right triangle, c = 2.5 miles, sin^ = 0.6, cos -4 =
0.8 ; compute the legs.
29. Construct (with a protractor) the A 20**, 40°, and 70**;
determine their functions by measuring the necessary lines,
and compare the values obtained in this way with the more
correct values given in the following table :
20°
40°
70°
sin
COi
tan
co<
«ec
ac
0.342
0.643
0.940
0.940
0.766
0.342
0.364
0.839
2.747
2.747
1.192
0.364
1.064
1.305
2.924
2.924
1.556
1.064
30. Find, by means of the above table, the legs of a right
triangle if ^ = 20^ c = 1 ; also, if ^ = 20^ c = 4.
31. In a right triangle, given a = 3 and c = 5; find the
hypotenuse of a similar triangle in which a = 240,000 miles.
32. By dividing the length of a vertical rod by the length
of its horizontal shadow, the tangent of the angle of elevation
of the sun at the time of observation was found to be 0.82.
How high is a tower, if the length of its horizontal shadow at
the same time is 174.3 yards?
TEIQONOMETBIC FUNCTIONS.
8
f:
^fi
T
^
\
vj
u
A
§ 2. Repbesentation op Functions by Lines.
The functions of an angle, being ratios, are numbers; but
we may represent them by lines if we first choose a unit of
length, and then construct right triangles, such that the de-
nominators of the ratios shall be equal to this unit. The most
convenient way to do this is as follows :
About a point (Fig. 3) as a
centre, with a radius equal to one
unit of length, describe a circle
and draw two diameters A A' and
JBJff perpendicular to each other.
The circle with radius equal to
1 is called a unit circle, AA' the
horizonialt and JSJff the vertical
diameter.
Let AOP be an acute angle,
and let its value (in degrees, etc.)
be denoted by x. We may regard
the Z X SA generated by a radius OP that revolves about
from the position OA to the position shown in the figure;
viewed in this way, OP is called the moving radius.
Draw PM± to OA. In the rt. A 0PM the hypotenuse
0P= 1 ; therefore, sin a: = PM\ cos a: = OM.
Since PMis equal to 01^^ and OiV is the projection of OP
on Pjff^ and since OMia the projection of OP on AA\ there-
fore, in a unit circle^
sijia: = projection of moving radius on vertical diameter;
cos a: = projection of moving radius on horizontal diameter.
Through A and £ draw tangents to the circle meeting OP,
produced in T and S, respectively; then, in the rt. A OAT,
the leg OA = 1, and in the rt. A OPS, the leg OP = l] while
the Z OSP = Zx (why ?). Therefore,
riff. a.
ia,TLx = AT; Becx = OT; cotx = £S] c8cx = 0S.
8 TBIGONOMETEY.
These six line values (as they may be termed) of the func-
tions are all expressed in terms of the radius of the circle as a
unit ; and it is cle^x that as the angle varies in value the line
values of the functions will always remain equal numerically
to the ratio values. Hence, in studying the changes in the
functions as the angle is supposed to vary, we may employ the
simpler line values instead of the ratio values.
Exercise II.
1. Eepresent by lines the fanctions of a larger angle than
that shown in Fig. 3.
2. Show that sin a: is less than tana:.
8. Show that seca? is greater than tana?.
4. Show that esc a; is greater than cot a?.
Construct the angle x if,
5. tana; = 3. 7. cosa? = -J-. 9. sina? = 2cosa?.
6. CSC a; = 2. 8. sina: = cosa:. 10. 4sina? = tanar.
11. Show that the sine of an angle is equal to one-half the
chord of twice the angle.
12. Find a; if sin a; is equal to one-half the side of a regular
inscribed decagon.
13. Given x and y^x+y being less than 90®) ; construct the
value of sin {x+y) — sin x,
14. Given x and yix + y being less than 90°) ; construct the
value of tan (x + y) — sin (a: + y) + tana? — sin a?.
Given an angle x ; construct an angle y such that,
15. siny = 2sina;. 17. tany = 3tana;.
16. cosy = -J-cosa?. 18. secy = csca?.
19. Show by construction that 2 sin -4 > sin 2 A,
20. Given two angles A and J5(A + J3 being less than 90**) ;
show that sin ( J. -f 5) < sin A + sin JB.
21. Given sina? in a unit circle ; find the length of a line
corresponding in position to sina? in a circle whose radius is r ?
22. In a right triangle, given the hypotenuse c, and also
sin J. = m, C08-4 = n ; find the "
TBIGOKOMETEIO PUHOTIONS.
FI8.4-
§ 3. Changes in the Functions as the Angle Changes.
If we suppose the A AOP^ or x (Fig. 4) to increase gradu-
ally by the revolution of the moving
radius OP about 0, the point P will
move along the arc AB towards B,
Twill move along the tangent AT
away from A, S will move along
the tangent BS towards P, and M
will move along the radius OA
towards 0.
Hence, the lines PM, AT, OT
will gradually increase in length,
and the lines OM, BS, 03, will
gradually decrease. That is,
As an acute angle increases, its
sine, tang&ni, and secant also in-
crease, while its cosine, cotangent, and cosecant decrease.
On the other hand, if we suppose x to decrease gradually,
the reverse changes in its functions will occur.
If we suppose x to decrease to 0®, OP will coincide with OA
and be parallel to BS. Therefore, PJf and J. Twill vanish,
OJf will become equal to OA, while BS and OS will each be
infinitely long, and be represented in value by the symbol oo.
And if we suppose x to increase to 90°, OP will coincide
with OB and be parallel to AT. Therefore, PJf and 08 will
each be equal to OB, OMbhS. P/?will vanish, while ATsjii
Twill each be infinite in length.
Hence, as the angle x increases from 0® to 90®,
sin X increases from to 1,
cos X decreases from 1 to 0,
tana; increases from to oo,
cot X decreases from oo to 0,
sec X increases from 1 to oo,
esc X decreases from oo to 1.
10
TRIGK)NOMETEY.
The values of the functions of 0** and of 90** are the limiting
values of the functions of an acute angle. It is evident, that
(disregarding the limiting values),
Sines and cosines are always less than 1 ;
Secants and cosecants are always greater than 1 ;
Tangents and cotangents have all values between and oo.
Kemabe. We are now able to underBtand why the sine, cosine, etc.,
of an angle are called /wnciicms of the angle. By d>, function of any mag-
nitude is meant another magnitude which remains the same so long as
the first magnitude remains the same, but changes in value for every
change in the value of the first magnitude. This, as we now see, is the
relation in which the sine, cosine, etc., of an angle stand to the angle.
§4. Functions of Complementary Angles.
The general form of two complementary angles is A and
90*'- A
In the rt. A ^5(7 (Fig. 5)
^ + ^ = 90°, hence ^ = 90° - A,
Therefore (§ 1),
sin A = co^B = cos (90° — A),
cos ^ = sin 5 = sin (90° — A\
tan -4 = cot 5 = cot (90° - A),
cot ^ = tan ^ = tan (90° - A\
sec -4 = CSC -B = esc (90° — A)^
V CSC J. = sec 5 = sec (90° — ^).
Therefore,
Each function of an acute angle is equal to the co-named
function of the com^plementary angle.
Note. Cosine, cotangent, and cosecant are sometimes called co-functions;
the words are simply abbreviated forms of complement's sine^ complement's
tangent, and complement's secant.
Hence, also.
Any function of an angle between 45° and 90° may be found
by taking the co-named function of the complementary angle
between 0° and 45°.
TEIGONOMETRIC FUNCTIONS. 11
EXEBCISE III.
1. Express the following functions as functions of the com-
plementaiy angle :
sinSO^. tan89^ csclSnC/. cot 82° 19'.
co845°. cotl5^ 00837^24'. C8c64°46'.
2. Express the following functions as functions of an angle
less than 45** :
sin 60°. tan67°. cscGQ^^. cot89° 59^.
cos 75°. cot 84°. cos85°39'. esc 45° 1'.
3. Given tan 30° =■- i V3 ; find cot 60**.
4. Given tan -4 = cot ^ ; find A,
5. Given cos ^ = sin 2-4 ; find A.
6. Given sin -4 = cos 2-4 ; find A.
7. Given cos A = sin (45° — J -4) ; find -4.
8. Given cot i ^ = tan A ; find A.
9. Given tan (45° + -4) = cot ^ ; find A.
10. Find A if sin -4 = cos 4 -4.
11. Find -4 if cot -4 = tan 8 A
12. Find -4 if cot -4 = tan nA,
§ 5. Kelations of the Functions of an Angle.
Since (Fig. 5) a^ + b^ = c*, therefore,
?+?- - ©■+©■-■
Therefore (§ 1), (sin Ay + (cos -4)* = 1 ;
or, as usually written for convenience,
Bin*A + co8»A = l. [1]
That is : The sum of the squares of the sine and the cosine of
an angle is equal to unity.
12 TEIGONOMETEY.
Formula [1] enables us to find the cosine of an angle when
the sine is known, and vice versa. The values of sin A and of
cos A deduced from [1] are :
sin ^ = Vl — co8*-4, cos-4 = Vl — sinfZ.
c,. a b a ^c a
Since — i.-. = ^x7 = T»
c c c o
therefore (§ 1), *»^A=Sl' [^^
That is : The tangent of an angle is equal to tfie sine divided
hy the cosine.
Formula [2] enables us to find the tangent of an angle when
the sine and the cosine are known.
Since -x- = l, -Xt = 1i and •=• X - = 1,
each a
therefore (§ 1), sin A X esc A = 1
cos A X sec A = 1
tanAxootA = l
[3]
That is : The sine and the cosecant of an angle, the cosine
and secant, and the tangent and cotangent, pair hy pair, are
reciprocals.
The equations in [3] enable us to find an unknown function
contained in any pair of these reciprocals when the other func-
tion in this pair is known. 9'-^ ^ 4^ — ^t« ' /V
Exercise IV.
1. Prove Formulas [1] - [3], using for the functions the line
values in unit circle given in § 3.
2. Prove that 1 + tan"^ = sec'-4.
!Eint. Divide the terms of the equation o* + 6* «= c* by J*.
3. Prove that 1 + cot'-4 = csc*-4.
cos -4.
4. Prove that cot A = --' — r*
sm A.
T&IGONOMETBIG TO|7CTI0NS. 13
§6. Application of Formulas [1]-[3].
Fonnulas [1], [2], and [3] enable us, when any one function
of an angle is given, to find all the others. A given value of
any one function, therefore, determines all the others.
Example 1. Given sin ^ = f ; find the other functions.
By[l], cos^ = Vr=|=V| = iV5.
By [21 tan^ = ?-i--V5 = -X— = — •
/LJ. ^^^ 8 3^^ 3 V5 V6
By [3], cot^ = ^, 8ec^ = -4^, csc^ = |
2 v5 2
Example 2. Given tan ^ = 3 ; find the other functions.
By [2], ^ = 8.
COS^
And by [1], sin*^ + cos*^ = 1.
If we solve these equations (regarding sin A and cos A as
two unknown quantities), we find that,
sin ^ = 3 VS» cos A = V^.
Tlienby[3], cot^ = |, secJ[ = VIO, csc^=iVIO.
Example 3. Given sec -4 = m ; find the other functions.
By [3], cos^ = l.
By [1], 8in^= JI31,=-^^l=lv^?:ri.
By [2], [3], tanA = ^/7^?^, cot^= — i
Vm* — 1
cacA = —^ .
Vm'-l
14 TEIOONOMETRY.
EXEKCISE V.
Find the values of the other functions when;
1. sin ^ = If. 5. ia.nA^^. 9. csc^=sV2.
2. sin ^=s 0.8, 6. cot^s=l. 10. ainA^m,
3. co8-4 = «. 7. cot ^ = 0.5. 11. Bin^ = --?^.
1 + m*
4. cos -4 = 0.28. 8. sec -4 = 2. 2mn
12. cos A = — T — -'
13. Given tan 45® = 1 ; find the other functions of 45°.
14. Given sin 30** = -J- ; find the other functions of 80°.
15. Given esc 60° = -f V3 ; find the other functions of 60°.
16. Given tan 15° = 2 - V3 ; find the other functions of 15°.
17. Given cot 22° 30'= V2 + 1 ; find the other functions
of 22° 80'.
18. Given sin 0° = ; find the other functions of 0°.
19. Given sin 90° = 1 ; find the other functions of 90°.
20. Given tan 90° = oo ; find the other functions of 90°.
^^21. Express the values of all the other functions in terms
of sin A,
y^ 22. Express the values of all the other functions in terms
of cos -4.
x^3. Express the values of all the other functions in terms
of tan A.
^ 24. Express the values of all the other functions in terms
of cot A,
25. Given 2 sin -4 = cos^ ; find sin^ and cos^l.
26. Given 4 sin A =^ tan A ; find sin A and tan A.
27. If sinu4 : co8j4 s= 9 : 40, find sin^ and cos -4,
28. Transform the quantity tan^J. + cot* J. — sin*-4 — cosM
into a form containing only cos^.
29. Prove that sin -4 + cos -4 = (1 + tan -4) cos A,
30. Prove that tan-4 + cot-4 = sec^ X esc Jl.
TBiaONOMETBIC FUNCTIONS.
15
§ 7. Functions op 45**.
Let ABC (Fig. 6) be an isosceles right triangle, in which
the length of the hypotenuse AB
is equal to 1 ; then AC\& equal to
BO, andjhe angle A is equal to 45**.
Sin^ j1C* + J5C^=1, therefore
2^^ = 1, and J[(7 = Vi = iV2.
Therefore (§ 1),
sin 45** = cos 45** = J V2.
tan45** = cot45** = l.
sec 45® = esc 45** = V2,
§8. Functions of 30** and 60**.
Let ABC be an equilateral triangle in which the length
of each side is equal to 1 ; and let CD bisect the angle C
Then CD is perpendicular to AB and bisects AB\ hence,
^2)= J, and aZ) = Vr^=vT=iV3j
In the right triangle ADC, the angle JtQD^Z^f, and the
angle CAD = m\ '
Whence (§ 1),
sin 30** = cos 60** = f
cos 30*" = sin 60** = ^ V3.
tan 30* = cot 60* = — = i V3-
V3
cot 30** = tan 60* = V3.
Bec30** = C8c60** = A=|V3.
V3
CSC 30** = sec 60** = 2.
The results for sine and cosine of 30**, 45**, and 60** may be
easily remembered by arranging them in the following form :
Angle . . .
Sine
Cosine. . .
30»
45»
JV2
60°
} = 0.5
JV2 = 0.70711
i }V3 = 0.86603
CHAPTER 11.
THE RIGHT TRIANGLE.
§ 9. The Given Paets.
In order to solve a right triangle, two parts besides the right
angle must be given, one of them at least being a side.
The two given parts may be :
I. An acute angle and the hypotenuse.
II. An acute angle and the opposite leg.
III. An acute angle and the adjacent leg.
IV. The hypotenuse and a leg.
V. The two legs.
§ 10. Case I.
Given A == 84** 28' and c = 18.76 ; required B, a, i,
^ 1. £ = 90*»-^ = 55*»32'.
2. - =Bin-4; .'. a = (?sin^.
h
Fig. 8.
log a = log c + log sin A
logc = 1.27300
log8inJ[= 9.75276 *
log a = 1.02576
a = 10.61
3. - =003-4; .•. & = ccos-4.
c
log J =logc + logcosjl
logc = 1.27300
log cos -4= 9.91617
logJ = 1.18917
I = 15.459
* For Logarithms, and directions how to use them, see Wentworth
and Hill's Five-place Tiibles.
When —10 belongs to a logarithm or cologarithm, and is not written,
it mnst be remembered that the logarithm or cologarithm is 10 too large.
THE RIGHT TBIANOLE.
17
§ 11. Case II.
Given ^ = 62* 10', a = 78 ; find B, J, e.
1. £ = 90**-^ = 27^50'.
2. - =cot-4; .*. J = acot-4.
a
3. - =8in-4.
c
:, a =CBin-4, and c =
Bin^
log 6 = log a + log cot A
log a = 1.89209
log cot ^= 9.72262
log* = 1.61471
b =41.182
log c = log a + colog sin A
loga = 1.89209
colog sin ^= 0.05340
logc = 1.94549
c =88.204
§ 12. Case III.
Given ^ = 50** 2', 6 = 88 ; find B, a, c.
1. B = W-A = Z9fb&.
, a = itanX
2. f = tan^ ;
o. - =cos-4.
.'. b =e;cos^, and (? =
cos^
loga = log i + log tan -4
log J = 1.94448
logtan^= 10.0767
loga = 2.02118
a = 105.0
log c = log b + colog cos A
logb = 1.94448
cologco8u4= 0.19223
logc = 2.13671
c = 137.0
18
TBIGONOMETRY.
§ 13. Case IV. *
Given c = 58.40, a = 47.55 ; find
A, B, b.
1. sinul= —
c
2. ^ = 90^-^.
3. - =cot^;
a
.*. 5 = acot-4.
log sin -4 = log a + colog c
log a =1.67715
colog c =8.23359
log h = log a + log cot A
loga = 1.67715
log cot -4= 9.85300
log sin ^ = 9.91074
logb = 1.53015
^ = 54^31'
b =33.896
^ = 35^29'
§14.
Case V.
Given a = 40, J = 27 ; find A,
B
. B.c,
^jT
1. tan^=^.
a
2. J5=90"-A
3. - =sin-4.
^ h
c •
Pur. 12.
A a
**•• •**•
Bin A
log tan A = log a + col
ogb
logc = log a+ colog sin J.
loga = 1.60206
loga = 1.60206
colog J = 8.56864
colog sin ji= 0.08152
logtan^ = 10.17070
logc = 1.68358
J. = 55*^59'
c =48.259
i = 34°l'
THE RIGHT TRIAKOLE. 19
j § 15. General Method of SoLviKa the Right Triangle.
From these fire cases it appears that the general method of
finding an unknown part in a right triangle is as follows :
Choosefrom the equ/oticm A + B = 9(f, and the eqiujUioTie tfuit
define the functicma of the angles^ an equation in which the re-
quired part only is unknovm; solve this equation^ if necessary,
to find the value cf the unknown part'; then compute the vainly
using logarithms whenever convenient.
Note 1. In Cases IV. and V. the unknown side may also be found
by Geomefcry, from the equation o^ + ^ — c^ ; whence we obtain
(for Case IV.) 6- v/?Tr^«-. y/(^Ta){c^)i
(for Case V.) c-V?T9.
These equations express the yalues of h and e directly in terms of the
two given sides ; and if the values of the sides are simple numbers (e.g.
5, 12, 13), it is often easier to find 5 or e in this way. But this value of
c is not adapted to logarithms, and this value of h is not so readily worked
out by logarithms as the value of h given in { 13.
Note 2. In Case IV. if the given sides (here a and c) are nearly alike
in value, then A is near 90^, and its value cannot be accurately found
from the tables, because the sines of large angles differ little in value (iis
is evident from Fig. 4). In this case it is better to find B first, by means
of a formula proved later (see page 47) ; viz.,
tan}-B--V^— ^'
and to find h by the method given in Note 1, since the same logarithms
are used in both cases.
Example. Given a — 49, c — 50 ; find A, B, b.
log tan J-ff - i [log (c — a)
log 6 =- } [log (C'-'a)-\- log (c + a)]
e^a —1
e+a -99
log (c- a) = 0.00000
log (c + a) = 1,99564
2)1.99564
log 6 -0.99782
h -9.95
+ colog (c + a)]
log(c-a) -0.00000
colog (c -ho) -8.00436
2)8.00436
log tan }5 -9.00218
iB -5044' 21"
B -11^29'
A -780 31'
20
TRIOONOMETEY.
§ 16. Aeea op the Eight Triangle.
It is shown in Geometry that the area of a triangle is equal
to one-half the product of the base by the altitude.
Therefore, if a and b denote the legs of a right triangle, and
J' the area, F^iab,
By means of this formula the area may always be found
when a and b are given or have been computed.
For example : Find the area, having given :
Case I. (§ 10).
^=34*'28', c = 18.75.
First find (as in § 10) log a
and log 5.
log (F) = log a +log b + colog 2
loga = 1.02578
logJ = 1.18915
colog 2 = 9.69897
log(J^= 1.91390
F =82.016
Case IV. (§ 18).
a = 47.54, (? = 58.40.
First find (as in § 18) log a
and log 5.
log (-F) = log a+log b + colog 2
loga =1.67715
log J =1.53025
colog 2 = 9.69897
log (J') = 2.90637
F =806.06
Exercise VI.
1. In Case II. give another way of finding c, after b has been
found.
2. In Case III. give another way of finding c, after a has
been found.
3. In Case IV. give another way of finding 6, iifter the
angles have been found.
4. In Case V. give another way of finding c, after the angles
have been found.
5. Given £ and c ; find A, a, b.
6. Given J3 and b ; find A, a, c.
7. Given B and a ; find A, b, c.
8. Given b and c ; find A, B, a.
THE BIGHT TBIAKGLE.
21
Solre the following triangles :
9
QivZH:
Bequiukd :
a-6,
c-12.
^-300, 5-60°.
6-10.392.
10
A^ecp,
6-4.
J?»30O c-8.
a -6.9282.
11
il-30O,
a -3.
5-60°, c-6,
6-6.1961.
12
a-i
5-4.
il - 5-450, c- 6.6668.
13
a-2,
c- 2.82843.
il- 5-450. 6 -2.
14
<j-627.
-4-23^30',
5-6603O'. a -260.02,
6-675.0.
15
<j-2280,
il-280 5'.
5-6I055'. a -1073.3.
6-2011.6.
16
c- 72.15,
il-39°34'.
5-50O26', a -45.958,
6-56.620.
17
c-1.
il-360.
5-540, 0- 0.58779.
6-0.80902.
18
<j-200,
^-21047'.
il-680 13', a -185.72,
6-74.219.
19
c-93.4.
J5-76°25'.
ii-13035'. a -21.936,
6-90.788.
20
a -637,
il- 4036'.
5 -850 25'. 6-7946,
c- 7971.5.
21
a - 48.532, ii- 36° 44'.
5 -530 16', 6-65.033,
c- 81.144.
22
a =.0.0008,
il-860.
5- 40. 6-0.0000559. C-O.OOO8O2I
23
J- 50.937,
5-43048'.
il-460 12'. a -63.116,
c- 73.59.
24
&-2.
.5- 3038'.
il-860 22'. a -31.497.
c- 31.560.
25
a -992,
5 -76° 19'.
ii-13041'. 6-4074.46.
c- 4193.55.
26
a -73,
5 -68° 52'.
il-2l0 8, 6-188.86.
(J -202.47.
27
a -2.189,
5-45026'.
il-44036'. 6-2.2211.
c- 3.1186. •
28
6-4,
A^zr>5&.
5-6204', a -3.1176.
c- 6.0714.
29
c-8590,
a -4476.
il-310 24'.5-580 36',
6-7332.8.
30
c- 86.53,
a -71.78.
il-6eo3', 5-33057',
6 - 48.324.
31
c-9.35,
a -8.49.
il-660l4',5-240 46'.
6-3.917.
32
c-2194,
6-1312.7.
il-630 16'.5-360 45',
a -1758.
33
c- 30.69.
6 - 18.256.
il-53O30'.5-36O30',
a -24.67.
34
a -38.313.
6-19.522.
il-630. 5-270.
c-48.
35
a -1.2291.
6-14.960.
^- 40 42'.5-860 18',
c-15.
36
a -415.38,
6-62.080.
ii-81O30'.5- 803O',
e-420.
37
a -13.690.
6-16.926.
^-380 58',5-610 2',
c- 21.77.
38
c- 91.92.
a -2.19.
A - lo21'55". 5 - 88038'5", 6 - 91.894.
Compute the unknown parts and also the area, having given :
89. a = 5, i = 6.
40. a = 0.615, c = 70^
41. b = -</2, c = VS.
44. c = 68, ^ = 69" 54'.
45. c = 27, .5 = 44° 4'.
46. a = 47, 5 = 48*49'.
42. a = 7. ^ = 18° 14'. 47. 6 = 9, 5 = 34' 44'.
43. 6 = 12, ^ = 29" 8'. 48. c = 8.462, 5 = 86° 4'.
22 TBIGONOlffiTET.
49. Find the value of Fin terms of e and A.
50. Find the value of -Fin terms of a and A,
51. Find the value of i^in terms of b and A.
52. Find the value of -fin terms of a and c,
53. Given F= 58, a = 10 ; solve the triangle.
54. Given F= 18, J = 5 ; solve the triangle.
55. Given -F= 12, .4 = 29** ; solve the triangle.
56. Given F= 100, i?= 22; solve the triangle.
57. Find the angles of a right triangle if the hypotenuse is
equal to three times one of the legs.
58. Find the legs of a right triangle if the hypotenuse = 6,
and one angle is twice the other.
59. In a right triangle given c, and A = nJB ; find a and b,
60. In a right triangle the difference between the hypote*
nuse and the greater leg is equal to the difference between the
two legs ; find the angles.
The angle of elevation of an object (or angle of depression,
if the object is below the level of the observer) is the angle
•which a line from the eye to ihe object makes with a horizon-
tal line in the same vertical plane.
61. At a horizontal distance of 120 feet from the foot of a
steeple, the angle of elevation of the top was found to be 60** 30';
find the height of the steeple.
62. From the top of a rock that rises vertically 326 feet out
of the water, the angle of depression of a boat was found to be
24® ; find the distance of the boat from the foot of the rock.
63. How far is a monument, in a level plain, from the eye,
if the height of the monument is 200 feet and the angle of ele-
vation of the top 3** 30'?
64. In order to find the breadth of a river a distance AB
was measured along the bank, the point A being directly op-
posite a tree C on the other side. The angle ABOwaB also
measured. If AB = 9Q feet, and AB0=2rU\ find the
breadth of the river.
If ABC= 45*^, what would be the breadth of the nver ?
THE BIGHT TBIAKGLE. 23
65. Find the angle of elevation of the sun when a tower
a feet high castB a horizontal shadow b feet long. Find the
angle when a = 120, b = 70.
66. How high is a tree that casts a horizontal shadow b feet
in length when the angle of elevation of the son is ^^ ? Find
the height of the tree when b^80,A = 5(f.
67. What is the angle of elevation of an inclined plane if it
rises 1 foot in a horizontal distance of 40 feet?
68. A ship is sailing due north-east with a velocity of 10
miles an hour. Find the rate at which she is moving due
north, and also due east.
69. In front of a window 20 feet high is a flower-bed 6 feet
wide. How long must a ladder be to readi from the edge of
the bed to the window ?
70. A ladder 40 feet long may be so placed that it will reach
a window 33 feet high on one side of the street, and by turn-
ing it over without moving its foot it will reach a window 21
feet high on the other side. Find the.breadth of the street.
71. From the top of a hill the angles of depression of two
successive milestones, on a straight level road leading to the
hill, are observed to be 5° and 15®. Find the height of the
hill.
72. A fort stands on a horizontal plain. The angle of ele-
vation at a certain point on the plain is 30®, and at a point 100
feet nearer the fort it is 45®. How high is the fort?
73. 'Frook a certain point on the ground the angles <^ eleva-
tion of the belfry of a church and of the top of the steeple were
found to be 40® and 51® respectively. From a point 300 feet
farther off, on a horizontal line, the angle of elevation of the
top of the steeple is found to be 33® 45'. Find the distance
frt>m the belfry to the top of the steeple.
74. The angle of elevation of the top of an inaccessible fort
(7, observed from a point A, is 12®. At a point B, 219 feet
frcfm A and on a line AB perpendicular to .^C, the angle ABC
is 61® 45'. Find the height of the fort.
24
TEIGONOMETRY.
§ 17. The Isosceles Triangle.
An isosceles triangle is divided by the perpendicular from
the vertex to the base into two equal right triangles.
Therefore, an isosceles triangle is determined by any two
parts that determine one of these right triangles.
Let the parts of an isosceles triangle ABC (Fig. 13), among
which the altitude CD is to be in-
cluded, be denoted as follows :
a = one of the equal sides.
c = the base.
h = the altitude.
A = one of the equal angles.
C= the angle at the vertex.
For example: Given a and (?; re-
quired A^ CJ A.
1. C08^=i? = -f-.
a 2a
2. 0^+2^ = 180^ .•.(7=180*'-2^ = 2(90^-^).
3. h may be found directly in terms of a and c from tlie
equation •
A« + f = a^
which gives A = V(a — ic) (a + 1 c).
But it is better to find the angles first, and then find h from
either one of the two equations,
^ = sin^,
a
or
■^ = tanJ[.
ic
whence
A = asin-4, or h =icia,nA,
The numerical values of A, C, and h may be computed by
the aid of logarithms, as in the case of the right triangle.
The area F of the triangle may be found, when c and h are
given or have been found, by means of the formula
F=ich,
THE ISOSCELES TRIAK6LE. 25
Exercise VIL
In an isosceles triangle :
1. Given a and A ; find (7, c, A.
2. Given a and C\ find A, (?, A.
3. Given c and -4 ; find C, a, A.
4. Given c and C; find -4, a, A.
5. Given A and ^ ; find C, a, c.
6. Given A and C; find A, a, c.
7. Given a and A ; find A, C, c.
8. Given c and A ; find A, 0, a.
9. Given a =14.3, c = ll; find ^, (7, A.
10. Given a = 0.295, A= 68** 10' ; find c, A, J!
11. Given c = 2.352, C= 69** 49' ; find a, A. F.
12. Given A = 7.4847, A = 76** 14' ; find a, c, R
13. Given a = 6.71, A = 6.60; find -4, (7, c.
14. Given (? = 9, A = 20 ; find A, C, a.
15. Given c = 147, i^= 2572.5 ; find -4, (7, a, A.
16. Given A = 16.8, F^ 43.68 ; find -4, (7, a, (?.
17. Find the value of F\u terms of a and c.
18. Find the value of Fin terms of a and CI
19. Find the value of Fm terms of a and A,
20. Find the value of F'm terms of A and C.
21. A barn is 40 X 80 feet, the pitch of the roof is 45° ; find
the length of the rafters and the area of both sides of the roof,
22. In a unit circle what is the length of the chord corre-
sponding to the angle 45° at the centre?
23. If the radius of a circle = 30, and the length of a chord
= 44, find the angle at the centre.
24. Find the radius of a circle if a chord whose length is 5
subtends at the centre an angle of 133°.
25. What is the angle at the centre of a circle if the corre-
sponding chord is equal to f of the radius ?
26. Find the area of a circular sector if the radius of the
circle = 12, and the angle of the sector = 30°.
26 TBIGONOMETEY.
§ 18. The Regular Polygon.
Lines drawn from the centre of a regular polygon (Fig. 14)
to the vertices are radii of the circumscribed circle ; and lines
drawn from the centre to the middle points of the sides are
radii of the inscribed circle. These lines divide the polygon
into equal right triangles. Therefore, a regular polygon is
determined by a right triangle whose sides are the radius of
the circumscribed circle, the radius of the inscribed circle, and
half of one side of the polygon.
If the polygon has n sides, the angle of this right triangle at
the centre is equal to
1/360^ ^^ 180^
If, also, a side of the polygon, or one of the above-mentioned
radii, is given, this triangle may be solved, and the solution
gives the unknown parts of the polygon.
Let,
n = number of sides.
c = length of one side.
r = radius of circumscribed circle.
k = radius of inscribed circle.
p = the perimeter.
F= the area.
Then, by Geometry,
F=ihp.
Tig. 14.
EXEECISE VIII.
1. Given n = 10, c = l; find r, A, F.
2. Given n = 12, p=70; find r, A, F.
3. Given w = 18, r = 1 ; find A, p, F
4. Given w = 20, r = 20; find A, c,F.
5. Given n = 8, A = 1 ; find r, c, F,
6. Given n = 11, F= 20 ; find r, A, c.
7. Given n = 7, F= 7 ; find r, A, p.
THE EEGULAE POLYGON. 27
8. Find the side of a regular decagon inscribed in a unit
circle.
9. Find the side of a regular decagon circumscribed about
a unit circle.
10. If the side of an inscribed regular hexagon is equal to 1,
find the side of an inscribed regular dodecagon.
11. Given n and c^ and let b denote the side of the inscribed
regular polygon having 2n sides; find b in terms of n and <?.
12. Compute the difierence between the areas of a regular
octagon and a regular nonagon if the perimeter of each is 16.
13. Compute the difierence between the perimeters of a
regular pentagon and a regular hexagon if the area of each is 12.
14. From a square whose side is equal to 1 the comers are
cut away so that a regular octagon is left. Find the area of
this octagon.
15. Find the area of a regular pentagon if its diagonals are
each equal to 12.
16. The area of an inscribed regular pentagon is 331.8;
find the area of a regular polygon of 11 sides inscribed in the
same circle.
17. The perimeter of an equilateral triangle is 20; find
the area of the inscribed circle.
18. The area of a regular polygon of 16 sides, inscribed in
a circle, is 100 ; find the area of a r^ular polygon of 15 sides,
inscribed in the same circle.
19. A regular dodecagon is circumscribed about a circle,
the circumference of which is equal to 1 ; find the perimeter
of the dodecagon*
20. The area of a regular polygon of 25 sides is equal tc
40 ; find the area of the ring comprised between the circum-
ferences of the inscribed and the circumscribed circles.
CHAPTER III.
GOKIOMETRY.
§ 21. Definition op Goniometry.
In order to prepare the way for the solution of an oblique
triangle, we now proceed to extend the definitions of the
trigonometric functions to angles of all magnitudes, and to
deduce certain useful relations of the functions of different
angles.
That branch of Trigonometry which treats of trigonometric
functions in general, and of their relations, is called Ghmiometry.
§ 22. Angles of any Magnitude.
Let the radius OP of a circle (Fig. 16) generate an angle by
turning about the centre 0. This
angle will be measured by the
arc described by the point P;
and it may have any magnitude,
because the arc described by P
may have any magnitude.
Let the horizontal line OA bo
the initial position of OP, and
let OP revolve in the direction
shown by the arrow, or opposite
to the way clock-hands revolve.
Let, also, the four quadrants into
which the circle is divided by the horizontal and vertical
diameters AA\ BB\ be numbered I., II., III., IV., in the
direction of the motion.
GONIOMETEY. 29
During one revolution OP will form with OA all angles from
0® to 360". Any particular angle is said to be an angle of the
quadrant in which OP lies ; so that,
Angles between 0" and 90" are angles of Quadrant I.
Angles between 90" and 180" are angles of Quadrant 11.
Angles between 180" and 270" are angles of Quadrant III.
Angles between 270" and 360" are angles of Quadrant IV.
If OP make another revolution, it will describe all angles
from 360" to 720", and so on.
If OP^ instead of making another revolution in the direction
of the arrow, be supposed to revolve hcLchvards about 0, this
backward motion tends to undo or cancel the original forward
motion. Hence, the angle thus generated must be regarded
as a negative angle ; and this negative angle may obviously
have any magnitude. Thus we arrive at the conception of an
angle of any magnitude, positive or negative.
§ 23. General Definitions op the Functions.
The definitions of the trigonometric functions may be ex-
tended to all angles, by mjiing the functions of any angle
equal to the line values in a unit circle drawn for the angle
in question, as explained in § 3. But the lines that represent
the sine, cosine, tangent, and cotangent must be regarded as
negative, if they are opposite in direction to the lines that repre-
sent the corresponding functions of an angle in the first qv^ad-
rant ; and the lines that represent the secant and cosecant miLst
be regarded as negative, if they are opposite in direction to tlie
moving radius.
Figs. 17-20 show the functions drawn for an angle AOP in
each quadrant taken in order. In constructing them, it must
be remembered that the tangents to the circle are always
drawn through A and P, never through A* or £\
Let the angle -4 OP be denoted by x; then, in each figure
the absolute values of the functions, that is, their values with-
out regard to the signs + and —, are as follows :
30
TRIGONOMETRY.
sin X = MP,
cosa; = OM,
B
ioxix^ AT,
cot X = B8,
seca; = Or.
cscar= 08,
B
Keeping in mind the position of the points A and B, we may
define in words the first four functions of the angle x thus :
sin X = the vertical projection of the moving radius ;
cos X =■ the horizontal projection of the moving radius ;
' the distance measured along a tangent to the circle
tana? = '< from the beginning of the first quadrant to the
moving radius produced ;
' the distance measured along a tangent to the circle
cota? = S from the end of the first quadrant to the moving
radius produced.
Sec a? and esc a: are the distances from the centre of the
circle measured along the moving radius produced to the tan-
gent and cotangent respectively.
GONIOMETEY.
31
§24. Algebbaic Signs op the Functions.
The lengtlis of the lines, defined above as the fanctions of
any angle, are expressed numerically in terms of the radius
of the circle as the unit. But, before these lengths can be
treated as algebraic quantities, they must have the sign -f or
— prefixed, according to the condition stated in § 23.
The reason for this condition lies in that fundamental rela-
tion between algebraic and geometric magnitudes, in virtue of
which contrary signs in Algebra correspond to opposite direc-
tions in Oeometry,
The sine JfPand the tangent ^T always extend from the
horizontal diameter, but sometimes upwards and sometimes
dovmwards; the cosine OM and the cotangent B8 always
extend from the vertical diameter, but sometimes towards the
right and sometimes towards the left. The functions of an angle
in the first quadrant are assumed to be positive. Therefore,
1. Sines and tangents extending from the horizontal diam-
eter upwards^ are positive ; downwards^ negative.
2. Cosines and cotangents extending from the vertical diame-
ter towards the rights are positive ; towards t?ie left, are nega-
tive.
The signs of the secant and cosecant are always made to
agree with those of the cosine and sine respectively. This
agreement is secured if secants and cosecants extending from
the centre, in the direction of the moving radius, are consid-
ered positive ; in the opposite direction, negative.
Hence, the signs of the functions for each quadrant are :
Sine and cosecant
Cosine and secant
Tangent and cotangent ....
I.
II.
III.
IV.
+
+
+
+
+
+*
32
TRIGONOMETEY.
§25. Functions of a Variable Angle.
Let the angle x increase continuously from (f to 360®;
what changes will the values of its functions undergo ?
It is easy, by reference to Figs. 21-24, to trace these
changes throughout all the quadrants.
Fig. 23.
1. The Sine, In the first quadrant, the sine MP increases
from to 1 ; in the second, it remains positive, and decreases
fn)m 1 to ; in the third, it is negative, and increases in abso^
lute value from to 1 ; in the fourth, it is negative, and
decreases in absolute value from 1 to 0.
GONIOMETEY. 38
2. The Cosine. In the first quadrant, the cosine OM de-
creases from 1 to 0; in the second, it becomes negative and
increases in absolute value from to 1; in the third, it is
negative and decreases in absolute value from 1 to 0; in the
fourth, it is positive and increases from to 1.
3. The Tanffent. In the first quadrant, the tangent AT
increases from to oo; in the second quadrant, as soon as
the angle exceeds 90^ hj the smallest conceivable amount, the
moving radius OP, prolonged in the direction opposite to that
of OF, will cut ATsi & point T situated very far below A ;
hence, the tangents of angles near 90** in the second quad-
rant have very large negative values. As the angle increases,
the tangent -4 T continues negative but diminishes in absolute
value. When a: = 180*, then T coincides with -4, and tan 180*
= 0. In the third quadrant, the tangent is positive and in-
creases from to 00 ; in the fourth, it is negative and decreases
in absolute value from oo to 0.
4. The Cotangent, In the first quadrant, the cotangent £S
decreases from oo to ; in the second quadrant, it is negative
and increases in absolute value from to oo ; in the third and
fourth quadrants, it has the same sign, and undergoes the same
changes as in the first and second quadrants respectively.
5. The Secant, In the first quadrant, the secant OT in-
creases from 1. to oo; in the second quadrant, it becomes
negative (being measured in the direction opposite to that of
OP), and decreases in absolute value from oo to 1, so that
sec 180* = — 1 ; in the third quadrant, it continues negative,
and increases in absolute value from 1 to oo ; in the fourth
quadrant, it is positive, and decreases from oo to 1.
6. The Cosecant, In the first quadrant, the cosecant OS
decreases from oo to 1 ; in the second quadrant, it remains
positive, and increases from 1 to oo ; in the third quadrant, it
becomes negative, and decreases in absolute value from oo to
1, so that CSC 270*=— 1; in the fourth quadrant, it is nega-
tive, and increases in absolute value from 1 to oo.
34
TEIGONOMETET.
The limiting values of the functions
are as follows ;
Sine
0-
90-
180*
27(r
360*
±0
1
±0
-1
±0
Cosine
1
±0
-1
±0
1
Tangent
±0
±00
±0
±00
±0
Cotangent
±00
±0
±00
±0
±00
Secant
1
±00
-1
±00
1
Cosecant
±00
1
±00
-1
±00
Sines and cosines extend from +1 to — 1 ; tangents and co-
tangents from +00 to — oo ; secants and cosecants from + oo
to + 1, and from — 1 to — oo.
In the table given above the doable sign db is placed before and
00. From the preceding investigation it appears that the functions always
change sign in passing through and oo ; and the sign + or — prefixed
to or 00 simply shows the direction from which the value is reached.
Take, for example, tan 90° : The nearer an acute angle is to 90°, the
greater the positive value of its tangent ; and the nearer an obtuse angle v
is to 90°, the greater the negative value of its tangent. When the angle
M 90°, OP (Fig. 21) is parallel to AT, and cannot meet it. But tan 90°
may be regarded as extending either in the positive or in the negative
direction ; and according to the view taken, it will be + oo or — oo.
§ 26. Functions op Angles Larger than 360**.
It is obvious that the functions of 360** + a: are the same
both in sign and in absolute value as those of x ; for the mov-
ing radius has the same position in both cases. In general, if
n denote any positive whole number,
The functions of(nX 360® + x) are the same as those qfx.
For example : the functions of 2200** = the functions of
(6 X 360** + 40**) = the functions of 40^
QONIOMETBT. 35
§ 27. Extension of Fokmulas [1]-[3] to all Angles.
The Formulas established for cuniie angles in § 5 hold true
for all angles. Thus, Formula [1],
sin'a: + coslr = 1,
is universally true ; for, whether JfP and O-Sf (Figs. 21-24)
are positive or negati ve, M JP ^ and OM* are always positive,
and in each quadrant MI^ + 0M^ =iOP =1.
Also, Formulas
r-., , sin X
[2] tanar= ,
(sin :r X CSC 2: = 1,
cos a: X sec ar = 1,
tana; X cot ar = l,
are universally true ; for the algebraic signs of the functions,
as given in the table at the end of § 24, agree with those in
Formulas [2] and [3] ; and with regard to the absolute values,
we have in each quadrant from the similar triangles OMF,
OAT, 0B8, (Figs. 21-24) the proportions
AT :OA=MP'.OM,
MP\OP^OB :0S,
OJf:OF=OA :0T,
AT :OA = OjB :£S,
which, by substituting 1 for the radius, and the right names
for the other lines, are easily reduced to the above formulas.
Formulas [l]-[3] enable us, from a given value of one func-
tion, to find the absolute values of the other five functions, and
also the sign of the reciprocal function. But in order to deter-
mine the proper signs to be placed before the other four
functions, we must know the quadrant to which the angle in
question belongs; or what amounts to the same thing, the sign
of any one of these four functions ; for, by reference to the
Table of Signs (§ 24) it will be seen that the signs of any two
functions that are not reciprocals determine the quadrant to
which the angle belongs.
36 TEIGONOMETRY.
Example. Given sin a: = + f, and tana: negative; find the
values of the other functions.
Since sin a: is positive, x must be an angle in Quadrant I. or
in Quadrant II.; but, since also tana: is negative, Quadrant I.
is inadmissable.
By [1], cosa: = ±Vl— if = ± f .
Since the angle is in Quadrant II. the minus sign must be
taken, and we have
cosa:'=--f.
By [2] and [3],
tana: = — -J, cota; = — f, seca: = — f, csca: = f.
EXEEOISE IX.
1. Construct the functions of an angle in Quadrant II.
What are their signs ?
2. Construct the functions of an angle in Quadrant III.
What are their signs ?
3. Construct the functions of an angle in Quadrant IV.
What are their signs ?
4. What are the signs of the fumjtions of the following
angles: 840^ 239^ 145^ 400^ 700^ 1200^ 3800°?
5. How many angles less than 360® have the value of the
sine equal to +^, and in what quadrants do they lie ?
6. How many values less than 720° can the angle x have
if cosa: = + -J, and in what quadrants do they lie ?
7. If we take into account only angles less than 180°, how
many values can x have if sin a; = ■}■ ? if cosa: = -J- ? if cosa: =
-I? if tana: = |? if cota: = -7?
8. Within what limits must the angle x lie if cos a: = — -J ?
ifcota: = 4? ifseca: = 80? if csca: = — 3? (a: to be less than
360°.)
9. In what quadrant does an angle lie if sine and cosine
are both negative ? if cosine and tangent are both negative ?
if the cotangent is positive and the sine negative ?
OONIOMETBY.
10. Between 0® and 3600® how many angles are there whose
sines have the absolute value f ? Of these sines how many
are positive and how many negative ?
11. In finding cos a: by means of the equation C08a: =
—^/l — sin*a;, when must we choose the positive sign and when
the negative sign ?
12. Given cos x = — VJ ; find the other functions when x ia
an angle in Quadrant II.
13. Given tan x = V3 ; find the other functions when x is
an angle in Quadrant III.
14. Given sec a; = + 7, and tan x negative ; find the other
functions of x,
15. Given cot a; = — 3 ; find all the possible values of the
other functions.
16. What functions of an angle of a triangle may be nega-
tive ? In what case are they negative ?
17. What functions of an angle *of a triangle determine the
angle, and what functions fail to do so ?
18. Why may cot 360° be considered equal either to + oo
or to — 00 ?
19. Obtain by means of Formulas [l]-[3] the other func-
tions of the angles given :
(i.) tan 90^ = 00. (iii.) cot 270° = 0.
(ii.) cos 180° = — 1. (iv.) esc 360° = - oo.
20. Find the values of sin 450°, tan 540°, cos 630°, cot 720°,
sin 810°, CSC 900°.
21. For what angle in each quadrant are the absolute values
of the sine and cosine alike ?
Compute the values of the following expressions:
22. a sin 0° + 6 cos 90° - c tan 180°.
23. a cos 90° - 6 tan 180° -f- 1? cot 90°.
24. a sin 90° — b cos 360° + (a - 5) cos 180°.
25. (c^-6»)co8360°-4a6sin270°.
38
TRIGONOMETEY.
§ 28. Reduction op Functions to the Fibst Quadrant.
In a unit circle (Fig. 25) draw two diameters Pit and Q8
equally inclined to the horizon-
tal diameter AA\ or bo that the
angles AOP, A'OQ, A'OE, and
A08 shall be equal. From the
points P, Q, JK, S let fall per-
pendiculars to AA^] the four
right triangles thus formed, with
a common vertex at 0, are equal ;
because they have equal hypote-
nuses (radii of the circle) and
equal acute angles at 0. There-
fore, the perpendiculars PM,
QN, BN, SM, are equal. Now these four lines are the sines
of the angles AOP, AOQ, AOP, and AOS, respectively.
Therefore, in ahsolute valuer
sin ^OP = sin AOQ = Bin AOP = BinAOS.
And from § 27 it follows that in absolute valtce the cosines
of these angles are also equal ; and likewise the tangents, the
cotangents, the secants, and the cosecants.*
Hence, /or every acute angle (AOP) there is an angle in each
of the higher quadrants whose functums^ in absolute value, are
equal to those of this acute angle.
Let Z AOP =x, APOB=^y\ then a: + y = 90^ and the
functions of x are equal to the co-named functions of y (§ 4) ;
further,
Z ^OQ (in Quadrant II.) ==180^-a?= 90* + y,
AAOR (in Quadrant III.) = 180* + a; = 270* - y,
A AGS (in Quadrant IV.) = 360* - a; = 270* + y.
Hence, if we prefix in each case the proper sign (§ 24), we
have the two following series of Formulas :
* In future, secants and cosecants will be disregarded. They may be found
by [3] if wanted, but are seldom used in computations.
GONIOMETBT. 89
Angle in QuadrofrU II.
"T sin (180° — ar) = sin x, sin (90° + y) = cos y.
cos (180° — a:) = — cos ar. cos (90° + y) = — sjn y.
tan (180° — a:) = — tanar. tan (90° + y) = — cot y.
cot (180° - a:) = — cot z. cot (90° + y) = — tany.
Angle in Qtmdrant III
sin (180° + a:) = — sin a;. sin (270° - y) = — cos y.
cos (180° + ar) = — cos ar. cos (270° — y) = — sin y.
tan (180° + a:) = tanar. tan(270°-y)= coty.
cot (180° + a?) = cotar. cot(270°-y)= tany.
Angle in Quadrant IV.
sin (360° — a;) = — sin x. sin (270° + y) = - cos y.
cos(360°--ar)= cosar. • cos (270° + y) == siny.
tan (360° - a:) = - tana:. tan (270° + y) = - cot y.
cot (360° - a;) = - cot a;. cot (270° + y) = - tany.
Hence, by selecting the right formulas,
Thefunctvms of all angles can be reduced to the functions of
angles not greater than 45°. Thus, to find the functions of
220° and 230°, we should consider 220° as (180° + 40°), but
230° 85(270° -40°).
It is evident from these formulas that,
If an acute angle he added to or subtracted from 180° or 8flO°i
the functions of the resulting angle are equal in absolute value
to the Vik^namei functions of the acute angle; but if an acute
angle be added to or subtracted from 90° or 270°, the functions
of the resulting angle are equal in absolute value to the oo-nomed
functUms of the curute angle.
It is evident from the formulas for (180°— a:) that,
A given value of a sine determines two supplementary angles,
one acute^ the other obtuse ; a given value of any other function
{except the cosecant) determines only one angle: acute if the
value is positive^ obtuse if the value is negative.
40
TRIGONOMETRY.
§29. Angles whose Difference is 90".
The general form of two such angles is x and 90" + ^\ a-i^d
they must lie in adjoining quadrants. The relations between
their functions were found in
§ 28, but only for the case when
X is acute. These relations, how-
ever, may be shown to hold true
for all values of x.
In a unit circle (Fig. 26) draw
two diameters Pi? and QS per-
pendicular to each other, and
let fall to A A' the perpendicu-
lars FM, QH, RK, and SN,
The right triangles OMP, OHQ,
OKR^ and ONS are equal, because they have equal hypote-
nuses and equal acute angles POM, OQH, ROK, and 08N,
Therefore, OJIf = QH= OK = N8,
and PJf = 0E= KR = ON,
Hence, taking also into account the algebraic sign,
sin^OQ= cos -4 OP; ^vclAOS = cos^OP;
cos^OQ = — sin-iOP; cos^O^S = — sin ^OP;
sin ^0P= cos^OQ; sin (360" + ^ OP) = cos^O/S;
cos -4 OP = - sin -40(2 ; cos (360" + A OP) = - sin ^OaS.
In all these equations, if a; denote the angle on the right-hand
side, the angle on the left-hand side will be 90"+ a:. There-
fore, if X be an angle in any one of the four quadrants,
sin (90" + x)— cos x,
cos (90" + x) = — sin x.
And, by § 27, tan (90" + x)=- cot x,
cot (90" + a;) = — tan x.
In like manner, it can be shown that all the formulas of
§ 28 hold true, whatever be the value of the angle x.
GONIOMETBY. 41
§ 30. Functions of a Negative Angle.
If the angle -4 OP (Fig. 25) is denoted by z, the equal angle
A08, generated by a backward rotation of the moving radius
from the initial position 0-4, will be denoted by — x. It is
obvious that the position OS of the moving radius for this
angle is identical with its position for the angle 360®— a:.
Therefore, the functions of the angle —z are the same as those
of the angle 360** - a? ; or (§ 28),
sin (— a:) = — sin z, tan (— ar) = — tan z,
cos (— ar) = cos a:, cot (— a:) = — cot z.
EZEBCISE X.
1. Express sin 250® in terms of the functions of an acute
angle greater than 45®, and also in terms of the functions of
an acute angle less than 45®.
Ans. 1. sin 250^ = sin (180^ +70^) = -sin 70®.
2. sin250® = 8in(270®-20®)=-cos20®.
Express the following functions in terms of the functions of
angles less than 45® :
2. sin 172®. 8. sin 204®. 14. sin 163® 49'.
3. cos 100®. 9. cos 359®. 15. cos 195® 33'.
4. tanl25®. 10. tan300®. 16. tan269®15'.
5. cot 91®. 11. cot 264®. 17. cot 139® 17'.
6. sec 110®. 12. sec 244®. 18. sec 299® 45'.
7. CSC 157®. 13. CSC 271®. 19. esc 92® 25'.
Express all the functions of the following negative angles in
terms of those of positive angles less than 45® ;
20. -75®. 22. -200®. 24. -52® 37'.
21. -127®. 23. -345®. 25. -196® 54'.
26. Find the functions of 120®,
HiiTT. 120° - 180° - 60°, or, 120° - 9(y> + 30° ; then apply { 28.
J
f
Firifl Ui« valnes i
1.
MIlX — {{.
2.
Mti ^ =- 0.8
8,
C(Mljl = lt
4,
OWtji-O.L'
13.
Oivon tail
14.
Oivon Mti
1&.
Oivon cm
10.
Oivcn tai.
IT.
„f ;>2*
Oivon CO'
80'.
18.
Given «•
10.
flivon ei
yt.
Oivi'ti til
of sin
.4.
of run
Ksprei-
.4.
of ivl .4. t
e^.
liiven !
'JA.
Given
i-
Ifsiu .
is.
f<ytm 1 1
Trov.
:vi
■• TtTr
i
H »
r r
;r
• - .1 *
&\ IVov^
* **''J.
[4]
[5]
assumed
If the sum
ngles X and
Fig. 28, the
rd for word,
, the only dif-
.t the sign of
live, as DO is
, therefore, hold
42 TEIGONOMETRY.
Find the functions of the following angles :
27. 135^ 29. 210^ 31. 240^ 33. -30^
28. 15(r. 30. 226^ 32. 300^ 34. -225^
35. Given sin a: = — Vj, and cos a? negative ; find the other
functions of x, and the value of x,
36. Given cota; = — V3, and x in Quadrant II.; find the
other functions of a;, and the value of x,
37. Find the functions of 3540^.
38. What angles less than 360® have a sine equal to — } ?
a tangent equal to — V3?
39. Which of the angles mentioned in Examples 27-34 have
a cosine equal to — -Vj? a cotangent equal to ~V3?
40. What values of x between 0® and 720" will satisfy the
equation sin a; = + ^ ?
41. In each of the following cases find the other angle be-
tween 0** and 360* for which the corresponding function (sign
included) has the same value: sinl2*, cos 26®, tan 45®, cot 72®;
sinl91®, cosl20^ tan244®, cot357®.
42. Given tan 238® = 1.6; find sin 122®.
43. Given cos 333® = 0.89 ; find tan 117®.
Simplify the following expressions :
44. a cos (90® -x) + b cos (90® + x).
45. m cos (90® -a:) sin (90® -a;).
46. (a - b) tan (90® -x) + (a + b) cot (90® + x).
47. a* + 6»-2aJcos(180®-a:).
48. sin(90®+a:)sin(180®+ar)+cos(90®+ar)cos(180®-a:).
49. cos (180® +x) cos (270® -y) - sin (180® +ar) sin (270® -y).
50. tana:+tan(-y)-tan(180®-y).
51. For what values of x is the expression sin a: + cos a:
positive, and for what values negative ? Represent the result
by a drawing in which the sectors corresponding to the nega-
tive values are shaded.
52. Answer the question of last example for sin a;— cos a:.
53. Find the functions of (x — 90®) in terms of the functions
of a;.
54. Find the functions of (x — 180®) in terms of the functions
of a:.
GOKIOMETBY.
43
§31. Functions of the Sum of Two Angles.
In a unit circle (Fig. 27) let the angle AOB = x^ tte angle
BOC^ y ; then the angle AOC^
x + y.
In order to express 8in(a? + y)
and co8(a? + y) in terms of the
sines and cosines of x and y, draw
CFJL OA, CD JL OB, DE± OA,
DG±CF; then CD = Any, OD
= cosy, and the angle DCO =
the angle ODO = x. Also,
sin (x + y)= CF= DE+ CO.
DE
— hence, DE = sin a: X OD = sina; cosy.
OD
CO
-— = cosa: ; hence, CO = cosa: X CD = cos a: siny.
Cx/
Therefore, Bm(x + y) = sinz coey + coex Biny.
Again, co8(ar + y) = 0F= OE- DO,
OE
[4]
OD
= cosa: ; hence, 0E= cosar X OD = cos a? cosy.
DO^
CD
Therefore,
^^ = 8mx;
hence, DO = sin a: X CD = sinar siny.
oog(x + y) = 008X cosy — sinx einy.
[5]
In this proof x and y, and also the sum x-^-y, are assumed
to be acute angles. If the sum
ar4-y of the acute angles x and
y is obtuse, as in Fig. 28, the
proof remains, word for word,
the same as above, the only dif-
ference being that the sign of
0-Pwill be negative, as -Dff is
now greater than OE. Tlie above formulas, therefore, hold
true for all acute angles x and y.
44 TRIGONOMETRY.
If these formulafl hold true for any two acute angles x and
y, they hold true when one of the angles is increased by 90®.
Thus, if for x we write x^ = 90® + x, then, by § 29,
sin {x^ +y) = sin (90° + x + y)= cos (a; + y)t
cos(a:'+ y) = cos (90® + ar + y) = — sin (a: + y).
Hence, by [5], sin (a;'+ y) = cos a; cosy — sin x siny,
by [4], cos(a?'+ y) = — sin a; cosy — cos a; sin y.
Now, by § 29, cos x = sin (90® + a:) = sin x\
sin a: = — cos (90® + a:) = -— cos a:'.
Therefore, by putting sin a/ for cos a?, and —-cos a/ for sin a;,
in the right-hand members of the above equations,
sin (a;' + y) = sin a;' cosy + cosar' sin y,
co8(a;'+ y) = cos a:' cosy — sin a?' sin y.
Hence, it follows that Formulas [4] and [5] are universally
true. For they have been proved true for any two acute
angles, and also true when one of these angles is increased by
90® ; hence they are true for each repeated increase of one or
the other angle by 90®, and therefore true for the sum of any
two angles whatever.
By §27,
, , , V sin(a: + y) sin a; cos y + cos a; siny
tan(a; + y)= — ) T \ = ^^ -'
cos (x + y) cosa; cosy — sm x sin y
If we divide each term of the numerator and denominator of
the last fraction^y cosa; cosy, and again apply § 27, we obtain
In like manner, by dividing each term of the numerator and
denominator of the value of cot (:p + y) by sin a: siny, we obtain
oot(z + y)=??^^^yfl. [7]
. •'^ ootx + coty
QOKIOMETBT.
45
§ 32. Functions op the Difference of Two Angles.
In a unit circle (Fig. 29) let the angle AO£ = x, the angle
COB = y ; then the angle A00==
ar — y.
In order to express 8in(a? — y)
and co8(a? — y) in terms of the
sines and cosines of x and y, draw
CF± OA, CD ± OB, DEJL OA,
i>ff J. i^C prolonged ; then CZ)=
siny, 0D = cosy f and the angle
DCG=ilie Angle UJDC^x. And,
sin (a: - y) = CF= DE- CO.
BE
ng. 9.
OB
= sin a: ; hence, BE'=^ sin a? X OB = sin a: cosy.
CO
CB '
hence, CO = cosa? X CB = cosa: siny.
Therefore, Bm(x — y) = sinx ooey — oobx siny. [8]
Again, cos (x-^y)^ 0F= 0E+ BO.
= cosa; ; hence, OH = cosa; X OD = cosa; cosy.
hence, DG = Bin xx CD = sin x sin y.
Therefore, oo» (i — y) = oosx cosy + ikix Bay. [9]
OE
OD
DG
— = sma;;
In this proof, both x and y are assumed to be acute angles ;
but, whatever be the values of x and y, the same method of
proof will always lead to Formulas [8] and [9], when due
regard is paid to the algebraic signs.
The general application of these formulas may be at once
shown by deducing them from the general formulas established
in § 31, as follows :
It is obvious that (x — y^ + y^^x. If we apply Formulas
[4] and [5] to {x - y) + y, then
46 TEIGONOMETRY.
sin { (^ — y) + y 1 or sin a: = sin {x — y) cosy + cos {x — y) siny,
cos{(a; — y) + y\ or cosa? = cos {x — y) cosy — sin {x — y) siny.
Multiply the first equation by cosy, the second by siny,
sin a: cos y = sin (x — y) cos' y + cos (x — y) sin y cos y ,
cosa: siny = — sin (a? — y) sin^y + cos (x — y) siny cosy ;
whence, by subtraction,
sin a; cosy — cosa? sin y = sin (a; — y) (sin*y + cos'y).
But sin'y + cos^y = 1 ; therefore, by transposing,
sin (^ — ■ y) = sin a? cosy — cosa? siny.
Again, if we multiply the first equation by sin y, the second
equation by cosy, and add the results, we obtain, by reducing,
cos (x — y) = cos a? cos y + sin a? sin y.
Therefore, Formulas [8] and [9], like [4] and [5], from which
they have been derived, are universally true.
From [8] and [9], by proceeding as in § 31, we obtain
taii(x-y) = ,^f-fy . [10]
^ •'^ 1 + tanxtany
^ oatxooty + l ^^^
^^ ooty — ootx ^ -'
Formulas [4] -[11] may be combined as follows:
sin (x dty)^= sin a? cosy db cosa: siny,
cos (xzizy) = cosa: cosy =F sin a: siny,
. / V tana:±tany
tanfa: ± y) = = — r — ^»
•^ l=^tana:tany
V
Y-
j. { N __ cota: coty =F 1
^ ^ coty dr cota:
GONIOMETRY. 47
§33. Functions of Twice an Angle.
If, in FormulaB [4] -[7], y = a:, they become :
8m2z = 2Bmxooez. [12] ooe2z = ooB*z— nn^z. [13]
tan2z = :i^. [14] oot2z = ?^^ [15]
By these formulas the functions of twice an angle are found
when the fanctions of the angle are given.
§ 34. Functions of Half an Angle.
Take the formulas
cos*a? + sin* a? = 1 [1]
cos'a? — sin* a: = cos 2a? [18]
Subtract, 2 sin* a: = 1 — cos 2ir
Add, 2 cos*a? = 1 + cos 2 a?
Whence
11 — co82a: 11
emx==dz'^ ^ 1 cosa:=±-o-
+ cos2a?
2
These values, if z is put for 2Xf and hence i z for x, become
rin}. = ±.^JL:^ [16] cciM=±^-+fi [17]
Hence, by division (§ 27),
tanjE = ±J?H^. [18] oot}E=±Jf^^ [19]
By these formulas the fanctions of half an angle may be
computed when the cosine of the entire angle is given.
The proper sign to be placed before the root in each case
depends on the quadrant in which the angle \z lies. (§ 24.)
Let the student show from Formula [18] that
tan } -B =xh^- (See page 19, Note 2.)
48 TEIGONOMETEY.
§36. Sums and Differences op Functions.
From [4], [5], [8], and [9], by addition and subtraction :
sin (a? + y) + sin (a; — y) = 2 sin a; cos y,
sin (ar + y) — sin (a; — y) = 2co8a? siny,
cos(a;-f-y) + cos(a? — y)= 2cosa;cosy,
cos (a: + y) — cos (a: — y) = — 2 sin a: sin y ;
or, by making x + y^A^ and x — y=^B,
and therefore, x = \{A + B), and y = ^ (j4 — £),
BinA + BinB= 28ini(A + B)ooBi(A-B). [20]
«inA-8inB=- 2cosi(A + B)Bini(A-B). [21]
cobA + cobB= 2coBi(A + B)oo8i(A-B). [22]
coBA-coBB = ~2Bini(A + B)Bin|(A-B). [23]
From [20] and [21], by division, we obtain
gj4±^ = tanK^ + 5)cotK^-£);
sm A — sin -o
or. since coti(^-5) = ^--^^L-^^.
sinA + sinB _ tan^(A + B) r^jn
8inA-BinB~"tani(A-By ^ ■'
(\^^ 1^ ' <?>•* EiERCISE XI.
1. Find the value of sin(a? + y) and cos(a: + y), when sin a:
= I, cosar = |, siny = ^, cosy = ||.
2. Find sin (90^ - y) and co8(90'-y)by making a; = 90°
in Formulas [8] and [9].
Find, by Formulas [4]-[ll], the first four functions of:
3. 90* + y. 8. 360* -y. 13. -y.
4. 180* -y. 9. 360* + y. 14. 45* ~y.
5. 180* + y. 10. a: -90*. 15. 45* + y.
6. 270* -y. 11. a: -180*. 16. 30* + y.
7. 270* + y. 12. a; -270*. 17. 60* -y.
GONIOMETBY. 49
18. Find sin 3ar in terms of sina?.
19. Find cos 3^ in terms of cos ar.
20. Given tan^a? = 1 ; find coax.
21. Given cot ^x = V3 ; find sina?.
22. Given sina; = 0.2 ; find sin^^a? and cos^a?.
23. Given cosa; = 0.5 ; find coe2a; and tan2a;.
24. Given tan 45** = 1 ; find the functions of 22* 30^.
26. Given sin 30** = 0.5 ; find the functions of 15'.
26. Prove that tanlS^ = ^^^ ^gl + '^" ^I '
C0833 +cos3**
Prove the following formulas :
27. 8in2ar=-^^5E^. 29. tanix = : «^°^
l+tan*a? ' l+cosa?
28. co82a;= ;-'^A 80. cotla;=: '^^^
l + tan*a? 1— cos*
31. sin^a: d= cos^a? = VI ± sin a:.
32. — -^ = ± tana: tan y.
cot a: db cot y ^
33. tan(45^-:r) = i:^^5E£.
1+tana;
If A, By (7 are the angles of a triangle, prove that :
34. sin jl + sin J? + sin C^=4cos^-4co8^^cos-J'(7.
35. cos-4 + co8^ + co8(7=l + 48in^J.8in-J'jBsin^C.
36. tan^ + tanj5 + tanC'=tan-4xtanJ?XtanC
37. coti^ + coti^ + cotiC=cot|^ X cotJJSx cotJC.
Change to forms more convenient for logarithmic computa-
tion:
38. cota? + tana;. 43. 1 + tana: tan y.
39. cota: — tana:. 44. 1 — tana: tan y.
40. cota; + tAny. 45. cota;coty+l.
41. cotar — tany. 46. cotarcoty — 1.
42 Lzi22^^. 47 tan a: + tan y
l + co82a: ' cot a; + cot y
CHAPTER IV.
THE OBLIQUE TRIANGLE.
§ 36. Law of Sines.
Let A, B, C denote the angles of a triangle -4^(7 (Figs. 30
and 31), and a, J, c, respectively, the lengths of the opposite
sides.
Draw 0D1.AB, and meeting AB (Fig. 30) or AB pro-
duced (Fig. 31) at D. Let CD = A.
B A
In both figures, - = sin -4.
In Fig. 30, *
In Fig. 31,
- = sin B,
= sin (180^ - jB) = sin B,
Therefore, whether h lies within or without the triangle,
we obtain, by division,
a sinA
b sinB
[26]
THE OBLIQUE TRIANGLE.
51
By drawing perpendiculars from the vertices A and B to
the opposite sides we may obtain, in the same way,
c sinC
a_
c
. sin^
sinC
Hence the Law of Sines, which may be thus stated :
T/ie sides of a triangle are proportioned to the sines of the
opposite angles.
If we regard these three equations as proportions, and take
them by alternation, it will be evident that they may be writ-
ten in the symmetrical form,
sin^ sin^ sin (7
Each of these equal ratios has a simple geometrical mean-
ing which will appear if the Law of Sines is proved as follows :
Circumscribe a circle about the triangle ABC (Fig. 32),
and draw the radii 0-4, OB^ 00;
these radii divide the triangle into
three isosceles triangles. Let H
denote the radius. Draw OM
1^ BO By Geometry, the angle
BOO =2A; hence, the angle
BOM=A,i}ienBM=IiBmBOM
= J2 sin A,
,\BOoTa = 2IiBmA.
In like manner, i = 2i?sin-B,
and c = 2-R sin CI Whence we
obtain
a h
Fig. 81
2i2=:
sin J. BinB sin (7
That is : The ratio of any side of a triangle to the sine of the
opposite angle is numerically/ eqv/xl to the diameter of the cir-
cumscribed circle.
52 TBIQONOMETRY.
§ 87. Law of Cosines.
This law gives the value of one side of a triangle in terms
of the other two aides and the angle included between them.
In Figs. 30 and 31, a* = }f + BU^.
In Fig. 30, BD =c-AD\
in Fig. 31, BD = AD — c ;
in both cases, BD^ = AI^ -2cxAD+(?,
Therefore, in all cases, a* = A* + AD^ + c* — 2 c X AD,
Now, V + AD^ = h\
and AD =icos-4.
Therefore, a« = V + c«-2bcooflA. [26]
In like manner, it may be proved that
i» = a« + c«-2accosJ?,
(? = c? + V-2dbQo%a
The three formulas have precisely the same form, and the
law may be §^ted as follows :
The square of any side of a triangle is equal to the sum of
the squares of the other two sides, diminished hy twice the
product of the sides and the cosine of the included angle,
§ 38. Law of Tangents.
By § 36, a\b = %iuA : sin JB ;
whence, by th^ Theory of Proportion,
g — 6 __ sin ^ -— sin B
a-\-b sin-4 + sin^
But by [24], page 48,
sin A — sin ^ _ tan \{A — B)
sin^ + sin^ tQ,ni{A + B)
Therefore,
a-b_ taaKA-B) ^g?]
a + b taii}(A + B)
THE OBLIQUE TBIAKGLE. 53
By merely changing the letters,
a + c tanJ(^+CO' b + c tanJ(jB+C)'
Hence the Law of Tangents :
The difference of two aides of a triangle is to their sum as the
tangent of half the difference of the opposite angles is to the tan-
gent of half their sum.
Note. If in [27] 6 > a, then B>A. The formula is Btill trne, but to
avoid negative quantities, the formula in this case should be written
6-0 tan}(^~^)
6 + a-tanl(-B + ^)"
EZEBCISE XII.
1. What do the formulas of § 86 become when one of the
angles is a right angle ?
2. Prove by means of the Law of Sines that the bisector of
an angle of a triangle divides the opposite side into parts pro*
portional to the adjacent sides.
3. What does Formula [26] become when ^ = 90**? when
-4 = 0**? when A = 180**? What does the triangle become in
each of these cases ?
Note. The case where A ~ 90^ explains why the theorem of { 37 is
sometimes termed the Generalized Theorem of Pyihagonu.
4. Prove (Figs. 30 and 31) that whether the angle B is
acute or obtuse c = a cos JB + J cos ^. What are the two sym-
metrical formulas obtained by changing the letters? What
does the formula become when B = 90** ?
5. From the three following equations (found in the last
exercise) prove the theorem of § 87 :
c =acos5+6cos-4,
i = a cos (7 + c cos J.,
a = i cos C + c cos jB.
HufTT. Multiply the first equation by e, the second by h, the third
by a ; then from the first subtract the mm of the second and third.
54
TEIGONOMETEY.
6. In Formula [27] what ia the maximum value of } {A--B) ?
oii(^A+B)?
7. Find the form to* which Formula [27] reduces, and
describe the nature of the triangle, when
(i.) C=90^ (ii.) X-jB = 90^,and5=0:
§ 39. The Given Paets op an Oblique Teiangle.
The formulas established in §§ 36-38, together with the
equation A + JB-\-C= 180®, are sufficient for solving every
case of an oblique triangle. The three parts that determine
an oblique triangle may be :
I. One side and two angles ;
II. Two sides and the angle opposite to one of these sides ;
III. Two sides and the included angle ;
IV. The three sides.
In all cases let A.B^O denote the angles, a, i, c the sides
opposite these angles respectively.
§ 40. Case I.
Given one side a, and two angles A and B; find the remain-
ing parts C, J, and c,
1. (7=180®-(^ + ^).
a sin 5 a
'o J__sinjB
^. ' 7 i
a am A
o £ _ sin (7 .
;.i =
>c = -
ainA
a sin (7
= ^^X8inji5.
sin^
■XsinC:
a sin-4 * sin J. sin J.
Example, a = 24.31, A = 45® 18', ^ = 22® 11'.
The work may be arranged as follows :
a= 24.31
A= 45® 18'
B== 22® 11'
A + B= 67®29'
C=112®31'
loga = 1.38578
cologsin^ = 0.14825
logsinjB = 9.57700
logJ = 1.11103
i = 12.913
= 1.38578
= 0.14825
logsin (7= 9.96556
logc = 1.49959
c = 31.593
THE OBLIQUE TRIANGLE. 55
Exercise XIII.
1. Givena = 500, X==10n2', J? = 46*36';
find C= 123^2', i = 2051.48, c = 2362.61.
2. Given a = 795, ^ = 79^59', 5 = 44^41';
find C= 55^ 20', i = 567.688, c = 663.986.
3. Given a = 804, ^ = 99*55', 5 = 45^';
find C= 35^ 4', b = 577.313, c = 468.933.
4. Given a = 820, ^ = 12*»49', jB = 141*59';
find 0= 25* 12', b = 2276.63, c = 1573.89.
5. Given c = 1005, ^ = 78*19', J? = 54* 27';
find C= 47* 14', a = 1340.6, b = 1113.8.
6. Given* = 13.57, .5=13*57', C=57*13';
find A = 108* 50', a = 53.276, c = 47.324.
7. Given a = 6412, X = 70*55', C=52*9';
find^ = 56* 56', b = 5685.9, c = 5357.5.
8. Given 6 = 999, .4 = 37*58', C=65*2';
find jB = 77*, a = 630.77, c = 929.48.
9. In order to determine the distance of a hostile fort A
from a place B, a line BC and the angles ABC and BCA
were measured, and found to be 322.55 yards, 60*34', and
56* 10', respectively. Find the distance AB.
10. In making a survey by triangulation, the angles B and
C of a triangle ABO were found to be 50*30' and 122*9',
respectively, and the length -BCis known to be 9 miles. Find
AB and AC,
11. Two observers 15 miles apart on a plain, and facing
each other, find that the angles of elevation of a balloon in
the same vertical plane with themselves are 55* and 58*,
respectively. Find the distance from the balloon to each
observer, and also the height of the balloon above the plain.
12. In a parallelogram given a diagonal d and the angles
X and y which this diagonal makes with the sides. Find the
sides. Compute the results when d= 11.237, x= 19* 1', and
y = 42*54.
56 TRIGONOMETRY.
13. A lighthouse was observed from a ship to bear N. 34® E. ;
after sailing due south 3 miles, it bore N. 23® E. Find the dis-
tance from the lighthouse to the ship in both positions.
Note. The phrase to bear N. 34° E. means that the line of sight to
the lighthouse is in the north-east quarter of the horizon, and makes,
ith a line due north, an angle of 34°.
14. In a trapezoid given the parallel sides a and J, and the
angles x and y at the ends of one of the parallel sides. Find
the non-parallel sides. Compute the results when a = 15,
J = 7, ar = 70®, y = 40®.
Solve the following examples without using logarithms :
15. Given i = 7.07107, ^ = 30®, (7= 105® ; find a and <?.
16. Given (7 = 9.562, ^ = 45®, J? = 60®; find a and i.
17. The base of a triangle is 600 feet, and the angles at the
base are 30® and 120®. Find the other sides and the altitude.
18. Two angles of a triangle are, the one 20®, the other 40®.
Find the ratio of the opposite sides.
19. The angles of a triangle are as 5 : 10 : 21, and the side
opposite the smallest angle is equal to 3. Find the other
sides.
20. Given one side of a triangle equal to 27, the adjacent
angles equal each to 30®. Find the radius of the circum-
scribed circle. (See § 36, Remark.)
§41. Case II.
Given two aides a and J, and the angle A opposite to the
side a; find the remaining parts jB, C, c.
This case, like the preceding case, is solved by means of
the Law of Sines.
Since ^=^, therefore 8in5 = i^^;
sin^ a a
C=180®-(^ + J5).
A J • c sin C rv p a sin (?
And since - = -: — 7, therefore c — —. — -•
a sin A sin -d
THE OBLIQUE TBIANOLE. 67
When an angle is determined by its sine it admits of two
values, which are supplements of each other (§ 28); hence,
either value of B may be taken unless excluded by the con-
ditions of the problem.
If a > J, then by Geometry A>B, and B most be acute
whatever be the value of A\ for a triangle can have only
one obtuse angle. Hence, there is one, and mly one, triangle
that will satisfy the given conditions.
If a = J, then by (Jeometry A = B; both A and B must be
acute, and the required triangle is iaoscelee.
If a < i, then by Geometry A<B, and A must be acute
in order that the triangle
may be possible. If A is ^/
acute, it is evident from JS:
Fig. 33, where Z BA C=A,
AC=b, OB = CB'=a,
that the two triangles ACB
and ACB' will satisfy the
given conditions, provided X
a is greater than the per- j^'"
pendicular CP\ that is,
provided a is greater than b sin A (§ 10). The angles ABC
and AB'C are supplementary (since Z ABC = Z BB^C)\
they are in isud the supplementary angles obtained from the
formula „• -n h^inA
sin Jo = •
a
If, however, a = i sin ^ = CP (Fig. 33), then sin 5 = 1,
B = 90°, and the triangle required is a right triangle,
li a<b sin A, that is, < CP, then sin B>\, and the tri-
angle is impossible.
These results, for convenience, may be thus stated :
If a > J, or a = 6, or if a = i sin A, One solution.
If a < 6, but > b sin A, and A < 90®, 2\oo solutions.
Ifa<6 and^>90*, or if a<i8in^ and -4<90^
JVb solution.
58
TRIGONOMETRY.
The number of solutions can often be determined by inspec-
tion. If there is any doubt, it may be removed by computing
the value of h sin A,
Or we may proceed to compute log sin B. If log sin -B = 0,
the triangle required is a right triangle. If log sin -B > 0, the
triangle is impossible. If log sin ^ < 0, there is <me solution
when a>h\ there are two solutions when a < i.
When there are two solutions, let B\ C\ d denote the un-
known parts of the second triangle ; then,
^' = 180^-5, (7' = 180*'-(^+jB') = ^--4,
f a sin C
sin -4
Examples.
1. Given a = 16, 5 = 20, ^ = 106**; find the remaining
parts.
In this case a < 6, and A > 90^ ; therefore the triangle is impossible.
2. Given a = 36, i = 80, A= 30° ; find the remaining
parts.
Here we have 5 sin A = 80 x } = 40 ; so that a < 6 sin A, and the
triangle is impossible.
3. Given a=72630, &=117480, ^=80°0'50"; find B, C, c.
a = 72630
h = 117480
il = 80°0'50"
colog a = 5.13888
log h = 5.06996
log sin J. = 9.99337
Here log sin jB > 0.
.'. no solution.
log sin .8 = 0.20221
4. Given a = 13.2, 5 = 15.7, .^ = 57° 13' 15.3"; find .5, C, c.
a = 13.2
6 = 15.7
^ = 57° 13' 15.3"
Here log sin jB = 0,
/. a right triangle.
colog a = 8.87943
log h = 1.19590
log sin J. = 9.92467
log sin J? = 0.00000
i? = 90<>
.-.(7 = 32° 46' 44.7"
c = 6 cos J.
log 6 = 1.19590
log cos -4 = 9.73352
log c = 0.92942
c=8.5
THE OBLIQUE TRIANGLE.
59
5. Given a=767, i = 242, ^ = 36^53'2"; find B, C, c.
a ^167
6-242
^ = 36° 53' 2"
colog a = 7.11520
log b = 2.38382
logsin-i- 9.77830
log sin 5 -9.27732
.5-10*>54'58"
.-. (7-132*12'0"
log a -2.88480
log sin C- 9.86970
cologsiuil- 0.22170
log c- 2.97620
c- 946.675
Here a>b,
and log sin ^ < 0.
.*. one solntion.
6. Given a = 177.01, 6 = 216.45, X=35**36'20"; find the
other parts.
a -177.01
5-216.45
^ = 35° 36' 20"
Here a < 6,
and log sin B<0.
•*• two solutions.
colog a -7.75200
log 5-2.33536
logsin A- 9.76507
log sin ^-9.85243
5-45«23'28"
orl34«36'32"
.••C-99«0'12"
or9«47'8"
log a -2.24800
cologsin^- 0.23493
log sinC- 9.99462
log c- 2.47755
c- 300.29 or
2.24800
0.23493
9.23034
1.71327
51.674
EXEECISE XIV.
1. Determine the number of solutions in each of the fol-
lowing cases :
(i.)a = 80, 5 = 100, ^ = 30*. i
(ii.)a = 50, i = 100, ^ = 30^
(iii.) a = 40, i = 100, A = 30^
(iv.) a = 13.4, b = 11.46, A = 77^ 20'.
(v.) a =70, i = 75, ^ = 60^
(vi.) a = 134.16, b = 84.54, £ = 52^ 9' 11".
2. Given a = 840, 6 = 485, ^ = 2^81';
find .3 = 12** 13' 34", C= 146^5' 26", c= 1272.15.
' 3. Given a = 9.399, b = 9.197, A = 120^ 35' ;
find 5 = 57^ 23' 40", C=2^1'20', (? = 0.88525.
4. Given a = 91.06, J = 77.04, ^ = 51** 9' 6";
find J5 = 4r 13', C= ST" 37' 54", c = 116.82.
5. Given a = 55.55, 6 = 66.66, J? = 77M4'40";
find A = 54** 31' 13", (7= 47** 44' 7", c = 50.481.
60 TBiaOWOMETBY.
6. Given = 309, 5 = 360, ^ = 21" 14' 25";
find 5 = 24' 57' 54", C= 133° 47' 41", c = 615.67,
^=155° 2' 6", C"=3-43'29", c'= 55.41.
7. Given o = 8.716, J = 9.787, ^ = 38' 14' 12";
find 5 = 44° 1' 28", C= 97° 44' 20", c = 13.954,
5'= 135° 58'32", C"= 5° 47' 16", c'= 1.4202.
8. Given a = 4.4, J = 5.21, ^ = 57° 37' 17";
find .5 = 90°, (7= 32° 22' 43", c = 2.79.
9. Given a = 34, 5 = 22, 5 = 30° 20';
find ^ = 51° 18' 27", C= 98° 21' 33", c = 43.098,
^'=128° 41' 33", C"=20°58'27", c'= 15.593.
10. Given 5 = 19, c = 18, C=15°49';
find JB = 16° 43' 12", ^ = 147° 27' 48", a = 35.52,
5'= 163° 16' 48", ^'=0°54'12", a'= 1.0412.
11. Given a = 75, 5 = 29, J? = 16° 15' 36" ; find the differ-
ence between the areas of the two corresponding triangles.
12. Given in a parallelogram the side a, a diagonal d, and
the angle A made bj the two diagonals ; find the other diag-
onal and the other side.
Special case: o = 35, rf=63, ^ = 21° 36' 30".
§ 42. Case III.
Oiven two sides a and b and the included angle C; find the
remaining parts A,B, and c.
Solution I. The angles A and B may both be found by
means of Formula [27], § 38, which may be written
tan } (^ - 5) = 5^ X tan } (^ -I- 5).
a-f-o
Since iiA + B) = }(180°- 0), the value of \{A + 5) is
known ; so that this equation enables us to find the value of
i(^A-B). We then have
i{A + B) + i{A-B) = A.
and i{A + B)-k{A-B) = B.
THE OBLIQUE TBIAKQLE.
61
After A and B are known, the side c may be found by the
Law of Sines, which gives its valae in two ways, as follows :
a sin (7 ^ ^ isinC
c = -
or c-
sin -4 ' "' " sin ^
Solution II. The third side c may be found directly from
the equation (§ 37)
c= Va" + i«-2aAco8C7;
and then, by the Law of Sines, the following equations for
computing the values of the angles A and B are obtained :
6in^ = aX
sinC
sin-B = 6x
sin (7
Solution III. If, in the triangle ABO (Fig. 34), BD is
drawn perpendicular to the side
AC, then
Now
and
/.tan -4 =
AD AC- DC
BD = aQinC (§10),
DC = acoQa
a sin (7
b — acoaC
By merely changing the letters,
J sin (7
tanjB = .
a — 6 cos C.
It is not necessary, however, to use both formulas. When
one angle, as -4, has been found, the other, -B, may be found
from the relation A + B+C= 180^
When the angles are known, the third side is found by the
Law of Sines, as in Solution I.
Note. When all three unknown parts are required, Solution I. is the
most convenient in practice. When only the third side e is desired, Solu-
tion II. may be used to advantage, provided the values of a* and 6* can
be readily obtained without the aid of logarithms. But Solutions II.
and III. are not adapted to logarithmic work.
62
TEIGONOMETEY.
Examples.
1. Given o = 748, i = 375. C= 63" 35' 30" ; find A, B,
and c.
a + 6-
1123
a-6-
373
(A+5)=
1160 24'3(y'|
\{A^B)^
58<»
12' 15''
}(^-5)-
28«
10' 52"
il»
86°
23' 7"
jB-
30°
1'23"
log(a- 6) -2.57171
colog(a + 6) « 6.94961
log tan J(^+-B)-a20766
log tan J (il-^)- 9.72898
J(^--B)-28n0'52"
log 6 =2.57403
log sin (7= 9.95214
colog sin 5 =0.30073
logc-2.82690
c- 671.27
Note. In the aboye Example we xue the angle B in finding the side
c, rather than the angle A, becaxue A is near 90°, and therefore its sine
should be avoided.
. 2. Given a = 4, c = 6, jB= 60** ; find the third side I.
Here Solution II. may be used to advantage. We have
6- Va» + c»- 2ac cos .5-V16 + 36-24 -V28;
log 28 = 1.44716, log V28 - 0.72358, \/28 « 5.2915 ;
that is, & = 5.2915.
2.
3.
4.
6.
EZEBCISE XV.
Given o = 77.99,
6 = 83.39, C
= 72° 15'; [J38.N.
find^ = 6ri5',
.3 = 56° 30',
= 95.24.
Given 6 = 872.5,
c = 632.7,
^ = 80°;
find5 = 60»45'.
C=:39n5',
a = 984.8.
Given a = 17,
6 = 12,
C=59°17';
find ^ = 77° 12' 53",
5 = 43° 30' 7",
c = 14.987.
Given J = V5,
c = V3,
^ = 35°53';
find5 = 93»28'36".
(7=50° 38' 24",
a = 1.318.
Given a = 0.917,
6 = 0.312,
C=33°7'9";
find ^ = 132° 18' 27",
^ = 14°34'23",
c = 0.67748.
Given a = 13.715,
c = 11.214,
5 =15° 22' 36";
find ^ = 118' 55'49",
C= 45° 41' 35",
6 = 4.1552.
Given h = 3000.9,
c = 1587.2,
.4 = 86° 4' 4";
find 5 =65° 13' 52",
C=28°42'4",
a = 8297.2.
THE OBLIQUE TRIANGLE. 63
8. Given a = 4627, J = 3465, C=66**6'27";
find A = 68^ 29' 15", £ = 45** 24' 18", c = 4449.
9. Given a = 55.14, i = 33.09, C=30^24';
find ^ = 117^ 24' 33", jB = 32^ 11' 27", c = 31.431.
10. Given a = 47.99, i = 33.14, C= 175^9' 10" , •
find ^ = 2^ 46' 8", J? = r54'42", c = 81.064.
11. If two sides of a triangle are each equal to 6, and the
included angle is 60^, find the third side.
12. If two sides of a triangle are each equal to 6, and the
included angle is 120°, find the third side.
13. Apply Solution I. to the case in which a = 5 or the
triangle is isosceles.
14. If two sides of a triangle are 10 and 11, and the in-
cluded angle is 50**, find the third side.
15. If two sides of a triangle are 43.301 and 25, and ^he
included angle is 30°, find the third side.
16. In order to find the distance between two objects A
and £ separated by a swamp, a station C was chosen, and the
distances CA = 3825 yards, CB = 8475.6 yards, together with
the angle ACJ3 = 62° 31', were measured. Find the distance
from -4 to -B.
17. Two inaccessible objects A and B are each viewed
from two stations Cand D 562 yards apart. The angle ACJB
is 62° 12', BCD 41° 8', ADB 60° 49', and ADO 34° 51';
required the distance AB,
18. Two trains start at the same time from the same station,
and move along straight tracks that form an angle of 30°, one
train at the rate of 30 miles an hour, the other at the rate of
40 miles an hour. How far apart are the trains at the end
of half an hour?
19. In a parallelogram given the two diagonals 5 and 6,
and the angle that they form 49° 18'. Find the sides.
20. In a triangle one angle = 139° 54', and the sides form-
ing the angle have the ratio 5 : 9. Find the other two angles.
64 TRIGONOHCETRY.
§ 43. Case IV.
Oiven the three sides a, J, c; find the angle^ A, jB, C.
The angles may be found directly from the formulas estab-
lished in § 37. Thus, from the formula
we have cos A = ■ "*" ^""^ ■'
From this equation formulas adapted to logarithmic work
are deduced as follows :
For the sake of brevity, let a + J + c = 25 ; then J + e? — a
= 2(5 — a), a — J + c = 2(5 — i), and a + i — c=2(5— c).
Then the value of 1 — cos -4 is
2bc 2bc 2hc
2bc be
and the value of 1 + cos A is
2bc 2bc 2bc
__ (b + c + a){b + C'-d) _ 28{8'-a)
2bc be '
But from Formulas [16J and [17], § 34, it follows that
1 — cos -4 = 2 sin* J ^, and l + cos-4 = 2cos*}^.
.'.^AnHA^ ^i'-^)^'-") , and 2cos»M = 5%l^.
be be
whence sin J A = -J (' ~ ^) (' ~ °) , [28]
008 U = -^"^^, [29]
and therefore tan } A = J(B-b)(a-o) j-g^ ^
^ S ( B ""~ ft I
THE OBLIQUE TRIANaLE. 65
By merely changing the letters,
. - „ l(8 — a)(8 — c) . , ^ f(» — a)(« — 6)
^i^ = f S -' '^i<^=^\- ^ -'•
tanJ5 = J5^S^ tanJ(7=Ji^SIES.
There is then a choice of three different formulas for finding
the value of each angle. If half the angle is very near 0®,
the formula for the cosine will not give a very accurate result,
because the cosines of angles near 0^ differ little in value ; and
the same holds true of the formula for the sine when half
the angle is very near 90^. Hence, in the first case the
formula for the sine, in the second that for the cosine, should
be used.
But, in general, the formulas for the tangent are to be
preferred.
It is not necessary to compute by the formulas more than
two angles ; for the third may then be found from the equation
A + £ + C=lSO^.
There is this advantage, however, in computing all three
angles by the formulas, that we may then use the sum of the
angles as a test of the accuracy of the results.
In case it is desired to compute all the angles, the formulas
for the tangent may be put in a more convenient form.
The value of tan i A may be written
l (s-a)(s-b)(8-c) 1 Us - a) {3 -b) {,-<>)
\ «(«-«)• a-aS s
Hence, if we put
f(g-a)(a-b)(B-o) pjj
4
8
we have tan } A = • [B2]
8 — a
66
TRIGONOMETRY.
In like manner,
tanjj5 = -
^^*
tanJC=-
8 — C
Examples.
1. Given a = 3.41, i = 2.60, c = 1.58; find the angles.
Using Formula [30], and the corresponding formula for tan ^^, we
may arrange the work as follows :
a = 3.41
, b = 2.60
c = 1.58
28-Y.59
8 = 3795
a -a = 0.385
a- J = 1.195
a- c = 2.215
colog a = 9.42079
colog(a- a) = 0.41454
log(a- 6) -0.07737
log (a -c) = 0.34537
2 )0.25807
log tan Jil = 0.12903
Jil= 53^23' 20"
il- 106° 46' 40"
colog a =
log(a-a) =
colog (a — J) =
log(a-c) =
9.42079-10
9.58546-10
9.92263 - 10
0.34537
2 )19.27425 - 20
logtan}P= 9.63713-10
iB= 23° 26' 37"
£=- 46° 53' 14"
.A + B^ 153° 39' 54". and C- 26° 20^ 6".
2. Solve Example 1 by finding all three angles by the use
of Formulas [31] and [32].
Here the work may be compactly arranged as follows, if we find
logtanjil, etc., by subtracting log (a — a), etc., from logr instead of
adding the cologarithm :
a = 3.41
b = 2.60
c = 1.58
2a = 7.59
a = 3.795
s -a =0.385
8-J = 1.195
*-c = 2.215
log(8- a) = 9.58546
log(a- J) = 0.07737
log(a-c) = 0.34537
colog a = 9.42079
log r» = 9.42899
logr -9.71450
logtanJJ. = 10.12903
log tan} J? = 9.63713
log tan } C =
}5 =
9.36912
53° 23' 20"
23° 26' 37"
13° 10' 3"
^ = 106° 46' 40"
.5= 46° 53' 14"
(7= 26° 20' 6"
2a = 7.590 (proof).
Proof, il + P + a = 180° 0' 0"
Note. Even if no mistakes are made in the work the sum of the
three angles found as above may dififer very slightly from 180° in conse-
quence of the fact that logarithmic computation is at best only a method
of close approximation. When a difference of this kind exists it should
be divided among the angles according to the probable amount of error
for each angle.
THE OBLIQUE TEIANGLE.
67
EXEECISE XVI.
Solve the following triangles, taking tlie three sides as the
given parts :
1
a
b
e
A
B
51
65
20
38*62' 48"
126* 52' 12"
14* 15*
2
78
101
29
32*10' 54"
136* 23' 50"
11*25^16"
3
111
145
40
27*20^32"
143* 7' 48"
9* 31' 40"
4
21
26
31
42* 6' 13"
56* 6' 36"
81* 47' 11"
6
19
34
49
16* 25' 36"
30*24'
133* 10' 24"
6
43
50
57
46*49*34"
5r59'44"
75* 10* 42"
7
37
58
79
26* 0*28"
43* 25' 20"
110* 34' 12"
8
73
82
91
49* 34' 58"
58* 46' 58"
71*38' 4"
9
14.493
55.4363
66.9129
8*20*
33*40'
138*
10
v/5
V6
V7
51* 53' 12"
59*31*48"
68*35'
11. Given a = 6f i = 8, c = 10 ; find the angles.
12. Given a = 6, i = 6, c = 10 ; find the angles.
13. Given a = 6, J = 6, c = 6 ; find the angles.
14. Given a = 6, J = 5, <? = 12 ; find the angles.
15. Given a = 2, J = V6, c==VS-l; find the angles.
16. Given a = 2, J = V6, c = VS + 1 ; find the angles.
17. The distances between three cities A, j5, and C are as
follows : AB = 165 miles, A C= 72 miles, and £0= 185 miles.
JS is dvCQ east from A, In what direction is (7 from A ? What
two answers are admissible ?
18. Under what visual angle is an object 7 feet long seen
by an observer whose eye is 5 feet from one end of the object
and 8 feet from the other end ?
19. When Formula [28] is used for finding the value of an
angle, why does the ambiguity that occurs in Case II. not
exist?
20. If the sides of a triangle are 3, 4, and 6, find the sine
of the largest angle.
21. Of three towns A, B, and C, A is 200 miles from B
and 184 miles from C, B is 150 miles due north from C\ how
far is A north of (7?
TRIGONOMETr.Y.
§44. Area op a Triangle.
If F denote the area of the triangle ABQ (Fig. 30 or 31,
page 50), then, by Geometry,
F=\cy.OB.
By §10, aZ) = asin^.
Therefore, F = JaoBinB. [33]
And, in like manner,
jP=}a5sin(7 and -P=}J(?sin^.
That is : The area of a triangle is equal to half the product
of two sides and the sine of the included angle.
By Formula [33] the area of a triangle may be found directly
when two sides and the included angle are given ; in the other
cases the formula may be used when these parts have been
computed.
When the three sides of a triangle are given, as in Case IV.,
a formula for its area may be found as follows :
By §33, sin^=^2sin}^XcosJ^.
By substituting for sin } B and cos } B their values in terms
of the sides given in § 43,
sin jB = —-s/sis — a) (s — 5) (s — c).
ac
By substituting this value of sin 5 in [33],
F = Vs(B-a)(s-b)(s~o). [34]
If a denote (as in § 36) the radius of the circumscribed
circle, we have, from § 36,
8inj5 = -^.
2jB
By substituting this value of sin B in [33],
F = ||. [35]
THE OBLIQUE TRIANGLE. 69
If r denote the radius of the inscribed circle, and we divide
the triangle into three triangles by lines from the centre of
this circle to the vertices, the altitude of each of the three tri-
angles is equal to r. Therefore,
F = }r(a + b + c) = rs. [36]
By substituting in this formula the value of i^ given in [34],
_l (s-a)(s-b)(s-c) .
wheoce r, in [31] § 43, is equal to the radius of the inscribed
circle.
Exercise XVII.
Find the area :
1. Given a = 4474.5, b = 2164.5, C= 116^ 30' 20".
2. Given 5 = 21.66, c = 36.94, ^ = 66^ 4' 19".
3. Given a = 510, c=173, .5 = 162^ 30' 28".
4. Given a = 408, J = 41, (? = 401.
5. Given a = 40, i = 13, e = 37.
6. Given a =624, i = 205, e = 445.
7. Given b = 149, A = 70* 42' 30", JB = 39* 18' 28".
8. Given a = 215.9, c = 307.7, ^ = 25* 9' 31".
9. Given 5 = 8, c = 5, ^ = 60^
10. Given a =7, c = 3, ^ = 60^
11. Given a =60, .B = 40* 35' 12", area = 12; find the
radius of the inscribed circle.
12. Obtain a formula for the area of a parallelogram in
terms of two adjacent sides and the included angle.
13. Obtain a formula for the area of an isosceles trapezoid
in terms of the two parallel sides and an acute angle.
14. The sides of a quadrilateral, taken in order, are 2416,
1712, 1948, and 2848 ; the angle between the first two is 30*,
and that between the last two 150* ; find its area.
15. The base of an isosceles triangle is 20 and its area is
100 -^ V3 ; find its angles.
70 TRIGONOMETBY.
Exercise XVIII.
1. From a ship sailing down the English Channel the Eddy-
stone was observed to bear N. 33® 45' W. ; and after the ship
had sailed 18 miles S. 67® 30' W. it bore N. IV 15' E. Find
its distance from each position of the ship.
2. Two objects, A and jB, were observed from a ship to
be at the same instant in a line bearing N. 15® E. The ship
then sailed north-west 5 miles, when it was found that A bore
due east and JB bore north-east. Find the distance from A
to 5.
3. A castle and a monument stand on the same horizontal
plane. The angles of depression of the top and the bottom of
the monument viewed from the top of the castle are 40® and
80® ; the height of the castle is 140 feet. Find the height of
the monument.
4. If the sun*s altitude is 60®, what angle must a stick make
with the horizon in order that its shadow in a horizontal
plane may be the longest possible?
5. If the sun's altitude is 30®, find the length of the longest
shadow cast on a horizontal plane by a stick 10 feet in length.
6. In a circle with the radius 3 find the area of the part
comprised between parallel chords whose lengths are 4 and 5.
(Two solutions.)
7. A and B, two inaccessible objects in the same horizontal
plane, are observed from a balloon at C and from a point D
directly under the balloon, and in the same horizontal plane
with A and £. If CD = 2000 yards, Z ACD = 10® 15' 10",
Z BCD = 6® 7' 20", Z ADD = 49® 34' 50", find AB.
8. A and B are two objects whose distance, on account of
intervening obstacles, cannot be directly measured. At the
summit C of a hill, whose height above the common horizontal
plane of the objects is known to be 517.3 yards, Z ACB is
found to be 15® 13' 15". The angles of elevation of C viewed
from A and B are 21® 9' 18" and 23® 15' 34" respectively.
Find the distance from A to B,
SPHERICAL TRIGONOMETRY.
CHAPTER V.
THE BIOHT SPHERICAL TBIANOLE.
§ 45. Inteoduotion.
The object of Spherical Tngonometry is to show how spheri-
cal triangles are solved. To solve a spherical triangle is to
compute any three of its parts when the other three parts are
given.
The sides of a spherical triangle are arcs of great circles.
They are measured in degrees, minutes, and seconds, and
therefore by the plane angles formed by radii of the sphere
drawn to the vertices of the triangle. Hence, their measures
are independent of the length of the radius, which may be
assumed to have any convenient numerical value; as, for
example, unity.
The angles of the triangle are measured by the angles made
by the planes of the sides. Each angle is also measured by
the number of degrees in the arc of a great circle, described
from the vertex of the angle as a pole, and included between
its sides.
The sides may have any values from 0** to 360® ; but in this
work only sides that are less than 180® will be considered.
The angles may have any values from 0® to 180®.
If any two parts of a spherical triangle are either both less
than 90® or both greater than 90®, they are said to be alike in
kind; but if one part is less than 90®, and the other part
greater than 90®, they are said to be unlike in kind.
72
TRIGONOMETRY.
Spherical triangles are said to be isosceles, equilateral, equi-
angular, right, and oblique, under the same conditions as
plane triangles. A right spherical triangle, however, may
have one, two, or three right angles.
When a spherical triangle has one or more of its sides equal
to a quadrant, it is called a quadiantal triangle.
It is shown in Solid Geometry, that in every spherical tri-
angle
I. I%e sum of the sides is less than 360^.
II. The sum of the angles is greater than 180®.
III. Iff from the vertices as poles, arcs of great circles are
described, another spherical triangle is formed so related to the
first triangle thai the sides of each triangle are supplements of
the angles opposite to them in the other triangle.
Two such triangles are said to be polar with respect to each
other.
Let A, B, C (Fig. 35) denote the angles of one triangle ; a,b,c
the sides opposite these angles
respectively; and let ^^-B'jC*
and a\ b\ c' denote the cor-
responding sides and angles
of the polar triangle. Then
the above theorem gives the
six following equations :
A +a' = 180^
5+J' = 180^
C +c' = 180^
^' + a=180^
B' + b =180^
cr+c =180^
Exercise XIX.
1. The angles of a triangle are 70", 80*", and 100^ find the
sides of the polar triangle.
THE BIGHT SPHEKICAL TBIANGLE.
73
2. The sides of a triangle are 40**, 90**, and 125*; find the
angles of the polar triangle.
3. Prove that the polar of a qnadrantal triangle is a right
triangle.
4. Prove that, if a triangle have three right angles, the sides
of the triangle are quadrants.
5. Prove that, if a triangle have two right angles, the sides
opposite these angles are quadrants, and the third angle is
measured by the number of degrees in the opposite side.
6. How can the sides of a spherical triangle be found in
units of length, when the length of the radius of the sphere is
known ?
7. Find the lengths of the sides of the triangle in Example
2, if the radius of the sphere is 4 feet.
§46. FoBMULAs Relatiko TO Right Tbiangles.
As is evident from § 45, Examples 4 and 5, the only kind of
right triangle requiring further investigation is that which
contains mly (me right angle.
Let ABC (Fig. 36) be such a right triangle, and let A, B, C
denote the angles of the tri
angle ; a, J, c, respectively, the
opposite sides.
Let C be the right angle, and
for the present suppose that
each of the other parts is less
than 90*.
Let planes be passed through
the sides, intersecting in the
radii OA, OB, and 0C\ and
for the sake of simplicity let
the radius of the sphere be nc. as.
taken equal to 1.
Also, let a plane perpendicular to OA be passed through
B, cutting OA at -ffand OCat D, Join BE, BD, and DE,
74
TRIGONOMETRY.
BE and DE are each ±0-4 (Geom. § 454) ; therefore ZBEL
= -4. The plane 5Z)^ is ± the
plane ^OC(Geoni. §472); hence
BD, which is the intersection
of the planes BBEa^nd BOC, is
± the plane ^OC(Geom. § 475),
therefore ± OCand BE. Now
coBc = 0E= OB X cosS,
^ and OB = cos a.
.-. coBO = coBaooBb. [37]
sina = BB = BEx sin A,
Fig. 36 (»<«).
Therefore,
[•88]
and BE=^Bmc,
Bina = smo8inA
changing letters, gin b = sin o sin B .
Again, BE = BE x cos A,
and also BE = OB X sin J ;
hence, BE XcoaA = OB X sin b ;
that is, sin <? cos -4 = cos a sin J ;
whence cos ^ = cos a sin b esc c*
By substituting in this formula the value of cos a, obtained
from [37], we obtain
008 A = tan b cot
In like manner, oos B = tan a oot c.
[39]
And by substituting in the same formula the valueuof sin 5,
as given in [38], we obtain
008 A =008 a BinB 1
In like manner, oos B = oos b Bin A /
Also, BB = BE X tan A,
BB = sin a,
BE=cosa sin J;
therefore, sin a = cos a sin b tan A ;
[40]
* To avoid fractions, cscc is written in place of its equal, ■
THE RIGHT SPHEEICAL TRIANGLE.
75
8i]ib = taiiaootA
Bina=taiibootB
whence,
and, similarly, Bina=taiibootBJ ^ -^
If in [87] we substitute for cos a and cosb their values from
[40], we obtain
oo8o = ootAootB. [42]
In deducing these formulas, it has been assumed that all
the parts of the triangle, except the right angle, are less than
90**. But the formulas also hold true when this hypothesis is
not fulfilled.
Let one of the legs a be greater than 90^, and construct a
figure for this case (Fig. 37) in the same manner as Fig. 36.
B
The auxiliary plane JSDE will now cut both CO and AO
produced beyond the centre ; and we have
cos(180 — = OJS= OB X cos6 = cos(180 - a) cos J,
a result which reduces to [37] if we substitute ~ cose in place
of cos (180 — c) and — cosa in place of cos(180--a). Like-
wise, the other formulas, [38] -[42], hold true in this case.
Again, suppose that both the legs a and h are greater than
90^ In this case the plane BDE (Fig. 38) will cut GO pro-
duced beyond 0, and AO between A and 0; and we have
cose ^OE^OBx cos (180 - h)
= cos (180 - a) cos (180 - h)
= cosa cos 6,
76 TRIGONOMETRY.
a result agreeing with [37], And the remaining formulas
may be easily shown to hold true.
Like results follow in all cases; in other words, Formulas
[37] -[42] are universally true.
Exercise XX.
1. Prove, by aid of Formula [37], that the hypotenuse of a
right triangle is less than or greater than 90®, according as the
two legs are alike or unlike in kind.
2. Prove, by aid of Formula [40], that in a right spherical
triangle each leg and the opposite angle are always alike in
kind.
3. What inferences may be drawn respecting the values of
the other parte : (i.) \ic = 90** ; (ii.) if a = 90® ; (iii.) if c = 90°
and a = 90® ; (iv.) if a = 90® and J = 90® ?
Deduce from [37]- [42] the following formulas:
4. tan'iJ = tan}((?--a)tan}(c + a).
6. tan' (45® — J-4) = tan i ('^ "~ ^) cot} (c + a).
6. tan' } -B = sin (c — a) esc {c + a).
7. tan»}c = -cos(^ + ^)8ec(^-J?).
8. tan»Ja=tan[J(^ + .S)-45®]tan[}(^-J5) + 45®].
§47. Napier's Rules.
The formulas deduced in § 46 express the relations between
five parte of a right triangle, — the three sides and the two
oblique angles. All these relations may be shown to follow
from two very useful Rules, devised by Baron Napier, the
inventor of Logarithms.
For this purpose the right a»gle (not entering the formulas)
is left out of account, and instead of the hypotenuse and the
two oblique angles, their respective complements are employed ;
so that the five parts considered by the Rules are : a, i, co. e,
CO, A, CO, B, Any one of these parte may be called a middle
part ; and then the two parte immediately adjacent are called
adjacent parte, and the other two are called opposite parte.
THE EIGHT SPHERICAL TRIANGLE. 77
Rule I. The sine of the middle part is equal to the product
of the tBJigents of the tAjacerU" parts.
Rule II. The sine of the middle part is equal to the product
of the cosines of the opposite parts.
These Rules are easily remembered by tbe expressions,
taxL acL and oos. op.
Tbe correctness of these Rules may be shown by taking each
of the five parts as middle
part, and comparing the *
resulting equations with the
equations contained in For-
mulas [37] -[42].
For example, let co.c h^
taken as middle part, then ^'^'
CO. A and co. B are the ad-
jacent parts, and a and b
the opposite parts, — as is
very plainly seen in Fig. 39,
in which the order of the
parts is shown by arranging them around the circumferenco
of a circle. Then, by Napier's Rules :
sin {co, c) = tan (co. A) tan {co. -B),
or cosc = cot-4 cotJS;
sin(co. c) = cos a cos J,
or cosc = co6a C06&;
results which agree with Formulas [37] and [42] respectively.
Exercise XXI.
1. Show that Napier's Rules lead to the equations contained
in Formulas [38], [39], [40], and [41].
2. What will Napier's Rules become, if we take as the five
parts of the triangle, the hypotenuse, the two oblique angles,
and the complements of tKe two legs ?
78 TRIGONOMETRY.
§48. Solution of Right Triangles.
By means of Formulas [37] -[4^] we can solve a right tri-
angle in all possible cases. In every case two parts besides
the right angle must be given.
Case I. Oiven the two legs a and b.
The solution is contained in Formulas [37] and [41]; viz:
cos c = cosa cos J,
tan J. == tan a esc J,
tan 5= tan b esc a.
tor example, let a = 27^ 28' 36", b = &V IZ 8" ; then the
solution by logarithms is as follows :
logcosa = 9.94802
logcos£ = 9.79697
logcosc = 9.74499
c = 66^ 13' 40"
log tan a =9.71604
log esc J =0.10826
logtan^ = 9.82430
.^ = 33M2'60"
logtan b = 10.09477
log CSC a = 0.33593
logtan .5 = 10.43070
5 ==69° 38' 54"
Case II. Given the hypotenuse c and the leg a.
From Formulas [37], [38], and [39] we obtain
COS b =■ cos c sec a,
sin J. = sin a cscc,
cos .5 = tan a cote.
Although two angles in general correspond to sin ^, one
acute the other obtuse, yet in this case the indetermination is
removed by the fact that A and a must be alike in kind (see
Exercise XX., Example 2). *
THE RIGHT SPHERICAL TRIANGLE. 79
Case III. Given the leg a and the opposite angle A.
By means of Formulas [38], [40], and [41], we find, that
Bine = sin a CSC ^,
sini =tana cot^,
sin J? = sec a cos ^ ;
or, from [37] and [39],
COS& = cos ^ sec a,
cos^ = tana cote.
When c has been computed, b and JS are determined by
these values of their cosines; but, since c must be found from
its sine, c may have in general two values which are supple-
ments of one another. This case, therefore, really admits of
two solutions.
Case IV. GKven the leg a and the adjacent angle B.
Formulas [39], [40], and [41] give
tanc = tana sec j5,
t^n b = sin a tan B,
coSwtl=cosasin£.
Case V. Oiven the hypotenuse c and the oblique angle A,
From Formulas [38], [39], and [42] it follows that
sin a = sin c sin A^
tan J =tan<?co8-4,
cot B=co&e tan A.
Here a is determined by sin a, since a and A must be alike
in kind (see Exercise XX., Example 2).
Case VI. Oiven the two oblique angles A and B»
By means of Formulas [40] and [42] we obtain
cose =cot-4 cot 5,
cos a == cos -i. CSC B,
cos J =008 B CSC A.
80 TRIGONOMETRY.
Note 1. In Case I. (a and b given) the formula for compnting c fails
to give accurate results when c is very near 0® or 180** ; in this case it
may be found with greater accuracy by first computing B, and then com-
puting c, as in Case IV.
Note 2. In Case II. (c and a given), if 6 is very near 0® or 180**, it
may be computed more accurately by means of the derived formula
tan'}5 >- tan}(e + a) tan}(c — a).
And if ^ is so near 90^ that it cannot be found accurately in the Tables,
it may be computed from the derived formula
tan«(45« - Jil) - tan}(« - o)«)fc J(c + a>
In like manner, when B cannot be accurately found from its cosine we
may make use of the formula
tan' } S — sin (c — a) CSC (c + a).
Note 3. In Case III. (a and A given), when the formulas for the
required parts do not give accurate results, we may employ the derived
formulas
tan«(45<»-Jc) -tan}(^-a)cot}(4 + a),
tan« (45* - i &) « sin (A - a) cbc[A + a),
tan»(45« - J5)=- tan J(^ - o) tan J(-4 + a).
Note 4. In Case IV. (a and B given), if J. is near 0® or 180®, it may
be more accurately found by first computing b and then finding A,
Note 5. In Case V. (c and A given), if a is near 90®, it may be found
by first computing 6, and then computing a by means of Formula [41].
Note 6. In Case VI. (A and B given), for unfavorable values of the
sides greater accuracy may be obtained by means of the derived formulas
tan'Jc = — cos(il + B) Bec{A — B%
tan« Ja = tan[}(-A + -B) - 45®] tan[45o + }(^ - i?)],
tan« J 6 =- tan [i(A + ^) - 45®] tan [45® - J (^ - -B)].
Note 7. In Cases I., IV., and V., the solution is always possible; in
Case II., in order that the solution should be possible, it is necessary
and sufficient that sin a < sine; in Cases III. and VI., the conditions of
possibility follow obviously from the equations employed ; in Case III.,
it is also necessary that a and A should be alike in kind.
THE BIGHT SPHERICAL TBIANGLE. 81
Note 8. It is easy to trace analogies between the formulas for solving
right spherical triangles and those for solving rigiht plane triangles. The
former, in fiEkct, become identical with the latter if we suppose the radius
of the sphere to be infinite in length ; in which case the cosines of the
sides become each equal to 1, and the ratios of the sines of the sides and
of the tangents of the sides must be taken as equal to the ratios of the
sides themselves.
If the formula required for any case is not remembered, it is
always easy to find it by means of Napier's Rules. In apply-
ing these Rules we must choose for the middle part that one
of the three parts considered — the two given and the one
required — which will make the other two either adjacent
parts or opposite parts.
For example : ffiven a and B; solve the triangle.
First, represent the parts as in Fig. 40, and to prevent mis-
takes mark each of the given
parts with a cross. To find ^^^^^ "^*^'*'
i, take a as the middle part ;
then b and co, B are adja-
cent parts; and by Rule I.,
sina=tan£cot^; co.A
whence, tan i = sin a tan ^.
To find c, take co.B^a
middle part; then a and
CO, c are adjacent parts ; and
by Rule I.,
cos ^ = tana cote;
whence, tan<? = tan a sec -B.
To find -4, take co. A as middle part ; then a and co. B are
the opposite parts; and by Rule II.,
cos -4 = cos a sin B.
In like manner, every case of a right spherical triangle may
be solted.
82
TRIGONOMETRY.
Exercise XXII.
Solve the following right triangles, taking for the given
parts in each case those printed in columns I. and II. :
I.
II.
III.
IV.
V.
a
h
c
A
B
1
36*27'
43° 32' 31"
64° 20'
46° 59' 43.3"
57° 59' 19.2"
2
86*>4(y
32° 40'
87° 11' 39.8"
88° 11' 57.8"
32° 42' 37.8"
3
50°
36° 54' 49"
59° 4' 25.7"
63° 15' 13.2"
44° 26' 21.4"
4
120** i(y
150° 59' 44"
63° 55' 43.3"
105° 44' 21.25"
147° 19' 47.2"
c
a
5
A
B
5
55° 9' 32"
22° 15' 7"
51° 63'
27° 28' 37.5"
73° 27' 11.1"
6
23° 49' 61"
14° 16' 35"
19° 17'
37° 36' 49.3"
54° 49' 23.3"
7
44° 33' 17"
32° 9' 17"
32° 41'
49° 20' 16.7"
50° 19' 16"
8
97° 13' 4"
132° 14' 12"
79° 13' 38.2"
131° 43' 50"
81° 58' 53.3"
a
A
c
h
B
9
77° 21' 50"
83° 56' 40"
78° 53' 20"
28° 14' 31.1"
28° 49' 57.4"
101° 6' 40"
151° 45' 28.9"
151° 10' 2.6"
10
77° 21' 50"
40° 40' 40"
'impossible;
why?
a
B
c
5
A
11
92° 47' 32"
50° 2' 1"
91° 47' 40"
50°
92° 8' 23"
12
2° 0'55"
12° 40'
2° 3' 55.8"
0° 27' 10.2"
77° 20' 28.4"
13
20° 20^20"
38° 10* 10"
25° 14' 38.2"
15° 16' 50.4"
54° 35' 16.7"
14
54° 30'
35° 30'
59° 51' 20.8"
30° 8' 39.2"
70° 17' 35"
e
A
a
5
B
16
69° 25' 11"
54° 54' 42"
60°
66° 50' 49"
63° 25' 4"
16
112° 48'
56° 11' 56"
50°
127° 4' 32"
120° 3' 50"
17
46° 40' 12"
37° 46' 9"
26° 27' 23.8"
39° 57' 41.4"
62° 0' 4"
18
118° 40' 1"
128° 0' 4"
136° 15' 32.7"
48° 23' 38.6"
58° 27' 4.3"
,
A
B
a
6
c
19
63° 15' 12"
135° 33' 39"
49° 59' 56"
143° 5' 12"
120° 55' 34"
20
116° 43' 12"
116° 31' 25"
120° 10' 3"
119° 59' 49"
75° 26' 59"
21
46° 59' 42"
57° 59' 17"
36° 27'
43° 32' 37".
54° 20' 3"
22
90°
88° 24' 35"
90°
88° 24' 35"
90° .
THE BIGHT BPHEBICAL TBIANOLE. 88
23. Define a quadrantal triangle, and show how its solution
may be reduced to that of the right triangle.
24. Solve the quadrantal triangle whose sides are :
a = 174^12'49.1", J = 94^ 8' 20", c-=9(f.
25. Solve the quadrantal triangle in which
c = 9(r. ^ = 110^ 47' 50", -B = 135^85'34.5".
26. Given in a spherical triangle -4, C, and c = 90® ; solve
the triangle.
27. Given ^ = 60^ C=90^ and (? = 90®; solve the tri-
angle.
28. Given in a right spherical triangle, A == 42® 24' 9",
5 = 9® 4' 11"; solve the triangle.
29. In a right triangle, given a = 119® 11', J5 = 126® 54';
solve the triangle.
30. In a right triangle, given c = 50®, i = 44® 18' 39" ; solve
the triangle.
31. In a right triangle, given ^=156® 20' 30", a=65®15'45";
solve the triangle.
32. If the legs a and J of a ri ght s pherical triangle are
equal, prove that cos a = cot-4 = Vcosc.
33. In a right triangle prove that cos* A X ein* c = sin (c— a)
sin (c + cl)'
34. In a right triangle prove that tana cose = sini cot-B.
35. In a right triangle prove that
sin'-4 = C08*J5 + sin* a sin* A
36. In a right triangle prove that
sin(i + <?) = 2cos'}-4 cos J sine.
37. In a right triangle prove that
sin ((? — &) = 2Bm*iA cosS sin c,
38. If, in a right triangle,^ denote the arc of the great circle
passing through the vertex of the right angle and perpendic-
ular to the hypotenuse, m and n the segments of the hypote-
nuse made by this arc adjacent to the legs a and b respectively,
prove that (i.) tan*a = tan<:tan7n, (ii.) sin*/? = tan m tan n.
84 TRIGONOMETRY.
§49. Solution op the Isosceles Spherical Triangle.
If an arc of a great circle is passed through the vertex of
an isosceles spherical triangle and the middle point of its base,
the triangle will be divided into two symmetrical right tri-
angles. In this way the solution of an isosceles spherical
triangle may be reduced to that of a right spherical triangle.
In a similar manner the solution of a regular spherical
polygon may be reduced to that of a right spherical triangle.
Arcs of great circles, passed through the centre of the polygon
and its vertices, divide it into a series of equal isosceles tri-
angles; and each one of these may be divided into two equal
right triangles.
Exercise XXIII.
1. In an isosceles spherical triangle, given the base b and
the side a; find A the angle at the base, B the angle at the
vertex, and h the altitude.
2. In an equilateral spherical triangle, given the side a;
find the angle A.
8. Given the side a of a regular spherical polygon of n
sides; find the angle A of the polygon, the distance H from
the centre of the polygon to one of its vertices, and the dis-
tance r from the centre to the middle point of one of its sides.
4. Compute the dihedral angles made by the faces of the
five regular polyhedrons.
5. A spherical square is a spherical quadrilateral which
has equal sides and equal angles. Its two diagonals divide it
into four equal right triangles. Find the angle A of the
square, having given the side a.
CHAPTER VI.
THE OBLIQUE SPHEBICAL TRIANGLE.
§51. Fundamental Fobmulas.
Let ABO (Fig. 41) be an oblique spherical triangle, a, 5, c
its tbree sides, A^B.C the angles ^
opposite to them, respectively.
Through C draw an arc CD of
a great circle, perpendicular to
the side AB, meeting AB at D.
For brevity let CD=p, AD=m,
BD = n, AAGD^x, ABCD
1. By §46 [38], in the right
triangles JBDCand ADC,
sin2? = 8ina sin^,
and sin^? = sin 5 sin ^.
Therefore, sina sinB = sinb sin A 1
similarly, iina nnO = lino sinA \ • [43]
and sinb sinO = lino nnB J
These equations may of course also be written in the form
of proportions ; as, for example,
sina : sini = sin^ : sin B,
In Fig. 41 the arc of the great circle CD cuts the side AB
within the triangle. In case it cut AB produced without the
triangle, sin (180** - A\ sin (180** - J?), or sin (180** ~ C\ would
be employed in the above proof instead of sin -4, sin J?, or
sin C. These sines, however, are equal to sin A, sin B, and
sin C, respectively, so that the Formulas [43] hold true in all
cases.
86 TRIGONOMETRY.
2. In the right triangle BDC, by §46 [37],
cos a = cos^ cosn = cos^ cos(c — w),
or (§ 32) cos a = cos^ cose cosm + cos^ sine sinw.
Now, cos^? cos m = cos i ; (§ 46 [37])
whence, cos^ = cos J sec w,
and cosp sin m = cos & tan m
= cos S tan & cos A (§ 46 [39])
= 8in& cos^.
Substituting these values of cos^ cosm and cos^ sinm in
the value of cos a, we obtain
cosa = cosb 0080 4- sinb sine 008 A ^
and similarly, oosb = oosa ooso + sina sine 008 B > • [44]
0080 = oosa oosb + sina sinb oosO J
3. In the right triangle ADC, by § 40 [41],
cos A = cosp sin x = cosjp sin ( 0— y),
or (§ 32) cos -4 = cos^ sin (7 cosy — cos^ cos C siny.
Now, cos^? siny =cosjB; (§46 [40])
whence, cos^? =cosjB cscy,
and cos^ cosy = cos J? coty
= cos J5 tan ^ COB a ' (§ 46 [42])
= sin-B cos a.
Substituting these values of cos^ siny and cosp cosy in the
value of cos -4, we obtain
co8A = — cosB oosO +8inB sinO cosa^
and similarly, oosB = — oos A cosO + sinA sinO oosb > • [45]
oosO = — 008 A oosB + sinA sinB coso J
Formulas [44] and [45] are also universally true ; for the
same equations are obtained when the arc CD cuts the side
AD without the triangle.
THE OBLIQUE SPHERICAL TRIANGLE. 87
Exercise XXIV.
1. What do Formulas [43] become if ^ = 90^? if J5 = 90^?
if C=90^? ifa = 90''? if^ = JB = 90^? ifa = J = 90*»?
2. What does the first of [44] become if ^ = 0** ? if ^ = 90** ?
if ^ = 180^?
3. From Formulas [44] deduce Formulas [45], by means of
the relations between polar triangles (§ 45).
§52. Formulas for the Half Angles and Sides.
From the first equation of [44],
A cosa — cosJcosc
COS^= :— ; — : ;
8in6 sine
whence,
1 «^a J — 8in& sin g + cosS cosg — cosa
X "^ COS Jx — _
sino sine
_ COS (6 -~ c) — cosa
sin& sine
1 4- co8^ = siP^ sine — cosS cosc + cosa
sini sine
__ coso ~ cos (6 + e) ^
sini sine '
or, by § 34 [16] and [17], and § 35 [23],
sin' } ^ = sin } (a -f J— e) sin } (a— i+ e) esc J csce,
cos* } -4 = sin } (a+ J+ e) sin } (5+ e— a) esc h esc c.
Now, let J(« + ^ + = ^J
whence, } (S + e — a) = 5 — a,
} (a — J + e) = « — J,
} (a + 6 — e) = s ~ e.
88 TRIGONOMETRY.
Then, by substitution and extraction of the square root,
..[46]
Bin } A= VBin(a -— b) Bin(B — c) csob caoo
co8}A= Vsins 8in(s — a) csab csoo
tan}A= VosoB o8o(B--a) 8in(B — b) Bin(8--o) ^
In like manner it may be shown that
Bin }B = VBin(B — a) Bin(B — o) caca osco,
COB }B = Vsins Bin(B — b) csoaoBoo,
taniB = VoBOB o8o(b — b) «n(B — a) Bin(8 — o).
ainiO = VBin(8 — a) 8in(8 — b) csoa csob,
cob}0 = VsinB Bin(8 — c) csca csob,
tan i = Vcaoa c8o(a — o) 8in(B — a) 8in(B~b).
Again, from the first equation of [45],
__ cos B cos C + cos A .
vv/o i/ — ~
whence,
co8a= . ^ . ^
sinj5 smC/
1 sin B sin C— cos B cos (7— cos A
1 — cosa = ' T> ' n '
sm B sm G
1 , „ sin jB sin 0+ cos B cos O +qobA
1 + cosa= ■ ■■ — -- — : — ■ ■ ■■ ' ■■■ ; ■ « *
sin B BmG
If we place i(A + B + C) = S, and proceed in the same
manner as before, we obtain the following results :
Bin Ja =V— coaS coa(8— A) cacB caoO
008 ia = Vco8(8 — B) cob (8 — 0) csoB osoO
taaia =V~co88 co8(8-A) Bec(8 — B) Beo(8 — 0)
.[47]
THE OBLIQUE SPHEEICAL TBIANGLE. 89
And, in like manner,
sin }b = V— ooiB ooi(S— B) oboA oboO,
cos 1 b = V 008 (8 —A) 008(8^^0) omAomO,
tan}b = V-oo«Booi(8-B)ieo(8-A)ieo(8-0).
rinio =V— 0088 008(8 -0)080 A 080 B,
008 }o =Voob(8— A)ooB(8^^B)o8oAoBoB,
tanjo = V-0088 oo8(8"^^0) 8eo(B-A) 8eo(8-B).
§ 53. Gauss's Equatioks and Napieb's Analogies.
By §31 [5],
C08^(-4. + JB) = 008^-4 008-^5-- sin J-4 Bin^B;
or, by substitnting for cos|^j1, cos^^, sin^^, sin ^5, their
values given in § 52, and reducing,
cobUA+JB)==J'^''' 6in(^-a) Isin^ 8in(^-^}
\ sin&sinc N sinasinc
i sin (8 — t) sin (s — c) f sin (s — o) an (g— c)
^1 sin i sine \ sinasinc
_ nn g—sin (g— g) ^ Isin (a — a) sin (a — i j
sine if sin a sin 6
' This value, by applying §§ 33 [12], 35 [21], and observing
that the quantity under the radical is equal to sin } (7, becomes
cosK^ + 5) = g^-^ ,"^»(^7*") sinig,
*^ ^ 2sinJcco8|<? *
which, by cancelling common factors, multiplying by cos^c,
and observing that « — J(? = J(a + J), reduces to the form
cos^(^ + £) cos^c = co8'J-(a + i) sin J 0.
90
TRIGONOMETRY.
By proceeding in like manner with the values of
8ini(A + £), co8j(^-J5), and sin ^(^4- 5),
three analogous equations are obtained.
The four equations,
ooBj(A + B)coB}o = ooB}(a + b)Bm}0
sin } ( A + B ) ooB } = ooB } (a — b) cos }
cob}(A — B)Binio = sinJ(a + b)8iii}0
sin i(A — B)8in }o = Bin J(a--b)oo8}0-
[48]
are called Gauss's Equations.
By dividing the second of Gauss's Equations by the first,
the fourth by the third, the third by the first, and the fourth
by the second, we obtain
taai(A + B) = °°'^i*7^) ootiO]
008i(a + D)
sin } (a + b)
ootJO
[49]
These equations are called Napier's Analogies.
In the first equation the factors cos i(a — b) and cot } C are
always positive; therefore, tan-J-(u4 + JB) and cos -J- (a + ^)
must always have like signs. Hence, if a + i < 180**, and
therefore cosi(a + S) > 0, then, also, tan}(-4 + JB) > 0, and
therefore A + B< 180**. Similarly, it follows that if
a + J>180^ then, also, A + B>lSO\ If a + & = 180^
and therefore cos i (a + 5) = 0, then tan } (-4+ JB) = oo ;
whence i(A + B) = 90^ and" A + B = 180^
Conversely, it may be shown from the third equation, that
a + b ia less than, greater than, or equal to 180**, according as
-4 + J5 is less than, greater than, or equal to 180**,
THE OBLIQUE SPHERICAL TRIANGLE.
91
§54. Case I.
Given two aides, a and b, and the included angle C,
The angles A and J5 may be found hj the first two of
Napier's Analogies ; viz. :
tan J(^ + 5) = ^22ii«:=4) cot} C.
tanj(^-
sin t (a + b)
After A and J5 have been found, the side c may be found
by [43] ; but it is better to use for this purpose Gauss's Equa-
tions, because they involve functions of the same angles that
occur in working Napier's Analogies. Any one of the equa^
tions may be used ; for example^ from the first we have
Example.
cos
a-
b'
cos } (A+£)
logco8}(a — 5)
logaec J(a + b)
log cot} (7
. 73«58'54",
.38*46' 0",
.460 33'41",
. 9.97914
-0.25668
-0.36626
log tan }(^ + -B)- 0.60198
log sec }(il + J5)-
logcoB}(a + 6) -
logBin^C
. 0.61547
. 9.74342
.9.69686
log cos }c
. 9.95575
.25«25'40"
therefore, i(a — h)»
iO .
log8inJ(a — 5) »
logc8c}(a + 5) '
log cot }C
17* 36' 57"
.56* 21' 57^
-23* 16' 50.5"
' 9.48092
. 0.07956
r 0.36626
logtani(^--B)- 9.92674
A'
75° 57' 40.7"
40* 11' 25.6"
-116* 9' 6.3"
. 35* 46' 15.1"
. 50* 51' 20"
If the side c only is desired, it may be found from [44],
without previously computing A and J5. But the Formulas
[44] are not adapted to logarithmic work. Instead of chang-
ing them to forms suitable for logarithms, we may use the fol-
lowing method, which leads to the same results, and has the
advantage that, in applying it, nothing has to be rememberad
except Napier's Rules :
92
TEIOONOMETRY.
Make the triangle (Fig. 42), as in § 51, equal to the sam
(or the diflference) of two right
triangles. For this purpose,
through B (or A, but not C)
draw an arc of a great circle
perpendicular to AC^ cutting
AC&iD. Let BD=p,CD=m,
AD=n\ and mark with crosses
the given parts.
By Rule I.,
cos C = tan m cot a,
whence tanm = tan a cos C,
By Rule II.,
cos a = cos m cosp, whence cos^ = cos a sec m.
cos c = cos n cos^, whence cos^ = cos c seen.
Therefore,
or, since
It is evident that c may be computed, with the aid of loga-
rithms, from the two equations
tanm = tana cos 0,
coBC =cosasec7» cos(ft — m).
Example. Given a = 97^30'20", b = 55n2'10", C= 39**
58'; find^.
log tan a = 0.88025 (n)
cos (? sec n = cos a sec m ;
w = & — m, co8<? = cosaBecw cos(J — w).
logcosg= 9.88447
logtanm = 0.76472 (n)
m= 99^ 45' 14"
J-7» = -44^33' 4"
logcosa = 9.11602 (w)
logco8(J-7?i) = 9.85286
logsecm = 0.77103 (n)
logcos(? = 9.73991
c = 56'' 40' 20"
EXEBCISE XXV.
1. Write formulas for finding, by Napier's Rules, the side
a when 5, c^ and A are given, and for finding the side b when
a, c, and £ are given.
THE OBLIQUE SPHEBICAL TKIANGLE.
2. Given a = 88» 12' 20", h = 124» 7' 17", C = 50" 2' 1" ;
find ^ = 63" 15' 12", jB=132'17'59", c = 59''4'25".
3. Given a =120" 55' 35", 6 = 88" 12' 20", C=4r42'l";
find A = 129" 57'59", B = 63» 15' 12", c = 55" 52' 43".
4. Given i = 63''15'12",c = 47''42'l".^ = 59''4'25"; find
5 = 88" 12' 20", C= 55" 52' 43", o = 50°2'l".
5. Given i = 69" 25' 11", c = 109''46'19", ^ = 54''54'42"^
find J5 =56"' 11' 56", C=123'»2ri3", o = 67«'12'.
§55. Case II.
Owen two angles, A and B, and the included tide c.
The sides a and b may be found by the third or fourtih of
Napier's Analogies,
afid then the angle C may be found hj one of Gauss's equa-
tions ; as, for instance, the second, which gives
cos
JC=Sn^^±fcoBj..
COS i(a — b)
Example. -^'
e>
\ogcoBi{A-B)^
logBec}(il4--B).
log tan} 6
.10^47' 7"
. 38<>58'27"
. 51«41'14"
.9.91648
. 0.54369
. 9.68517
logtan}(a + 5) » 0.14524
logBin}(^ + 5) =
logBec}(a — 5) '
log cos }e
. 9.98146
: 0.01703
. 9.95423
logcosJC
iC"
9.95272
26*»14'62.5"
JM + 5).
logBin}(il~-^).
l0gC8C}(^ + -B)-
logtftn}c F
.34«24'20"
.730 22' 47"
25** 50' 37"
9.75203
0.01854
' 9.68517
log ton J (a -6) -9.45579
J(a-5) =
!i;
54® 24' 24.4"
15«56'25.6"
70*»20'50"
38° 27' 59"
52^20' 45"
94
TRIGONOMETRY.
If the angle Q alone is wanted, the best way is to decompose
the triangle into two right triangles, and then apply Napier's
Eules, as in Case I., when the side c alone is desired.
Let (Fig. 43) Z ABD = x, ACBD = y, BD =p ; then,
By Kule I.,
cose = cot a: cot -4,
whence cotar = tan^ cose.
By Kule II.,
cos-4 = cos^8inrr,
whence cos ^ = cosul esc a?.
cos(7 = cos^siny,
whence cos p = cos Ocacy,
Therefore,
cos = cos -4 CSC a? siny = cos A esc a: sin (B — x).
It is clear that (7 may be computed from the equations
cot a: = tan -4 cose,
cos (7= cos A CSC X sin (B — x).
Example. Given ^ = 35*»46'15", JB=115*»9'7", (?=5r 2';
findC.
logtanu4 = 9.85759
log cose =9.79856
logcota; =9.65615
X =65** 37' 37.2"
B-x =49**31'29.8"
log cos ^ =9.90922
log sin {B-x) = 9.88120
log CSC a: = 6.04154
log cosC= 9.83196
(7= 47^3*26"
Exercise XXVI.
1. What are the formulas for computing A when jB, C, and
a are given; and for cotnputing B when -4, C, and h are
given?
2. Given A = 26^58'46", B = 39**45'10", c = 154°46'48'\-
finda = 37n4'9", i = 12r28'9", (7=16r22'10"
THE OBLIQUE SPHERICAL TRIANGLE. 95
3. Given ^ = 128*»4r49", ^ = 107^33'20", = 124^2'
31"; finda = 126^41'44", & = 82*»47'35", C= 127^22'?".
4. Given J5 = 153^7' 6". C=78«43'36", a = 86n5'15";
find J = 152*' 43' 52", c = 88^2' 19". ^ = 78**15'41".
5. Given ^ = 125M1'44", e=82*47'35", i = 52^37' 57";
find a = 128*41'47", c = 107^ 33' 20", B = 55^47'29".
§56. Case III.
Oiven two sides a and b, and the angle A opposite to one of
them.
The angle S ia found from [43], whence we have
sin ^ = sin ^ sin & CSC a.
When £ has been found, C and c may be found from the
fourth and the second of Napier's Analogies, from which we
obtain
8inJ(^-j5) ^ ^'
8in}(a — 6)
The third and the first of Napier's Analogies may also be
used for the same purpose.
Note 1. Since B is determined from its sine, the problem in general
has two solutions ; and, moreover, in case sin ^> 1, the problem is impos-
sible. By geometric constmction it may be shown, as in the correspond-
ing case in Plane Trigonometry, nnder what conditions the problem
really has two solntions, one solution, and no solution. But in practical
applications a general knowledge of the shape of the triangle is known
beforehand ; so that it is easy to see, without special investigation, which
solution (if any) corresponds to the circumstances of the question.
It can be shown that there are two solutions, when
ii<90<», a + 6<180*, anda<ft,
^r.when A>9QP, a + 6>18(y», anda>6.
96
TEIGONGMETRY.
Note 2. The side c or the angle C may be computed, without first
finding B, by means of the formulas
tanm «" cos J. tani, and co8(<? — m) » cos a X sec b X cosm,
cot a; » tan A cos b, and cos {0— x) » cot a X tan b x cos x.
These formulas may be obtained by resolution of the triangle into
right triangles, and applying Napier's Rules ; m is equal to that part of
the side c included between tiie foot of the perpendicular from C and the
vertex A, and x is equal to the corresponding portion of the angle C.
Example. Given a - 57*> 36', h - 3P 12', A - 104«> 25' 30".
In this case A> 90®,
and o + 6<180*>;
therefore, -4h-^<180**;
hence, B< 90**,
and only one solution.
a- 6 «26'»2e'
-4 + -B=-140«>51'53"
-4-^- 67^59' 7"
logBin}(il + P)- 9.97416
log CSC }(-A- -5) -0.25252
logt an^(a-6) «=» 9.37080
logtanic = 9.59748
}c« 21° 35^38"
c- 43m' 16''
logsin^»9.9
log sin 6 -9.71435
log CSC g =0.07349
log sin 5 « 9.77393
5 « 36*^27' 20"
i(a + b):
iia^b)-.
log8in}(a + 6) «
logC8cJ(a — 6) =
logtan|(il-^ =
.44«>25'
. 13° 13'
= 70° 26' 25"
»33°59' 5"
- 9.84502
. 0.64086
9.82873
log cot} C«
0.31461
25° 51' 15"
51° 42' 30"
Exercise XXYII.
1. Givena = 73^49'38", i=:120^53'35", ^ = 88*'52'42";
find.S = 116M4'48", c = 120°55'35", (7= 116M4'48".
2. Given a = 150^57'5", i = 134*»15'54", .4 = 144°22'42";
find J5i=120M7'44", Ci=55°42' 7.5", a = 97*»42'55";
£»= 59*^12' 16", c,= 23*»57'29.4", (^=29^ 9' 9.4".
3. Given a = 79^ 0' 54.5", 6 = 82*» 17' 4", ^ = 82^ 9' 25.8" ;
find5 = 90^ (? = 45°12'19", a=45M4'6".
4. Given a = 30° 52' 36.6", 6 = 3r9'16", ^ = 87"34'12";
show that the triangle is impossible.
THE OBLIQUE 8PHEBICAL TBIAKOLE. 97
§ 57. Case IV.
GUven two angles A and B, and the side a opposite to one of
them.
The side b is found from [43], whence
Bin& = 8ina einJScsc^.
The values of e and Cjbaj then be found hj means of Na-
pier's Analogies, the fourth and second of which give
cotia=SEiI« + »}tanJ(^-5).
sin} (a — 6)
Note 1. In this case, also, an unknown part is found from iti sine ;
and it may be shown thai, under certain conditions, the problem is im-
possible, or that it admits of two solutions. In practice, the ambiguity is
usually removed by the circumstances of the question. If sin5>l the
problem of course is imposisible; and it may be shown that there are two
solutions, when
a<90», A + BKISOP, and il<-B.
or a>9(y», il + -B>180». hnd A>B.
Note 2. By proceeding as indicated in Case III., Note 2, formulas for
computing e or C, independent of the side &, may be found ; vis.:
tanm » tan a cos .S, and sin (o ~ m) — cot ii tan ^ sin m,
cot X i- cos a tan ^, and sin(C— c) i- cosJ. sec .S sin x.
In these formulas m — BD, a » ^ BCD^ D being the foot of the per-
pendicular from the vertex (7.
EXBBOISB XXVIII.
1. Given ^ = 110^0', 5=133^8', a=147*5'32"; find
J = 165^5'18", (? = 33n'45", C=70^20'60^
2. Given^=113^39'21", 5=123M0'18", a = 65*»39'46";
find6 = 124^7'28", (?=159**50'4", e=159M3'20".
98
TEIOONOMETRY.
3. Given ^=100° 2' 11.3", ^=98°30'28", a = 95°20'38.7";
findi = 90^ (? = 147**41'43", C=148*»5'35".
4. Given ^=24° 33' 9", ^ = 38°0'12", a = 65** 20' 13" ;
show that the triangle is impossible.
§ 58. Case V.
Oioen the three sides^ a, 5, and c.
The angles are computed by means of Formulas [46], and
the corresponding formulas for the angles B and (7.
The formulas for the tangent are in general to be preferred.
If we multiply the equation
tan i -4. = Vcsc 8 esc (s — a) sin (« — b) sin (a — c)
by the equation
1 — ?i5i£jZ^
sin(«— -a)
and put
Vcsca sin(s — a) sin(s — b) sin(s — c) = tanr,
and also make analogous changes in the equations for tan } B
and tani(7, we obtain
tanj^ = tanr esc (a — a),
tan i-B = tanr csc(s — i),
tan } C = tanr esc (a — c),
which are the most convenient formulas to employ when all
three angles have to be computed.
Example 1.
o= 5(y»54'32"
6- 3^47' IS''
2« = 163« 33^40"
»= 8P46'5(y'
«-a= 30O 52^18''
,_J= 43059' 32"
«-c« 6°66' 0"
logc8c«» 0.00448
logcsc(«-a)= 0.28978
log8in(«-J)« 9.84171
log8in(»-c)= 9.08072
2)19.21669
log tan} -4= 9.60835
}A-22<> 5'20"-
-4 « 440 10^40"
THE OBLK^UE SPHERICAL TRIANGLE.
99
Example 2. a - 124i*» 12^ 31"
6- 54<»18'16"
c-9ri2^
2«-275«43'12''
log sin (« - a) - 9.37293
logBin(»- 6) -9.99725
log Bin (t - e) » 9.81390
logcsc «- 0.17331
logtanV- 9.35739
logtftnr- 9.67870
«- 137° 51' 36"
«-a- 13°39' 5"
,-6-. 83'»33'20''
,-c» 40* 39^11"
logtanjil"
logtan}^-
log tan} C-
0.30577
9.G8145
9.86480
63*>41'3.8"
25«39'5.G"
36° 13^20.1"
A-
127^22' 7"
51° 18' 11"
72° 26' 40"
Exercise XXIX.
1. Given a = 120^65'35", &=59M'25", c = 106n0'22";
find^ = 116M4'48", 5 = 63n5'12", C=9r7'18".
2. Given a = 50^2' 4", i = 116** 44' 48", <?= 129M1'42";
find ^ = 59*4' 25", 5=94*23'10", C= 120*4' 50".
3. Given a = 131*35' 4", i = 108*30'14", c = 84*46'34";
find ^=132* 14' 20", 5 = 110*10' 40", C=99*42'24".
4. Given a = 20* 16' 38", b = 56* 19' 40", c = 66* 20' 44" ;
find -4 = 20* 9' 55", £ = 55* 32' 30", C= 114* 20' 16".
§59. Case VI.
Oiven the three angles, Ay B, and C.
The sides are computed by means of Formulas [47], and
the corresponding formulas for the angles B and C The
formulas for the tangents are in general to be preferred.
If we multiply the equation
tan Ja= V- cos/Sco8(/8'- A) sec(/S— ^)sec(AS'- C)
by the equation 1 = ^^^^~^) ,
^{8 -A)
and put
V-cos/8'8ec(/S--4)Bec(/S'~ B) sec{S-C) = tan J2,
100
TRIGONOMETRY.
and also make analogous changes in the equations for tan J b
and tan \ c, we obtain
tan J a = tan Iicoa(S— A),
tan } 6 = tan ^ cos(/S- B),
tan i (? = tan ^ COS (/S— C),
which are the most convenient formulas to use in case all
three angles have to be computed.
Example I.
P-130*>
(7=^150^
2S='50(y*
S'^2B0P
/S- (7 « 100*>
log cos iSf=. 9.53405 (n)
logco8(/Sf-^) = 9.93753
logsec (iSf- 5) - 0.30103 (n)
log8ec(;8f- g) = 0.76033 (n)
2)0.53294
logtanJa= 0.26647
ia= 6I034' 6"
o-123*» 8' 12"
Note. Here the effect, as regards algebraic sign, of three negative
feictors, is cancelled by the negative sign belonging to the whole fraction.
Example 11. ^= 20° 9^56"
B^ 65° 62' 32"
(7«114*> 20^14"
2i8f= 190° 22' 42"
logcosi8f= 8.96638 (n)
log sec(/8f-^)- 0.58768
logsecCiS'- 5) -0.11143
log 8ec(5--CHa02472
log tan*JR- 9.68021
log tan i2« 9.84010
logtan^as
log tan} 6 «
logtan}e>
95° 11' 21"
75° 1'26"
39° 18' 49"
-19° 8' 53"
. 9.26242
. 9.72867
. 9.81538
10° 8' 18.9"
28° 9' 60.4"
33° 10' 21.3"
-20° 16^38"
.56° 19^41"
= 66° 20' 43"
Exercise XXX.
1. Given ^ = 130^ J5 = 110^ (7=80^
find a = 139^21' 22", i = 126'* 67' 52", (? = 56**51'49".
2. Given A = 59^ 55' 10", £ = 85° 36' 50", = 59** 55' 10" ;
find a = 129° 11' 40", b = 63° 15' 12", c = 129° 11' 40".
THE OBLIQUE SPHERICAL TBIAKOLE. 101
■
3. Given A = 102* 14' 12", 5 = 64* 32' 24", (7= 89* 5' 46" ;
find a = 104^ 25' 8", b = 53* 49' 25", c = 97** 44' 18".
4. Given ^ = 4^ 23' 35", 5 = 8^ 28' 20", (7= 172* 17' 56" ;
find a = SV 9' 13", i = 84* 18' 28", c = 115^ 10' 4".
§60. Abea of a Sfhebical Triangle.
I. When the three angles A, JB, C, are given.
Let a = radius of sphere,
J57= the spherical excess =^A + JS+ 0— 180*,
-F= area of triangle ;
then, by Solid Geometry,
r=A,,E.. [50]
II. When the three sides a, 6, <?, are given,
A formula for computing the area is deduced as follows :
From the first of [48],
co8K-^+^) _^ cosKq + &) .
008(90*^-1^) cosjc
whence, by the Theory of Proportions,
co8l(ji+^)~cos(90**-~-^(7) _ cosi(a+ft)-co8|<? r^\
cosi(jl+^) + cos(90*'-J6/) co8i(a+i) + cosic*
Now, in § 35, the division of [23] by [22] gives
5^4^^ = -tanl(^+^tanJ(^-^. (b)
cos A + cos JO
in which for A and B we may substitute any other two angu-
lar magnitudes, as for example, h(^A + E) and (90 — i C), or
i (a + J) and i c.
If we use in place of A and B the values \{A + B) and
(90® — ^ C), the first side of equation (b) becomes
co8i(^ + i?)-co8(90®-ia) .
cos J (^ + B) + cos (90® - i C) '
102 TEIGONOMETEY.
and tlie second side becomes
• -t&ni(iA+i£+90''-iC) ta.iii(iA+iB-90''+iO);
or,
- tan t (^ + ^ - (7+ 180") tan t (^ + -B + C- 180").
If we remember that 11= A+£+C— 180°, and observe
that
tani(^+£-C+180'')=tanK360"'-2C+^+J5+C-180')
=tant(360''-2C+J7)
=tan[90''-i(2C-j;)]
=cot}(2C-J7),
it will be evident that equation (b) may be written
25ii(4±|)^£-Mzi^=_eoti(2(7-JJ)tant^.(c)
If we substitute, in equation (b), for A and J?, the values
} (a + i) and } c, and also substitute 8 for J (a + J + c?) and
s — c for i{a-\-b — c)f equation (b) will become
cos J (a + 5) — cos} (? X i J. 1/ \ /j\
— ) ; ( f- = -tan}atani(5-c). (d)
cos}(a + i) + cos}c ^ ^
Comparing (a), (c), and (d), we obtain
cot t (20- ^) tan t ^ =? tan } 5 tan } (« - c). (e)
By beginning with the second equation of [48], and treating
it in the same way, we obtain as the result,
tan k(2a-E)ta,niE= t&ni(s - a) tan } (5 - J). (f)
By taking the product of (e) and (f), we obtain the elegant
formula,
tanHE = tan}Btani(B-a)tan}(B-b)tan}(s-c), [51]
which is known as I'Huilier's Formula.
By means of it U may be computed from the three sides,
and then the area of the triangle may be found by [50].
THE OBLIQUE SPHERICAL TBIAN6LE.
103
III. In all other caaea^ the area may be found by first solv-
ing the triangle so far as to obtain the angles or the sides,
whichever may be more convenient, and then applying [50]
or [51].
Example I.
C- 89° &^*
2i5^52f22f'
jE:- 65» 62^22''
- 237142^'
180°-64800(y'
Example IL a-l330 2e'i9''
6- 64*»6(y53"
c-U4^13^
2 g- 342° 30^ 57^^
«- 171° 16' 28.5"
»-a- 37° 49^ 9.5"
«- 6 -106° 24' 35.5"
,-c- 27° 1'43.5"
log -S- 5.37601
»log
648000
-4.68567-10
0.06058
J?'- 1.1497 i?
.85°3r44"
■ 18° 54' 36"
.63° 12' 18"
.13' 30' 52"
logtan}«.
logtan}(« — a)>
logtaii}(t~&) .
logtanJ(«— e) >
. 1.11669
9.53474
0.12612
. 9.38083
log tan* J 1^-
logtan \E'
E-
0.15838
0.07919
50° 11' 43"
200° 46* 62'
Exercise XXXI.
1. Given ^ = 84^20^19", ^=27^ 22' 40", C=75^33';
find ^57= 26159", 7^= 0.1268 i?.
2. Given a =69^5^6", 5 = 120° 42' 47", c=159°18'33";
find ^57= 216° 40' 23".
3. Given a = 33°l'45", 6 = 155° 6' 18", C=110°10';
find j;= 133° 48' 55".
4. Find the spherical excess of a triangle on the earth's
surface (regarded as spherical), if each side of the triangle is
equal to 1°.
* See Wentworth & Hill's Tables, page 20.
CHAPTER VII.
APPLICATIONS OF SPHERICAL TRIGONOMETRY.
§ 61. Peoblem.
To reduce an angle measured in space to the horizon.
Let (Fig. 44) be the position of the observer on the
ground, AOB = h the angle
measured in space (for example,
the angle between the tops of
two church spires), OA* and 0-B'
the projections of the sides of the
angle upon the horizontal plane
SB, AOA*=m and BOB'=n
the' angles of inclination of OA
and OB respectively to the
horizon. Required the angle
A'OB^=x made hj the projec-
tions on the horizon.
The planes of the angles of inclination AOA^ and BOB^
produced intersect in the line 0(7, which is perpendicular to
the horizontal plane (Geom. § 475).
From as a centre describe a sphere, and let its surface
cut the edges of the trihedral angle = ABC in the points
M, JV, and P. In the spherical triangle MNP the three
sides MN=h, JfP = 90 — w, JVP=90 — n, are known, and
the spherical angle P is equal to the required angle x.
From § 52 we obtain
Fig. 41
cos -J- a? = Vcos 8 cos (s — h) sec m sec n,
where -J- (m + n + A) = a.
APPLICATIONS. 105
§ 62. Pboblem.
To find the distance between two places on the earth's surface
(regarded as spherical), given the hxtUudes of the places and
the difference of their longitudes.
Let M and iV^(Fig. 45) be the places; then their distance
MN is an arc of the great cir-
cle passing through the places. -^
Let P be the pole, AB the y^"^^ \\^K
equator. The arcs MR and / \ \/J\
N8 are the latitudes of the / V/X \
places, and the arc 118, or the A'''" yir \\b
angle MPN, is the difference K^.^^^ y/l \g^
of their longitudes. Let MR \ y r ^ /
=a. JV»=6, Ra-=l\ then in \y^ /
the spherical triangle MNP ^^^^^ ^^y/^
two sides, JfP=90-a. NP ^^7^[li
= 90 — 5, and the included
angle MPN==^l, are given, and we have (from § 64)
tanm =cotacos/,
cos MIf= sin a sec m sin (b + m).
From these equations first find m, then the arc MIT, and
then reduce MNio geographical miles, of which there are 60
in each degree.
§ 63. The Celestial Sphebe.
The Oelestial Sphere is an imaginary sphere of indefinite
radius, upon the concave surface of which all the heavenlj
bodies appear to be situated.
The Oelestial Equator, or Equinoctial, is the great circle in
which the plane of the earth's equator produced intersects
the surface of the celestial sphere.
The Poles of the equinoctial are the points where the earth's
a3ds produced cuts the surface of the celestial sphere.
106 TEIGWNOMETRY.
The Odestial Hezidian of an observer is the great circle in
which the plane of his terrestrial meridian produced meets
the surface of the celestial sphere.
Hour CKideSy or Oiioles of Declination, are great circles passing
through the poles, and perpendicular to the equinoctial.
The Horizon of an observer is the great circle in which a
plane tangent to the earth's surface, at the place where he is,
meets the surface of the celestial sphere.
The Zenith of an observer is that pole of his horizon which
is exactly above his head.
Vertical Oiicles are great circles passing through the zenith
of an observer, and perpendicular to his horizon.
The vertical circle passing through the east and west points
of the horizon is called the Prime Vertical; that passing
through the north and south points coincides with the celestial
meridian.
The Ediptic is a great circle of the celestial sphere, appar-
ently traversed by the sun in one year from west to east, in
consequence of the motion of the earth around the sun.
The Equinoxes are the points where the ecliptic cuts the
equinoctial. They are distinguished as the Vernal equinox
and the AtUumnal equinox ; the sun in his annual journey
passes through the former on March 21, and through the.
latter on September 21.
Oiides of Latitude are great circles passing through the
poles of the ecliptic, and perpendicular to the plane of the
ecliptic.
The angle which the ecliptic makes with the equinoctial is
called the oUiqnity of the ecliptic; it is equal to 23® 27',
nearly, and is often denoted by the letter e.
These definitions are illustrated in Figs. 46 and 47. In
Fig. 46, AVBU is the equinoctial, F and F its poles, liFZjS
the celestial meridian of an observer, IfJSSWhia horizon, Z
his zenith, M a star, PMP the hour circle passing through
the star, ZMDZ^ the vertical through the star.
APPLICATIONS.
107
In Fig. 47, AVBU represents the equinoctial, EVFTJ the
ecliptic, F and Q their respective poles, Fthe vernal equinox,
TJ the autumnal equinox, M a star, PMR the hour circle
through the star, QJf^the circle of latitude through the star,
and ZrFi2 = «.
The earth's diurnal motion causes all the heavenly bodies
to appear to rotate from east to west at the uniform rate of
16® per hour. If in Fig. 46 we conceive the observer
placed at the centre 0, and his zenith, horizon, and celestial
meridian fixed in position, and all the heavenly bodies rotat-
ing around FP as an axis from east to west at the rate of 15**
per hour, we form a correct idea of the apparent diurnal
motions of these bodies. When the sun or a star in its diur-
nal motion crosses the meridian, it is said to make a iransd
across the meridian ; when it passes across the part NWS
of the horizon, it is said to Bet; and when it passes across the
part NE8, it is said to ri^e (the effect of refraction being
here neglected). Each star, as M, describes daily a small
circle of the sphere parallel to the equinoctial, and called the
Diurnal Oiide of the star. The diurnal circle is the smaller
the nearer the star is to the pole ; and if there were stars at
the poles F and F\ they would have no diurnal motion. To
108 TRIGONOMETRY.
an observer north of the equator, the north pole P is elevated
above the horizon (as shown in Fig. 46); to. an observer south
of the equator, the south pole -P is the elevated pole.
§64. Spherical Co-ordinates.
Several systems of fixing the position of a star on the sur-
face of the celestial sphere at any instant are in use. In each
system a great circle and its pole are taken as standards of
reference, and the position of the star is determined by means
of two quantities called its spherical co-ordinates.
I. If the horizon and the zenith are chosen, the co-ordinates
of the star are called its altitude and its azimuth.
The Altitude of a star is its angular distance, measured on
a vertical circle, above the horizon. The complement of the
altitude is called the Zenith Distanoe.
The Azimuth of a star is the angle at the zenith formed by
the meridian of the observer and the vertical circle passing
through the star, and is measured therefore by an arc of the
horizon. It ia usually reckoned from the north point of the
horizon in north latitudes, and from the south point in south
latitudes ; and east or west according as the star is east or
west of the meridian.
II. If the equinoctial and its pole are chosen, then the posi-
tion of the star may be fixed by means of its declination and
its hour angle.
The Declination of a star is its angular distance from the
equinoctial, measured on an hour circle. The angular dis-
tance of the star, measured on the hour circle, from the elevated
pole is called its Polar Distance.
The declination of a star, like the latitude of a place on the
earth's surface, may be either north or south ; but, in practical
problems, while latitude is always to be considered positive,
declination, if of a difierent name from the latitude, must be
regarded as negative.
APPLICATIONS. 109
If tbe declioation is negative, the polar distance is equal
numerically to 90^ + the declination.
The Hour Askf^ of a star is the angle at the pole formed hj
the meridian of the observer and the hour circle passing
through the star. On account of the diurnal rotation, it is
. constantly changing at the rate of 15^ per hour. Hour angles
are reckoned from the celestial meridian, positive towards the
west, and negative towards the east.
III. The equinoctial <md its pole being still retained, we
may employ as the co-ordinates of the star its declination and
its right ascension.
The Bight Asoension of a star is the arc of the equinoctial
included between the vernal equinox and the point where the
hour circle of the star cuts the equinoctial Right ascension is
reckoned from the vernal equinox eastward from 0^ to 360^
IV. The ecliptic and its pole may be taken as the standards
of reference. The corordinates of the star are then called its
latitude and its longitude.
The Latitude of a star is its angular distance from the eclip*
tic measured on a circle of latitude.
The LoDgitude of a star is the arc of the ecliptic included
between the vernal equinox and the point where the circle of
latitude through the star cuts the ecliptic.
In problems it is useful to employ certain letters to denote
these various co-ordinates. For a star Jf (Fig. 46), let
I = latitude of the observer,
h == DM = the altitude of the star,
z = ZM = the zenith distance of the star,
a = ZPZM= the azimuth of the star,
i = ZZFM= the hour angle of the star,
d = HM = the declination of the star,
p = PIT = the polar distance of the star,
r = VJR = the right ascension of the star,
u= MT (Fig. 47) = the latitude of the star,
v= Fr(Fig. 47) == the longitude of the star.
110
TRIGONOMETRY.
In many problems, a simple way of representing the mag-
nitudes involved, is to project the sphere on the plane of the
horizon, as shown in Fig. 48.
NE8W is the horizon, Z
the zenith, NZ8 the meridian,
TTZffthe prime vertical, Tr^J57
the equinoctial projected on the
plane of the horizon, P the ele-
vated pole, M a star, DM its
altitude, ZM its zenith dis-
tance, Z.FZM its azimuth,
MR its declination, FM its
polar distance, AZPM its hour
angle.
§65. The Astronomical Triangle.
The triangle ZPM (Figs. 46 and 48) is often called the
astronomical triangle^ on account of its importance in prob-
lems in Nautical Astronomy.
The side PZ is equal to the complement of the latitude of
the observer. For (Fig. 46) the angle ZOB between the
zenith of the observer and the celestial equator is obviously
equal to his latitude, and the angle POZ is the complement
of ZOB. The arc NP being the complement of PZ, it follows
that the aUitude of the elevaied pole ia equal to the latitude of
the place of observation.
The triangle ZPM then (however much it may vary in
shape for different positions of the star M), always contains
the following five magnitudes :
PZ= co-latitude of observer = 90**— Z,
ZM= zenith distance of star = 2,
PZM= azimuth of star = a,
PM=^ polar distance of star = jo,
ZPM= hour angle of star = t.
APPLICATIONS. Ill
A very simple relation exists between the hour angle of the
sun and the local (apparent) time of day. Since the hourly
rate at which the sun appears to move from east to west is
15^, and it is apparent noon when the sun is on the meridian
of a place, it is evident that if hour^angle = 0®, 15®, — 15*, etc.,
time of day is noon, 1 o'clock p.m., 11 o'clock a.m., etc.
In general, if t denote the absolute value of the hour angle,
time of day = -rP.M., or 12 — -- A.M.,
15 15
according as the sun is west or east of the meridian.
§ 66. Pboblem.
Oiven the latitude of the obeerver and the aUitude and azimuth
of a star, to find tie declination and its hour angle.
In the triangle ZPJf (Fig. 48),
given PZ= 90* — I = co-latitude,
ZJf = 90* - A = co-altitude,
Z.PZM'= a = azimuth ;
to find PM= 90* — rf = polar distance,
Z.ZPM=^ t = hour angle.
Draw MQl. N8, and put ZQ = m,
then, if a < 90^ PQ = 90® - (i -f w),
and if a>90®,PQ = 90®-(i-m);
and, by Napier's Rules,
cos a =d=tanmtanA,
sin (f = cos PQ, cos MQ^
sin A = cosm cos Jf Q;
whence, tan m = ± cot A cos a,
sin c? = sin A sin(^ di m) seem,
in which the — sign is to be used if a > 90*. The hour angle
may then be found by means of [43], whence we have
sin ^ = sin a cos A sec d.
112
TRIGONOMETRY.
§ 67. P&OBLEU.
lb find the hour omgU of a heavenly body when its declina-
tion, its aUUude, and the kui-
^ tvde of the place are known.
In the triangle ZPM (Fig.
49),
given PZ=90**-Z,
Zaf=90^-A;
required
If, in the first formula of [46],
sin } ^ = Vein (s -- b) sin (a — c) esc b esc c,
we put
A^t, o = 90**-A, *=;?, c^W-l,
we have
8-b = W-\(l+p + h\ e-c = \(l+p-h),
and the formula becomes
sin } < = ± [cos } {I +p + A) sin J (l+p — A) sec i cscp]*
in which the — sign is to be taken when the body is east of
the meridian.
If the body is the sun, how can the local time be found
when the hour angle has been computed ? (See § 65.)
APPIiIOATIOKS. 113
§68. Pboblem.
To find the aUUude and azimuth of a celestial body, when its
declination^ its hour angle, and the latitude of the place are
known.
In the triangle ZPM (Fig. 49),
' given PZ= 90* — i,
PM=-Wf-d^p,
AZPM^t\
required ZJf = 90* — A,
APZM^a.
Here there are given two sideB and the included angle.
Placing PQ = m, and proceeding as in § 66, we obtain
tan9n=:cotcf co8^,
an A =9in(2+^) sinef seem,
tana =8ec(2+^) tan^sinm,
in the last of which formulas a must be marked E. or W., to
agree with the hour angle*
§69. Pboblsm.
To find the IcUUude of the place when the altitude of a celes-
■ial body, its declination, and its hour angle are knoum.
In the triaugle ZPJtf (Fig. 49),
given ZJf=90"-A,
Pif= 90**- rf,
ZZPM=t;
required PZ=90*~t
Let PQ = m, ^Q = n.
114
TEIGONOMETBY.
Then, by Napier's Rules,
cos t = tan m tan d,
sin A = cosw cos MQ,
sin 6? = cosm cos MQ;
whence,
tanm = cotrfco8^,
cos n = cos m sin h esc (f ,
and it is evident from the fig-
ure that
?=90**--(m±w),
in which the sign + or the sign
— is to be taken according as
the body and the elevated pole
are on the same side of the prime vertical or on opposite
sides.
In fact, both values of I may be possible for the same alti-
tude and hour angle; but, unless n is very small, the two
values will difier largely from each other, so that the observer
has no difficulty in deciding which of them should be taken.
§70. Pboblem.
Given the declination^ the right ascension of a star^ and the
^ obliquity of the ecliptic^ to find
the latitude and the longitude of
^^"^H the star.
M
FIg.W.
required QJf=90**— w,
Let M (Fig. 50) be the star,
P be the pole of the equinoctial,
and Q the pole of the ecliptic.
Then, on the triangle PMQ,
given PQ = 6 = 23^27',
PJf=90^-rf,
ZJfPQ= 90«+r(see Fig. 47);
and ^PQM=^ 90** - v (see Fig. 47).
APPLICATIONS. 115
In this case, also, two sides and the included angle are given.
Draw MHA^PQ, and meeting it produced at jET, and let
Bt' Napier's Rules,
sin r =tanw tanrf,
sin u = cos(c + w) cos2fJ?|
sin c? = cos n cos MS,
Bin(« + «) = tant; tan^fZ?)
sin n = tanr tan MH\
whence, tan n = cot c? sin r,
sin w = sin 6? cos (tf + n) sec n,
tant7 = tanr 8in(tf + w) cscn.
To avoid obtaining u from its sine we may proceed as fol-
lows:
From the last two equations we have, by division,
sinu = tan t7 cot(tf + w) sine? cot r tann.
By taking JfSTas middle part, successively, in the triangles
JfQJTand MFH, we obtain
cosu co8V = cosc2co8r;
whence, cos w = sec v cos d cos r.
From these values of sinu and cosu we obtain, by division,
tanti = sin t7 cot(e + n) tan d esc r tann.
From the relation
sin r = tan n tan rf ,
it follows that tan c? esc r tan n = 1,
Therefore tan m = sin v cot {e + n),
a formula by which u can be easily found after t; has been
computed.
116 TBIGONOMETRY.
EXEBOISE XXXII.
1. Find the dihedral angle made by the lateral faces of a
regular ten-sided pyramid; given the angle -4 = 18°, made at
the vertex by two adjacent lateral edges.
2. Through the foot of a rod which makes the angle A with
a plane, a straight line is drawn in the plane. This line makes
the angle B with the projection of the rod upon the plane.
What angle does this line make with the rod ?
3. Find the volume V of an oblique parallelopipedon ;
given the three unequal edges a, b, c, and the three angles
I, m, 71, which the edges make with one another.
4. The continent of Asia has nearly the shape of an equi-
lateral triangle, the vertices being the East Cape, Oape Eomania,
and the Promontory of Baba. Assuming each side of this tri^
angle to be 4800 geographical miles, and the earth's radius to
be 8440 geographical miles, find the area of the triangle : (i.)
regarded as a plane triangle; (ii.) regarded as a spherical
triangle.
5. A ship sails from a harbor in latitude I, and keeps on
the arc of a great circle. Her course (or angle between the
dii^ection in which she sails and the meridian) at starting is a.
Find where she will cross the equator, her course at the equa-
tor, and the distance she has sailed.
6. Two places have the same latitude /, and their distance
apart, measured on an arc of a great circle, is d. How much
greater is the arc of the parallel of latitude between the places
than the arc of the great circle? Compute the results for
Z = 45^ rf=90^
7. The shortest distance d between two places and their
latitudes I and F are known. Find the difference between
their longitudes.
8. Given the latitudes and longitudes of three places oil the
earth's surface, and also the radius of the earth ; show how to
find the area of the spherical triangle formed by arcs of great
circles passing through the places.
APPLICATIONS* 117
9. The distance between Parie and Berlin (that is, the arc
of a great circle between these places) is eqnal to 472 geo-
graphical miles. The latitude of Paris is 48'' 30' 13"; that of
Berlin, 52^ 30' 16". When it is noon at Paris what time is it
at Berlin?
Note. Owing to the apparent motion of the son, the local time over
the earth's surface at any instant varies at the rate of one hour for 15®
of longitude ; and the more easterly the place, the later the local time.
10. The altitude of the pole being 45^, I see a star on the
horizon and observe its azimuth to be 45^; find its polar
distance.
11. Given the latitude I of the observer, and the declina*
tion d of the sun ; find the local time (apparent solar time) of
sunrise and sunset, and also the azimuth of the sun at these
times (refraction being neglected). When and where does the
sun rise on the longest day of the year (at which time d=
+23' 27') in Boston (l=^4Z*21% and what is the length of
the day from sunrise to sunset? Also, find when and where
the sun rises in Boston on the shortest day of the year (when
(2= —23'' 27'), and the length of this day.
12. When is the solution of the problem in Example 11 im*
possible, and for what places is the solution impossible ?
13. Given the latitude of a place and the sun's declination ;
find his altitude and azimuth at 6 o'clock a.m. (neglecting re-
fraction). Compute the results for the longest day of the year
at Munich (?= 48° 9').
14. How does the altitude of the sun at 6 a.m. on a given
day change as we go from the equator to the pole ? At what
time of the year is it a maximum at a given place f (Given
sin A = sin 2 sine?.)
15. Given the latitude of a place north of the equator, and
the declination of the sun ; find the time of day when the sun
bears due east and due west. Compute the results for the
longest day at St. Petersburg (i = 59** 56*).
118 TBiaONOMETEY.
16. Apply the general result in Example 15 (cost = coil
tan 6?) to the case when the days and nights are equal in
length (that is, when d = 0°). Why can the sun in summer
never be due east before 6 A.M., or due west after 6 p.m. ?
How does the time of bearing due east and due west change
with the declination of the sun ? Apply the general result to
the cases where Kd&nd l=d. What does it become at the
north pole ?
17. Given the sun's declination and his altitude when he
bears due east ; find the latitude of the observer.
18. At a point in a horizontal plane MN ^ staff OA is
fixed, so that its angle of inclination AOB with the plane
is equal to the latitude of the place, 51** 30' B"., and the direc-
tion OB is due north. What angle will OB make with the
shadow of OA on the plane, at 1 p.m. ?
19. What is the direction of a wall in latitude 52® 30' N.
which casts no shadow at 6 A.M. on the longest day of the
year?
20. At a certain place the sun is observed to rise exactly in
the north-east point on the longest day of the year ; find the
latitude of the place.
21. Find the latitude of the place at which the sun sets at
10 o'clock on the longest day.
22. To what does the general formula for the hour angle,
in § 67, reduce when (i.) A = 0^ (ii.) ? = 0'* and rf=0^ (iii.)
iorrf = 90®?
23. What does the general formula for the azimuth of a
celestial body, in § 68, become when < = 90® = 6 hours ?
24. Show that the formulas of § 69, if <=:90®, lead to the
equation sin^=sinAc8C6?; and that if 6?=0®, they lead to
the equation cos Z = sin A sec U
25. Given latitude of place 52® 30' 16", declination of star
38®, its hour angle 28® 17' 15" ; find its altitude.
APPLICATIONS. 119
26. Given latitude of place 51^ 19' 20", polar distance of
star 67^ 59' 5', its hour angle 15^ 8' 12" ; find its altitude and
its azimuth.
27. Given the declination of a star T 54', its altitude 22*
45' 12", its azimuth 129^ 45' 37" ; find its hour angle and the
latitude of the observer.
28. Given the longitude u of the sun, and the obliquity of
the ecliptic 6=23^27'; find the declination d, and the right
ascension r.
29. Given the obliquity of the ecliptic tf = 23® 27', the lati-
tude of a star 51^, its longitude 315^ ; find its declination and
its right ascension.
30. Given the latitude of place 44'' 50' 14", the azimuth of
a star 138* 58' 43", and its hour angle 20"* ; find its declination.
31. Given latitude of place 51* 31' 48", altitude of sun west
of the meridian 35* 14' 27", ite declination +21* 27' ; find the
local apparent time.
32. Given latitude of place I, the polar distance j? of a star,
and it49 altitude A ; find its azimuth a.
FOEMULAS.
1. 8in*-4 + cos'ul = 1.
2. J, Bin A
COSil
isin -4 X C8C il = 1.
cos -4 X sec ^ = 1.
tAn-4xcot-4 = l.
§6.
4 sin (a? + y) = aiia?cosy + coBa:8iny.
6. cos (a? + y) = cos a? cosy — sin a: sin y.
6. tan(a? + y)=,^^^ + ^°y>
ti«iv^-ry; i_tana?tany
7. cot(:r + y) = 22t£cg2r3i
^ ^^ eota? + ooty
8. sin (flf — y) = fiina; cosy — cosa? siny.
0. cos (ar— y)=co8a: cosy + sina: siny.
iA i / \ tana: — tany
10. tan(a? — y) = =-r-i r-^-^
^ ^^ 1 + tanartany
-- ./ \ cota?coty + l
IL cot(a? — y) = — r ^-7 —
^ ^^ cot y — cot a;
12. sin 2a; = 2 sin a; cos a;.
13. cos2ar
= 2Binarcosa;. ^
= co8*a? — sin'ar. J
;88.
§31.
§82.
122
FOBUULAS.
14
16.
16.
17.
18.
19.
sa
2L
22.
tan 2a; =
2tana;
cot 2x =
1-tan'a;
cofar — 1
2 cot a;
33.
. — cos z
sin } « = ±-vl—
««= 1 - _!_ / I + cos z
cosJz=±^— t:^
taniz
cot iz
=^
=.VB
— COS z
cosz
cosz
cosz
34,
sin ^ - sin ^ = 2 cos i (^ + J?) sin J (.4 - J5).
cos ^ + cos 5 = 2 cos i (-4 + £) cos i (.4 — ^.
cos ^ — cos J5 = — 2 sin } (^ + 5) sin } (-4 — JB).
sin ^ + sin ^ _ tan K-^ + •^)
sin ^ — sin JS tan } (4 — J5)'
§35.
26. ^ =
a sin^
27.
2&
b sinJS
a* = i« + c»-2i(?cosX §37.
g — 6 _ tan K-^ ~ ^) § o q
a + i tanj(^ + ^y ^
8ini^=^^|5EME^. §43.
FORMULAS.
123
coai A
\ be
ZL2l
80. tanJ^=*pE*)IiEi).
\ 8(8 — a)
8L J (»-fl)(«-ft)(«-c) _^.
82. tanj^ = -l-.
8— a
88. J'=ia<?sinA
34 ^=V«(«-o)(«-i)(»-c)
36.
86. F=ir(a + b + c) = r8.
§43.
§44.
Sphbbical Tkiqonombtey.
87. coac = coBaeoBb.
gg^ r sin a = sin c sin il.
I sin & = sin £r sin B,
jg r cos il = tan i cote.
* I cos -B == tanacot e.
^ r cos .^ = cos a sin B,
I cos £ = cos i sin .^.
^ fsini =
lsina =
tan a cot A.
tan b cot B,
42. cos c = cot il cot J?. .
§46.
isin a sin J? = sin ft sin A. 1
sin a sin {7 = sin c sin il. > § 51.
sin & sin (7 = sin c sin ^. J
124
FORMULAS.
r COS a = COS i cos <? + sin 6 sin c cos A.
44 -j cos 5 = cos a cos c + ^^ a sin c cos B.
I cos (? = cos a cos 6 + sin a sin b cos C,
{cos ^ = — <
cos ^ = — <
cos C = — <
' cos ^ = — cos -B cos C + sin 5 sin O' cos a.
■ cos ^ cos C + sin i4 sin C cos b,
• COS. A cos -B + sin -4 sin B cos <?.
§61.
47.
sin } ^ = Vsin (a — b) sin (a — c) esc 5 esc c,
46. • cos } -4 = Vsin « sin (s — a) esc i esc c,
tan J -4 =Vcsc « esc (« — a) sin (« — b) sin (s — c).
sin } a = V— cos S cos (JS — -4) esc B esc (7.
cos } a = Vcos (/S — -B) cos (JS — (7) esc B esc 01
tan } a = V- cos/Scos (/S--4) sec (/S-jB) sec (/S-O).
§511
48..
49.
cos J (-4 + -B) cos } = cos } (a + J) sin J C
sin } (-4 + -B) cos J tf « cos J (a— i) cos } C.
cos } (^ — 5) sin } {? = sin } (a + *) siii i 01
sin } (-4 — ^) sin } (? = sin } (a — i) cos } C.
tan i (^ + ^) = °°" ; (" 7 S cot i a
cos i (a + o)
tan J (4 - 5) = "°;i"-^) coti a
C08}(^-f-jB)
tani(a-J) = SBiMn^ tan i c.
50. F=
E
180°
■S*.
§53.
6L tan't^=tanj(«-a)tani(«-6)tftn}(«-c). .
60.
EXAMINATION PAPEES.'
PLANE TRIGOKOMETRY.
I.
{Haarvard CoUegt, AdmUnon, June^ 1881. Time, H Aours.)
1. Define a logarithm. What is the logarithm of i in the
system of which 16 is the base ? Find the logarithm of 25 in
the same system.
2. Compute the value ot^^j^^^^ by logarithms.
3. Find the functions of 127* 10' from your trigonometric
tables.
4. Prove the formula
(cos A - cos By + (sin A - sin J?)« = 4 sin«^^^^=^.
5. Two sides of a triangle are 243 feet and 188 feet, and
the angle opposite the second side is 42° 20'. Solve the tri-
angle completely.
6. A pine tree growing on the side of a mountain, which is
inclined to the horizontal at an angle of 20°, is broken by the
wind but not severed at a distance of 40 feet from the ground.
The top falls toward the foot of the mountain, and strikes the
ground 50 feet from the base of tha tree ; find the height of
the tree.
* Note. In theso papers, as in many text-books, the Greek letters a
(alpha), p {bayta), y {gamma), 6 (ddta), 6 {tJiayta), ^ (phec), are occasionally
used to denote angles.
126 TBIGONOMETBY.
II.
(Harvard College, Admistion, June^ 1882. Timej 1} Tiours.)
1. Explain the reason of the rule for finding the character-
istic (or integral part) of the logarithm of a number.
Show that (according to this rule) the mantissa (or frac-
tional part) is always positive.
In what cases is the logarithm, as a whole, positive, and in
what cases negative ?
Thus, state clearly the value of the logarithm of 36,270 ; of
0.003627. What decimal must be added to the latter loga-
rithm to produce the logarithm of 0.01 ?
2. Find the time required to increase a sum of money a
hundred fold, at ten per cent per annum, compound interest,
payable yearly.
3. Find the formulas for the trigonometric functions of
90^+ a.
4. Find by the tables the logarithms of the trigonometric
functions of 290° 38' (marking the signs).
6. An observer from a ship saw two headlands. The first
bore E.N.E. (i.e. 6T 30' from N. towards E.), and the second
N.W. by N. (i.e. 33^45' from N. towards W.). After he had
sailed 16.25 miles N. by W. (i.e. IV 15' from N. towards W.),
the first headland bore due E., and the second N.W. by W.
(i.e. 56** 15' from N. towards W.). Find the direction and dis-
tance of the second headland from the first.
6. Prove the formulas :
22£^^ = -tanKa + i8) tanKa- A
cosa + cos)8
l + tan'^d
EXAMINATION PAPERS. 127
III.
(Edrvard ColUgt^ Frahman Eascminaiion. April, 1879. lime, 3 hours.)
1. Prove the relations between the sine, cosine, and tan-
gent of 90^+^ and the functions of ^. Draw a figure for the
case where ^ is obtuse, and show that the proof still holds
good. •
Confirm your results by means of the formulas for the sine
and cosine of the sum of the two angles.
2. Deduce formulas for the sine, cosine, and tangent of 2 a
and Ja, in terms of functions of a.
3. Prove the formula :
cos (a + P) sin^S — cos(a + y) siny =
sin (a + P) cos j8 — sin (a + y) cosy.
4. Prove that in any triangle
a*=b*+c^—2bc cos A.
5. Deduce the formulas for the tangents of the half angles
of a triangle, in terms of the sides.
6. Solve the triangles :
C=35^ a = 600, <? = 250,
£ = 22** 22', a = 67.06, b = 60.03.
7. The Delta measures 241 yds. on Cambridge St. and 115
yds. on Quincy St., and the angle between these streets is
88** 52'. Find the other angles of the Delta.
8. Find the area of the Delta.
9. A person travelling east in a railroad train observes a
tower situated south of a station A^ and on the same horizontal
plane with the railroad. At the station £ he finds that the
distance of the tower is 2 miles ; and at C, 8 miles from J3, its
distance is 4 miles. Find the distance of the tower from A,
128 TEIGONOMETEY.
IV.
{Harvard CoUegt, Frethman ExamirMlion. April, 1880. Time, 3 hours.)
1. Deduce the formula sin (a + )8) = , drawing the figure
for the case in which a is in the second quadrant and a + fi
in the third.
2. Deduce the formulas for cos 2a, sin} a, and cos^a, in
terms of functions of a.
8. Prove the theorem of the sines.
4. From the fundamental formulas deduce the formula
tani(^-^) a-b
5. Prove that
^^^(^ + ^+y) =l-tanatanff-tanfftany-tanytana.
cos a coap cosy
6. In a triangle 5 = 4^13.4' and a = 2001, give all the
solutions in the following cases :
(1) J = 160,
(2) 4 = 200,
(3) 4 = 2001.
7. -4, -B, and Care the comers of a triangular field. A is
40 ft. W. of B and 400 ft. S.W. of O. What is the area of
the field ? What is the length of the fence which encloses it?
8. From two comers of the Delta, A and 5, lines which
make angles of 19** 52' and 57° 32' respectively with the side
AJB meet directly under Memorial Hall tower. The length
of AB is 345.1 ft., and the apgle of elevation of the tower at
A is 32'' 26'. Find the height of the tower, and its angle of
elevation at B,
EXAMINATION PAPEES. 129
V.
(Harvard College, Freshman Examination. April, 1881. Time, 3 hours.)
1. Deduce tlie formulas which connect the functions of
(90^ + <^)and^.
2. Prove the fundamental formula for cos (a + j8). i
3. From the formula just found, obtain three values for
cos 2 a.
4. Find the values of sin^ and cos^, in terms of coso.
2 2
5. Prove the formula
(cos A - cos By + (sin ^ - sin £)• = 4 sin* ^'I^
2
6. Solve the following triangles :
i = 2434, c = 1881, = 42^22'.
a = 0.00543, c = 0.07003, a == 4** 27'.
7. The sides of a triangle are 715, 541, and 868 ; find one
angle and the area.
8. The height of Memorial Hall tower is 190 feet. From its
top the angles of depression of the comers of the Delta which
lie on Cambridge St. are 57** 44' and 16** 59', and the angle
subtended by the line joining these corners is 99** 30'. Find
the length of the Delta on Cambridge St.
VI.
{Harvard College, Freshman Fxamination. April, 1882. Time, 3 hours.)
1. Obtain the formulas which connect the sine, cosine, and
tangent of (180* + 4d ''^ith the functions of ^.
2. Assuming the formulas for the sine and cosine of the
sum of two angles, prove that
(1) tan(a + ^)^^
(2) sin ^a = V^ (1 — cos a).
130 TRIGONOMETRY.
3. Find all tlie values of a;, between 0° and 360°, whicli will
satisfy the equations
(1) tan a; = 2 sin 2 a;,
(2) (sin X + cos xf = 2 sin 2 x.
4. The length of each side of a regular dodecagon is 24
feet ; find the radius of the inscribed circle and the area of the
polygon.
5. In a certain triangle, a = 20, 5 = 3** 24^ (7= 85** 31'.
Find t? by aid of the table containing the values of 8 and T,
G. Prove the Theorem of Sines, and solve the triangles
(1) J = 468, c = 327, (7=34*^15';
(2) a = 0.003641, c = 0.08091, ^= 6** 20'.
7. Given a^ = J^ + ^ — 2 Jc cos -4 ; obtain the formula
sini^ = ^(^-^H^-^ .
The sides of a triangle are a = 2408, J = 2028, c = 1884;
find the angle A,
S. Hingham is 12 miles south-east of Boston ; Quincy is 6f
miles west of Hingham. How far is Quincy from Boston ?
9. A person ascending Memorial Hall tower stops to rest
at a window, and notices that the angle of elevation of the
vane on Appleton Chapel is 3° 34'. When he reaches the top
of the tower, 190 feet above the ground, he finds that the
height of the Chapel subtends an angle of 13® 9'. The hori-
zontal distance between the two towers being 492 feet, find
tli6 height of Appleton Chapel and the distance of the window
above the ground.
EXAMINATION PAPEES. 131
VII.
{Cambridge, Eng, 2nd Frevums JSrom., Dec. 7, 1876. Time, 2} hours.)
1.* Assuming that the angle subtended at the centre of any
circle by an arc equal to its radius is a constant angle, show
that any angle may be expressed by the fraction — — — , the
constant angle being taken as the unit.
Find the length of the arc subtended by an angle of 60® in
a circle whose radius is 3 feet.
2. The sine of a certain angle is J ; find the other trigono-
metrical ratios of the angle.
8. Trace the change in sign and magnitude in the tangent
of an angle, as the angle increases from 0** to S60^,
4. Find, by a geometrical construction, the cosine of 60**
and of 45°, and deduce the value of cos 3360** and cos 2565°.
5. Prove the formulas :
(1) sin (A — B) = sin -4 cos -B — cos A sin -B,
(2) cos2^ = J-=^^.
(3) Bin2^+8in4^ ^^^3^^
cos2^+cos4-4
6. Express the cosine of half an angle in terms of the sine
of the angle, and explain the double sign.
Employ the formula to find the value of cos 75°, having
given sin 150° = J.
7. li A^ B^C be the angles of a triangle, and a, 5, c the
sides respectively opposite to them, show that
where 8 = one-half the sum of the sides.
8. Find the greatest angle in a triangle whose sides are 7
feet, 8 feet, and 9 feet.
*For aid in solving this and similar questions, see Wentworth & Hill's
Tables, pages xi and zii.
132 TEIGONOMETRY.
VIII.
{Carribridge, Eng. 2nd Previous Exam., Dec. 7, 1877. Time, 2 J hmn)
1. Define the cosine, cotangent, and cosecant of an angle,
and prove that these ratios remain unchanged so long as the
angle is the same.
Find the value of these three ratios for an angle of 45®.
2. Prove the formulas :
(1) sin A = VI — co8*-4,
(2) cos^ = - ^
Vl + tanM
If sec A = V2, find tan A, Am. 1.
3. Prove that
sin (90** + A) = cos A, and cos (90* + ^) = — sin -4.
Hence show that cos (180° + -4) = — cos A,
4. Show that cos* A tan' A + sin' A cot' -4 = 1.
5. Prove that cos (-4 + J5) = cos -4 cos J5 — sin J. sin B,
Hence show that
co8(^ + .g + =cot^cotJgcotg-cot^-cot.5~cot(7.
sin A sm B sm O
6. Given that sin -J -4 =^ -f, find the value of tan -4.
Ana.^.
7. Prove that the sides of any plane triangle are propor-
tional to the sines of the angles opposite to these sides.
If 2s = the sum of the three sides (a, J, c) of a triangle, and
if A be the angle opposite to the side a, prove that
2
sin A = ^Vs(s - a) (s — b) (s — c).
8. Prove that in any plane triangle
ta.ni(A-B) = ^^ cotiG
EXAMINATION PAPERS. 133
9. If the side a and the angles A and £ of a triangle be
known, prove that the side b ma,j be found by means of the
formula,
log b = log a + log sin ^ — log sin A.
Find J, having given that a=1000 yards, ^ = 50**, -B=64^
Ana. 1173.29 yards.
10. The minute-hand of a clock is 3 feet 6 inches in length ;
find how far its point will move in a quarter of an hour, it
being assumed that ir = ^. Ana, 5 feet 6 inches.
IX.
(Cambridge, Eng, 2nd Preoioxu Hxam,, Dec. 10, 1878. Time, 2J A(wr».)
1. Define sine, cotangent; and prove that sin*-4+cosF-4=l.
Express the other trigonometrical ratios in terms of the cosine.
2. What is meant by the circular measure of an angle?
How is the number of degrees in an angle found from its cir-
cular measure ? How many degrees are in the unit of circular
measure ?
3. Prove that
(1) sin(180»+^) = -sin^
(2) tan (90**+-4) = - cot^.
What is the use of these equations ?
4. Find the general form of all the angles whose sine is the
same as sin $,
Write down the sines of all the angles which are multiples
of 30^ and less than 360**.
184 TBIGONOMETEY.
5. Prove the following relations :
(1) cos (-4 — ^) = cos -4 cos -B + sin ^ sin -B,
(2) sin^ + sinJ5 = 28ini(^ + ^)cosi(^-5),
(8) tan«^==^~^^^^A
W taa^ l + cos2^
6. Find co830^ tan 45^ sin 15^
7. If tan -4 + sec -4 = 2, prove ttat sin -4 = |^ when A is less
than 90^ *
If sin -4 = ^, prove that tan-4 + sec -4 = 3 when A is less
than 90^.
8. Prove that cos 3 -4 = 4 cos* -4 — 3 cos ^ , and find tan 3 A
in terms of tan A,
9. In a triangle -4, £, (7, whose sides are respectively a, J, c,
prove that
(1) 8mi^=^ii35Ei.
/o\ sin A _ sin^ __ sin (7
10. Solve a triangle, having given two sides and the angle
opposite one of them.
Find -4, J?, 5, having given a = 25, <?= 24, (7= 65*59'.
X.
{Cambridge, Eng. 2nd Previous Exam., Dec. 10, 1879. Time, 2} Jmirs,)
1. Define V. Assuming that -^ is the circular measure of
two right angles, express the angle A** in circular measure.
2. Define the sine, secant, and cotangent of an angle, and
express any two of these ratios in terms of the third.
Find the trigonometrical ratios of the angle whose cosine
isf.
EXAMINATION PAPEES. 135
3. Prove that
(1) cos (180^ + ^) = cos (ISO**-^),
(2) tan (90** + ^) = cot (180**-^).
4. Express the cosine of the difference of two angles in
terms of the sines and cosines of these angles.
Prove that
* tan~^a: + tan'^y = tan"* ^^LjlJL.
l—xy
5. Prove the formulas :
(1) cosa: + cosy = 2co8 "^^ cos^^lJ^i
(2) %\n\x + cos-Ja? = db VT+smS,
(3) sin2r(2 cos2r — 1) = 2 sin^a: cosf ar.
6. Trace the changes in sign and magnitude of
2sintf — sin2g
2sin0+sin20'
as Q changes from to 2 tt.
. 7. Express the cosine of any angle of a triangle in terms of
the sides of the triangle.
If the angle opposite the side a be 60°, and if 5, <? be the
remaining sides of the triangle, prove that
(a+S+c)(J+c-a) = 35tf.
8. Solve a triangle, having given the three sides.
Given A — 36°, B = 72°, and a =f 1 ; solve the triangle.
9. The sides of a triangle are 2, 3, 4; find the least angle.
♦ tan-^a;— arc whose tangent is «.
136 TRIGONOMETEY.
XI.
{Cambridge, Eng. 2nd Prevuma Exam., Dec. 10, 1880. Tiw^, 2} Tumn)
1. Express in degrees, minutes, etc., (i.) the angle whose
circular measure is -^^ir; (ii.) the angle whose circular measure
is 5.
If the angle subtended at the centre of a circle by the side
of a regular pentagon be the unit of angular measurement, by
what number is a right angle represented?
2. Find, by geometrical constructions, the cosine of 45° and
the sine of 120**.
Prove that
(sin 30° + cos 30°) (sin 120° + cos 120°) = sin 30°.
3. If CSC -4 = 9, find cot -4 and sec -4.
4. Prove that
cos(180°+^) = -cos^.
Find the value of (i.) cot 840°; (ii.) sec37r.
5. Assuming the formula for the sine of the sum of two
angles in terms of the sines and cosines of the separate angles,
find (i.) sin 76° ; (ii.) sin 3 -4 in terms of sin -4.
6. Prove the formulas :
(1) cos«(^ - J5) - 8in«(^ + J5) = cos2u4 cos2 J?,
(2) l + tana; tanja? = seca?.
7. Prove that
cos-4+cosj5 = 2cos^(^+5) cos|(-4— -B),
1 cos 5 ^ + cos $ • 1 i
and express — --— ^ as a smcle term.
cos6^ — cos^.
8. Solve the equations :
(1) 5 tan' a; + sec*a; = 7,
(2) cos5 tf + co83tf = V2 cos4tf.
EXAMINATION PAPEES. 137
9. Proye that in any triangle cob -4 = — ^~-
Obtain the formula for tani j1 in terms of the sides.
10. Find an expression for the area of a triangle in terms
of its sides. The lengths of the sides of a triangle are 3 feet,
5 feet, and 6 feet; what is its area?
11. Given that
sin 38^26' = 0.6213757, sin 38*26' = 0.6216036;
find the angle whose sine is 0.6215000.
XII.
{Oombridge, Eng, 2nd Previous Exam., Dec, 10, 1881. Timey 2} houn.)
1. Define the unit of circular measure. The ratio of the
circumference of a circle to its diameter being 3.14159, find
the circular measure of an angle of 126°.
2. Define the tangent, cotangent, and cosecant of an angle.
Find the tangent and cotangent of an angle whose cosecant is
1.25.
3. Trace the changes in sign and magnitude of sin ^ as ^
changes from 90° to 270°.
4. Prove the following :
(1) tan(ir + ^)=tan^,
tan A — tan B
(2) tan(^-5)=, _ ,
^ ^ ^ ^ l + tan^tan5
,3x sin^ + sin^^^^^^-^^
^ ^ cos-4-cos-B 2
138 TEIGONOMETEY.
5. Determine the value of cos 18°, and prove that
cos 36° = cos 60° + cos 72°.
6. Show that for certain values of the angles
2cos}-4 = Vl + sin-4 — VI — sin ^.
Is this formula true for values of A lying between 200° and
220° ? and if not, how must it be modified ?
7. Prove that in any triangle, with the usual notation,
and that the area is equal to
V5(s — a) (« — b) (8 — c),
Show, also, that
sin*-4. = cos*-B + 008*0^+ 2 cos A cos £ cos 0,
8. "When one side of a triangle and the two adjacent angles
are given, show how to solve the triangle.
Find the greatest side of the triangle, of which one side is
2183 feet, and the adjacent angles are 78° 14' and 71° 24'.
AlsrSWEES.
Exercise I.
1. Bin-B-* C08-B-?, tan5-^, cot5-?, secJ-^, cacB^t.
c c a a b
3. (i.) Bin =- f , COB - 4, (ii.) sin = A- «*c- (v) sin - «, etc.
tan = I, cot = I, f iii.) sin = -^y, etc. (vi.) sin - JJf , eta
Bee — J, CBC =» f . (iv.) Bin = ^j, etc.
4. The required condition is that o' + 6* — c*. It is.
5. (i.) Bin« _4^. etc. (iii.) ein - 1. etc.
(ii.) Bin-" J^, etc. (iv.) Bin - "^^ etc.
7. In (iiiOp'j* + 2*** — I)**" ; in pv.) m^n^t? + m'p* r* — n^^i'.
8. c — 145; whence, BinJ. — -^ — cos^; cobAh- JJf — Bin-B;
tan^-^-cot^; cot-A- J^- tan5; Beo4-}Jf-CBc5; etc
9. &«0.023; whence, tan A- cot^-^; cotA«- tanj?-^, etc.
10. a - 16.8 ; whence, Bin il - JJf "■ cob 5, etc.
11. e='p + q: whence, Bin J. — —^ ^ — cofl5 ; etc.
p + q
>l-
12. 6 — V J (p + j) ; whence, tan J. « -vf^ — cot-B ; etc.
13. a « J) — g ; whence, Bin A — r-T-§ =■ cob 5 ; etc.
14. Bin-A = {V6« 0.89443; etc. 15. ainJ.-}; etc.
IG. Bin ^ - J (5 + V7) - 0.95572 ; etc.
17. cosil - J (>/3l-l)- 0.57097; Bin ^ - i(\/31 + 1)- 0.82097; etc.
18. a =12.3. 20. a -9. 22. c = 40.
19. 6 = 1.54. 21. 6-68. 23. c- 229.62.
24. Constnict a rt. A with legs equal to 3 and 2 respectively; then
construct a similar A with hypotenuse equal to 6.
In like manner, 25, 26, 27, may be solved.
28. a - 1.5 miles ; 6 » 2 miles. 31. 400,000 miles.
30. a -0.342, 6-0.940; a -1.368, 6-3.760. 32. 142.926 yards.
TRIGONOMETRY.
Exercise II.
5. Througb A (Fig. 3) draw a tangent, and take -4jr=3; the angle
A0T\& the required angle.
6. From (Fig. 3) as a centre, with a radios =» 2, describe an arc cut-
ting at i^the tangent drawn through B \ the angle BOA is the
required angle.
7. In Fig. 3, take Oif = }, and erect MP ± OA and intersecting the
circumference at P; the angle POif is the required angle.
8. Since sin a: — cos a, Oif— FM (Fig. 3), and x = 45® ; hence, construct
aj-45^
9. Construct a rt. A with one leg = twice the other ; the angle opposite
the longer leg is the required angle.
10. Divide OA (Fig. 3) into four equal parts ; at the first point of divi-
sion from erect a perpendicular to meet the circumference at
some point P. Join 0F\ the angle AOPis the required angle.
21. r sin x, 22. Leg adjacent to J. — tic, leg opposite to J. « mc.
Exercise III.
1. cosBO^ cot p. 8ec71»50'. tan T 41'.
8in46^ tan76«. 8in62«3G'. sec 36« 14'.
2. CO8 30O. cot33«. Bec20*»58'. tanO^l'.
sinl6«. tan6^ Bin4*»21'. sec 44° 59.
3. JV3.
4. tanil-cot-4-cot(90®--4); hence,-4«90®-il and -4=45<>.
5. 30*». 7. 90«. 9. 22«30'. 11. 10«.
6. 30*» 8. W. 10. 18«. j2 ^**
n+1
Exercise V.
1. cosil = 3f^, tanJ. = J^, cot-4«»-jf^, sec-4=J^, C8c-4=l^f.
2. cos ^ = 0.6, tan ^ = 1.3333, cot^=»0.76, sec ^ = 1.6667, esc ^-1.25.
3. sinil = Ji, tan^ = JJ, cotil=»}f, secil^f^, cscA^ff
4. sin ^ = 0.96, tan il « 3.42857, cot ^ - 0.29167, sec ^-3.5714.
5. sin A = 0.8, cos A = 0.6, cot A = 0.75, sec il =» 1.6667, esc A =- 1.25.
6. sinil-}V2, co8^=»}\/2, tan-4-1, sec-A=>/2, csc^=-V2^
7. tan ^ - 2, sin A » 0.90, cos ^ - 0.45, sec A » 2.22, cec A - 1.11.
AKSWEBS. 8
8. cosil-}, 8iii-A-}\/3, tan-A-VS, cot-A-}V5", csc-A-lVJ.
9. 8in-A-}V2^ C08-A-JV2. tan-A-1, cotil-1, Beoil-\/2.
10. cos-A-VI-to', tan-A-r-^Vr^*, cotil-ivrrr^i
1 — m' wi
11. coB^-i^ tanil--^. cot^-i^*. Mc.l-i±^'.
1+m' 1 — m* 2m 1— »>^
lasin^-^!^'. tan ^-2^ sec ^-5^'.
m' + n' 2mn 2mn
13. cot- 1, Bin-}V2. co8-}\/2. seo-V^ cse-iV^
14. cofl-}V5, tan-jVs". cot-V3, Bec-fV3, c8o-2.
16. 8m-}V3, cofl-J, tan->/3, cot-J>/3, seo-a
16. Bin-}V2-V3, C08-JV2+V3; cot-2+>/a[
17. Bin-}V2-V2, COB-JV2+V2, tan-V2-l.
18. CO8 — 1, tan>-0, cot««oo, secail, esc — oo.
19. co8««0, tan — 00, cot — 0, sec — oo, esc— 1.
20. sin — 1, cos —0, cot — 0, sec — oo, esc — 1.
21. coB^-Vl-8in«il, tan^ "°^ , c8c-4--L-.
Vl~8inM wn-4
22. sin^-Vl-co8»il, tanil-:i!^IHi^, cot il - -J22L4_,
Bec-A--^-, csc^- ±
^^^ Vl-cosM
23. Bin^-— ^4=. cos^-_^i__ cot^-.
Vl+tan«il ^/TT^2 tanil
sec-4-Vl+tanM. esc il - Ji^L±^!l.
tan J.
. . Vl+COt>ui
cos^^ ^^^"^ . ,^^- >^l+cotM
Vl + cotM cot il
25. sinil-jVg; cos^-fVS; 27. sin^-A. coe^-}f.
26. Bin^-J\/i6. tan^-VIB. ^ l-ScoeM + ScosM
cos' .4 — cos* J.
TEIGONOMETEY.
Exercise VI.
1. - — COB J. ; .'. c — J. , . o
39. c - 7.8102, A « 39<» 48'. 5 - 50<> 12'.
40. 6-69.997, il-30'12", -B « 89*> 29' 48'».
41. a« 1.1885. ^ - 43<» 20'. -B « 46° 40'.
42. 6-21.249. c- 22.372. 5-7P46'.
43. a -6.6882. c- 13.738, 5-60«>52'.
44. a- 63.86. 6-23.369. 5-20*' 6'.
45. o« 19.40. 6-18.778. il-450 56».
46. 6-53.719. c- 71.377, il-4Pll'.
47. a- 12.981, c- 15.796. il-55«16'.
48. a- 0.58046, 6-8.4418. il- 3^56'.
49.2?'-J(«'8in^co8il). 51. J* -}(6«tanX).
50. l^=}(a« cot ^). 52. F '^HaVc^-a^.
53. 6-11.6, c - 15.315. il-40*» 45' 48", -B - 49<> 14' 12".
54. a- 7.2. c- 8.766. J? - 34° 46' 42". il - 55° 13' 18".
55. a- 3.6473, 6 = 6.58, c- 7.5233, -B-61«.
56. a- 10.26, 6-19.42, ^-2?' 51', 5-62»9'.
57. 19° 28' and 70° 32'. 65. tan^--, ^-59° 45'.
6
58. 3 and 5.196.
66. a-6tan^, 95.34.
67. 1°25'56".
9QO 68. 7.071 miles in each direction.
QQO
59. o-ccoB -f 67. 1°25'56".
»+l
"^n + 1' 69. 20.88 feet.
60. 36° 52' 12" and 53° 7' 48". 70. 56.65 feet.
61. 212.1 feet 71. 228.63 yards.
62. 732.21 feet. 72. 136.6 feet.
63. 3270 feet. 73. 140 feet.
64. 37.3 feet, 96 feei 74. 84.74 feet.
Exercise VII.
1. C-2(90°~^), c-2aco8-4, A-asin-^.
2. il-}(180°-(7), c-2acoB^, A-asinA
3. C-2(90°-il), a-c-<-2coB^, A-asiniL
AKSWEBS.
4.
A^i(lS(y>-C!). a-c + 2coeil,
A — asinX.
6.
C-.2 (90<>--4), a-A + Binil.
e — 2 a cos .i.
6.
A^i(lSO^^C), o-A + sin^
c — 2 a cos il.
7.
flin^-A-i-a. C-2(90<>-^)
c — 2 a COS J..
8.
tanil-A-i-Jc. C-2(9a>-il)
a- A -H sin .4.
9.
^ - er 22' 60". C- 450 14' 20", h - 13.2.
10.
c- 0.21943, A -0.27384,
i^- 0.03004.
11.
a -2.0565, A -1.6862.
F" 1.9819.
12.
a -7.706, c- 3.667.
^- 13.73.
13.
il-79*>37', C- 20*46'.
c-2.42.
^14.
^-7r'19'. C-25«'22'.
a -20.5.
* 15.
il-25<»28'. C-129«4'.
a -81.4, A -35.
16.
il-8n2'. (7-ir'36'.
a- 17. 6-5.3-
17.
i?'-icV4a«-c«.
22. 0.76537.
18.
^-a'sinJCcoeJC.
23. 94*' 20'.
19.
F^a^smAcoeA,
24. 2.7261.
20.
i?-A«tanja
26. 38«66'33".
21.
28.284 feet. 4625.44 sq. feet.
26. 37.7
Exercise VIII.
1.
r - 1.618. A - 1.5388. i^- 7.694.
2.
r - 11.269. A - 10.885. ^- 380.9.
3.
A -0.9848, p- 6.2514. 2^-3.0781.
4.
A- 19.754, c- 6.2536, ^-1235.4.
5.
r - 1.0825, e - 0.8284, F^ 3.3137.
6.
r-2.5fe, A -2.488, c- 1.4615.
7.
r- 1.599. A -1.441. p- 9.716.
8.
0.618. 12. 0.2239.
17.
11.686
9.
0.6498. 13. 0.31.
18.
99.641
10.
0.51764. 14. 0.8284.
19.
1.0235.
11
J e 15. 94.63.
20.
0.635.
XA.
2co8?2! 16. 415.
TEIGOKOMETEY.
Exercise IX.
5. Two angles : one in Quadrant I., the other in Quadrant 11.
6. Four values : two in Quadrant I., two in Quadrant IV.
7. X may have two values in the first case, and one value in each of
the other cases.
8. If cos af — —J, oj is between 90® and 270® ; if cot a? — 4, » is between
0® and 90® or 180® and 270® ; if sec a? - 80, a? is between 0® and
90® or between 270® and 360®; ifcsca?--3, «is between 180®
and 360®.
9. In Quadrant III. ; in Quadrant II. ; in Quadrant III.
10. 40 angles ; 20 positive and 20 negative.
11. +, when X is known to be in Quadrant I. or IV. ; — , when x is known
to be in Quadrant II. or III.
14. sinaj — — fVS, tanoj — — 4\/3, cota — i— , csco!"- ^.
4V3 4V3
15. sinaj— ± . cossj — t 1 tanaj — — }, seca— t^^^^»
\/lO >/IO 3_
CSC « — ± VlO.
16. The cosine, the tangent, the cotangent, and the secant are negative
when the angle is obtuse.
17. Sine and cosecant leave it doubtful whether the angle is an acute
angle or an obtuse angle ; the other functions, if + determine an
acute angle, if — an obtuse angle.
20. sin450 -sin (360 + 90) -sin 90«1; tan 540® = tan 180® = 0;
cos 630® -cos 270® =-0; cot 720® -cot 0® = oo;
sin 810® - sin 90® - 1 ; esc 900® - esc 180® - oo.
22. 0. 23. 0. 24. 0.
Exercise X.
11. cot264®=^*tan6®.
12. sec 244®- -CSC 26®.
13. CSC 271® --seel®.
14. sin 163® 49'- sin 16® 11'.
15. cos 195® 33' --cos 15® 33'.
16. tan 269® 15'- cot 0®45'.
17. cot 139® 17' --cot 40® 43'.
18. sec 299® 45'- esc 29® 45'.
19. esc 92® 25'- sec 2® 25'.
21.
45®, 135®, 225®, 315
25.
a«-6« + 4a6.
2.
sin 172®- sin 8®.
3.
cos 100® --sin 10®.
4.
tan 125® cot 35®.
5.
cot 91® --tan 1®.
6.
sec 110®- -CSC 20®.
7.
esc 157® - CSC 23®.
8.
sin 204® --sin 24®.
9.
COS 359®- COS 1®.
10.
tan 300® --cot 30®.
ANSWERS.
20. 8in(- 750) — Bin 75»^ cob15«,cob(- 75V cos 75«- Binl6«,ete.
21. 8m(-127®)- -am 127®- -cos 37*, coe(-127®)- cos 127®- -sin 37®, etc.
22. 8in(-200®)- sin 160®- Bin20®,co8(-200®)-co8200®— cos20®.etc.
23. 8in(-345®)— sin345®- Binl5®, co8(-346®)-co8346®-co8l5®.etc.
24. sinf- 62® 370 --Bin 62®37'--coe37®23',
cos(
i(- 62®370--8in 62®37'^ coe37®23',
8 (- 62® 370- cos 62® 37'- sin 37® 23', etc.
25. sin (-196® 540- »in 16®54'- cosl6®64',
cos(- 196® 64') - cos 196® 64' - -cos 16® 54', etc.
26. sin 120® - } Vs, cos 120® - -}. etc.
27. sin 135® - + J\/2, co8l35® - - Jv^, etc.
28. sinl60®- + J, cosl50®--}V3. etc
29. sin 210®--}. cos210®--}V3. etc.
3a sin225®--jV2, cos226®-- JV?. etc.
31. sin 240® --JVJ, cos 240®- -J, etc.
32. sin300®--}V3, cos300®- +}, etc.
33. sin (-30®) --}. oos(-30®) - + jV3,eto
34. sin (- 225®) - + } VS", C08(- 226®) - - J\/2, etc
35. co8a?--jV2 or - Vj, «tc., »-225®.
36. tanx-- VJi sinaj-}, cosa?--}>/3, a?-150®.
37. sin 3540® -sin 300®-- sin 60®- -}V3. cos 3540® - + }. etc.
38. 210® and 330® ; 120® and 300®.
39. 135®, 225®, and -225® ; 150® and -30®.
40. 30®. 150®, 390®, and 510®.
41. sin 168®. cos 334®, tan 225®, cot 252®.
sin 349®, cos 240®. tan 64®. cot 177®.
42. 0.848. (Hint: tan 238® - tan 68®. sin 122® - sin 68®).
43. -1.956. 47. a* + V + 2ahc(»x.
44. (a — 6) sin a;. 48. 0.
45. msinxcoso;. 49. cosx sin y— sins cosy.
46. (a — i)cot« — (a + &)tanx. 60. tanx.
51. Positive between x — 0® and x — 135®, and between x - 315® and
X - 360® ; negative between x - 135® and x - 315®.
52. Positive between x-45® and x-225®; negative between x-0®
and X - 45®, and between x - 225® and x - 360®.
53. sin(x- 90®)--cosx, cos(x- 90®) - sin x, etc.
64. sin (x - 180®) - - sin x. cos(x - 180®) - - cos x. etc.
Exercises 53 and 54 sbould be solved by drawing suitable figures, and
employing a mode of proof similar to that used in f 28
TRIGONOMETBY.
Exercise XI.
1. 8iii(aj + y) = Jj, C08(ar + y) = J|.
3. Bin (90° + y) = cosy, cos ( 90 + y) = -sin y, etc,
4. 8m(180-y)= siny, cos (180 - y) =~ cosy, etc.
5. sin (180 +y)« -siny, cos (180 +y) =-- cosy, etc.
6. sin (270 - y) = -cosy, cos (270 - y) - -sin y, etc.
7. sin(27O4-30=~cosy, cos (270® + y)=: siny, etc.
8. sin (360° - y)=i -sin y, <30s (360 - y) =- cos y, etc.
9. sin(360 + y)« siny, cos(360° + y>= cosy, etc.
10. sin (a? - 90°) « -cos aj, cos (a? - 90°) - sin «, etc.
11. sin (a? - 180°)= -sin «, cos {x - 180°>= -cos a?, etc.
12. sin (a? - 270°)= cos a?, cofl(a:-270°)«-8ina;, etc.
13. sin (- y) = -sin y, cos (- y) - cos y, etc.
14. sin (45°-y)= J V2 (cosy-sin y), cos(45°-y)- J >/2(cosy+siny),etc.
15. 8in(45°+y)=. J V2(cosy+siny), cos(45°+y)=} V2 (cosy-sin y). etc.
16 8in(30°+y)=J(co8y+V3siny), co8(30°+y)=}(\/3cosy-Biny),etc.
17. 8in(60°-y) = }(V3cosy-fiiny), cos(60°-y)=J(co8y+V3"siny),etc.
18. 3 sin a? — 4 sm' a?. 1 9. 4cos»a?— 3cosa;. 20. 0. 21. jV3.
22. sin ia;= ^^-y^ = 0.10051; cos }:.-^l^tMvf« 0,99494.
23. co8 2a; = -}, tan2a?° .-V3.
24. sin 22}° = } V2 - V2 = 3827, cos 22}° - } V2 +V2-= 0.9239.
tan 22}° « V2 -1 = 0.4142, cot 22}° = \/2 + l = 2.4142.
25. sin 15° = }V2 -VS = 0.2588, cos 15° -}V2+V3 = 0.9659.
tan 15° «2-V3 =0.2679, cot 15° «2+V3 =3.7121.
27-33. The truth of these ec^nations is to be established by expressing
the given fanctions in terms of the same function of the same
angle. Thus, in Example 27,
sin 2ap = 2 sin x cosx,
and 2 tan a; = 2 ?15f . 1 + tan'aj = sec»aj V-
cos a; cos'a?
By making these substitutions in the given equation its truth
will be evident.
34. sin -4. + sin5 + sin (7= sin^l + sin-B + 8in[180 - (-4 + B)]
= siuil +.8in5 4- Bin.(A + B)
= 28in}(-4 + B) cosUA - -B) + 28in}(^ + B) cos}(-4 + B)
= 28in}(A + B) [cos}U - B) + co8}(^ + B)]
= 4sin } (a + B) cos } il cos } B, (see J2 34 and 35)
But cos } C = cos [90° - } (^ + B)] = sin }(il + B).
Therefore, sin 2L + 8in .B + sin &= 4cos } J. cos } .5 cos } 0.
AKSWEBS. 9
35. Proof similar to that for 34.
cobAoobB cobAcobB cosC
sinC sin (7 BinCcosC+coBilcofl^BinC
COB A GOB JB COB cobAcobBcobG
(cobAcobB + cob C) Bin C _^ [cob A cobB — cob(A-\- B)] bjp C
*" cos ^ cob S cob (7 cob^cob^cobC?
^Biniisin^Bina -tan^tan^taaC.
COB il cob ^ cos (7
37. Proof similar to that for 36.
38 2 42. tan' a. aq coB{X'^y)
sin 2a; ^o co B(g — y) * Binxsiny
39. 2cot29. ' coBOJcosy* - 47. tans tan y.
40. ^^i'-y) 44. cofl(ag+y)
Bin a; cosy ' cos x cosy
^ C0B(g-fyX 45^ coB(x-y)
mnxcoBy ' sinxainy
EXEBCISE XII.
1. Ii; for instance, B*^ 90°, [25] becomes ^ «i sin^.
3. a«-6» + c", a«-i6« + c»-26c, a«-6« + c« + 26a
6. 90° in each case.
7. j[i.) «zJ - tan (^ - 45°). and a right triangle.
(ii.) a'hb''{a-h)(2 — V3), an isoBcelee triangle with the angles 30°,
30°, 120°.
EZEBOISE XIIL
9. 300. 15. a -5, 0-9.6593.
10. AB - 59.564 milee ; 16. a - 7, 6 - 8.573.
AC - 54.285 miles. ^^ g.^^^ ^^ ^^^^ ^^^ ^^^^^ ^^^^ .
11. 13.819 miles, 13.348 milee, altitade, 519.6 feet
11.32 miles. jg 855:1607.
12. 4.15 and 8.67. 19. 5.438 and 6.857.
13. 6.143 miles and 8.792 milee. no 15 588
14. 8 and 5.4722.
Exercise XIV.
11. 420. 12. The other diagonal » 124.617 ; the other side » 92.325.
10
TEiaOHOMETBY.
•
Exercise XV.
11. 6.
15. 25. 18. 10.266.
12. 10.392.
16. 3800 yards. 19. a -5.0031. 6-2.3385.
14. 8.9212.
17. 729.7 yards. 20. 26^ 0' 10" and 14° 5' 50''.
Exercise XVI.
11. -4-36^52',
B - 630 8', C- 90». 16. 45o, 60o. 75o.
12. ^-jB-33«33'27",(7-112053'6". 17. 4° 23' W. of N.. or W. of S.
13. A^B^C^
60«. 18. 60°.
14. Impossible.
20. 0.88878.
15. 16^46^120*
21. 64.516 miles.
Exercise XVII.
1. 4333600.
2. 365.68.
3. 13260,
4. 8160.
5. 240.
6. 26208.
7. 15540.
8. 29450 or 6982.7.
9. lOVS- 17.3205.
10. 6\/3- 10.3923.
11. 0.20021.
12. ah BmA,
13. J(a«-J«)tanX
14. 2421024.
15. 30°, 30°, 120°.
Exercise XVIII.
1. 21.1655 miles
2. 6.3397 miles.
3. 119.29 feet.
4. 30°.
24.966 miles.
5. 20 feet.
6. 2.6248 or 21.4588.
7. 276.14 yards.
8. 383.56 yards.
ANSWERS. 11
SPHERICAL TRIOONOMETRY.
EXEBCISE XIX.
1. ll(y». 100*, 80». 2. 140». 90°. 55«. 7. fir.2ir, V^r.
EZEBCISE XX.
3. (i.) Either a or 5 mogt be equal to 90°; if a>-90*, then ii»-90f and
jB-6; if6-9a>, then 5-90*>. and il-a. (ii.) c-.90*>, il-.90*>.
B^b. (iii.)-A-90*>, J?-6. (iv.) c - 90», il - 90». ^ - 90».
EXEBOISE XXI.
2. I. The cosine of the middle part «i the product of the cotangents of
the adjacent parts.
II. The cosine ot the middle part ^ the product of the sines of the
opposite parts.
EZEROISE XXII.
24. -4-176<>5ri0", B-135»42'55". C-135»34/8".
25. C-.104<»41'37.2''. o - 104*' 63' 0.8^', 6 - 133« 39^ 47.7".
26. sina-isinilcscC, cosi — — tan^cotC, oos.5«i~seciicoe01
27. o-60», 6-90». jB-90*>.
28. The triangle is impossible ; why ?
29. 6-.130«41'42''. o-7P2r43". il - 112* SV 0.7^.
30. a-26<>3'5y', il-36« jB-65»4a'e^.
81. Impossible; why?
EXEEOISE XXIII.
1. cos^i«ootatan}&, sin} j9* esc a sin} 6, cos A — cos a sec} 5.
2. sin}il'-}sec}a, orcosil>«}cosasec'}a, ortan*}^ — sin}acsc}a.
ifiQo 180°
3. sin}il««sec}acoszzrL, sinJS-isiniacsc----,
n n
sin r «- tan } a cot =^.
n
4. Tetrahedron, 70° 32^; octahedron. 109° 28^; icosahedron, 138° 11';
cube. 90°; dodecahedron, 116° 34<.
6. cot}il»Vcosa.
12 TEIGONOMETRY,
Exercise XXV.
1. (i.) tan m » tan 5 cos A, (ii.) tan m » tan a cos B^
cos a»co8isecmcoB(e— m); cos 6»-cosasecmco8(c— m).
EXEECISB XXVI.
1. (i.) tan X — tan B cos c, (ii.) tan a; « tan A cos 5,
cosil — cos.Bcscaj8in(C7— »); coB.B«-co8ilcscajsin(C7— «).
Exercise XXXI.
4. 27.21".
Exercise XXXII.
1. If » denote the angle required, sin } a; « cos 18^ sec 9^, x « 148^ 42^.
2. co8a;»co0j.sec^.
3. Let w » the inclination of the edge e to the plane of a and h. Then
it is easily shown that V= ahc sin I sin w. Now, conceive a sphere
constructed having for centre the vertex of the trihedral angle
whose edges are a, 6, c. The spherical triangle, whose vertices are
ihe points where a, i, e meet tne snrfSEU^e of tnis sphere, has for its
sides l,m,n\ and w is equal to the nerpendicular arc from the side
Z to the opposite vertex. Let L, J/, iv, denote the angles of this
triangle. Then, by means of [38] and [47], we find that
sin ur a sin m sin jy^* 2 sin m sin Jjycos } JY*
2
— -7—- Vsin « sin (« — Z) sin (« — m) sin (» — n),
sinZ
where ««}(? + m + n) ;
hence, V«« 2 a5c Vsin s sin (« — ^^^ (* *" ^) ^^^ (* "~ ^)'
4. (i.) 9,976,612.8 square miles ; (ii.) 13,292,812.8 square miles.
5. Let m »> longitude of point where the ship crosses the equator, h ■»
her course at the equator, d = distance sailed. Then
tan m = sin £ tan a, cos&="COsZsina, cot (2 >= cot Z cos a.
6. Let k = arc of the parallel between the places, x » difference re-
quired; then sinjife = sinj(i8eci"-}, »« (i— ife = 30®.
7. cos(m — m') — (cos d® — sin Z sin V) sec Z sec Z' ; where m and m' are
the longitudes of the places.
9. 12 hours 44 minutes, very nearly. 10. 60®.
ANSWEBS. 13
11. COS < »- — tan d tan l; time of snnrise «i 12 — -- o'clock a.m. ; time of
snnfiet » -r o'clock P.M. ; cob a — sin d sec Z. For longest day at
15
Boston : time of snnrise, 4 hrs. 26 min. 52 sec. a.m. ; time of snnset,
7 hrs. 33 min. 8 sec. p.m. Azimuth of sun at these times, 57^ 25^ ;
length of day, 15 hrs. 6 min. 16 sec. ; for shortest day, times of
snnrise and snnset are 7 hrs. 33 min. 8 sec. a.m. and 4 nrs. 26 min.
52 sec. P.M.; azimnth of snn, 122° 35^ ; length of day, 8 hrs. 53 min.
44 sec.
12. The problem is impossible when cot d<iaai I; that is, for places in
the frigid zone.
13. For the northern hemisphere and positive declination,
sin A *" sin Z sin d, cot a >■ cos 2 tan d.
Example : A - 17° 14' 36", a - 73° 51' 34" E.
14. The farther the place from the equator, the greater the sun's altitude
at 6 A.M. in summer. At the equator it is 0°. At the north pole
it is equal to the sun's declination. At a given place, the sun's
altitude at 6 a.m. is a maximum on the longest day of the year,
and then sin A >a sin Z sin e (where e *» 23° 27').
15. cost — cot Z tan d Times of bearing due east and due west are
12— -L o'clock A.M., and ^ o'clock P.M., respectively.
15 15
Example : 6 hrs. 58 min. A.M. and 5 hrs. 2 min. p.m.
16. When the days and nights are equal, i- 0°, cost- 0, < - 90° ; that
is, sun is everywhere due east at 6 a.m., and due west at 6 p.m.
Since I and d must both be less than 90^. cos t cannot be negative,
therefore, t cannot be greater than 90°. As d increases, t decreases ;
that is, the times in question both approach noon. If Kd, then
cos t > 1 ; therefore this case is impossible. Ifl'^d, then cos t —l,
and t » 0° ; that is, the times both coincide with noon. The ex-
planation of this result is, that the sun at noon is in the zenith ;
nence on the prime vertical. At the pole I » 90°, coe t — 0,
<» 90°: therefore the sun in summer always bears due east at 6
A.M., and due west at 6 p.m.
17. sin I " Bind OK h. 18. 11° 5(/ 35".
19. The bearing of the wall, reckoned from the north point of the hori-
zon, is given by the equation cots — coeZ tan e; whence, for the
given case, x - 75° 12^ 8".
20. 65°45'6"N. 21. 63° 23^ 41" N. or S.
22. (i.) cost" — tan Z tan d; (ii.) <— s ; (iii.) the result is indeterminate.
23. cot a — cos Z tan c?. 28. sin c? *- sin e sin U| tan r — cose tan u.
25. A - 65° 37' 30". 29. rf - 32° 23' 47", r - 301° 48' 17".
26. A - 58° 25' 15". a- 152° 28'. 30. d - 20° 48' 14".
27. t -45° 42', Z- 67° 58' 55". 31. 3 hrs. 59 min. 27 sec. p.m.
32. cos } a — Vcos } (Z + A + 1)) cos J (Z + A — p)sec I sec K.
SURYEYII^G.
CHAPTER I.
DEFINmOKS. DnSITBUMEHTS AKD THEIR USBS.
§ 1. Definitions.
Btufyeying is the art of determining and representing dis-
tances, areas, and the relative position of points upon the
surface of the earth.
In plane surveying, the portion surveyed is considered as a
plane.
In geodetic surveying, the curvature of the earth is re-
garded.
A Flumb-Idne is a cord with a weight attached and freely
suspended.
A Yertioal Line is a line having the direction of the plumb-
line.
A Yertical Fkne is a plane embracing a vertical line.
A Horiiontal Flane is a plane perpendicular to a vertical line.
A Hon«mtal Line is a line in a horizontal plane.
A Horisontal Angle is an angle the sides of which are in a
horizontal plane.
A Yertical Angle is an angle the sides of which are in a
vertical plane. If one side of a vertical angle is horizontal,
and the other ascends, it is an angle of elevation ; if one side
is horizontal, and the other descends, it is an an^le of
depression.
The Magnetic Meridian is the direction which a bar-magnet
assumes when freely supported in a horizontal position.
SUEVEYINQ.
The Magnetic Bearing of a line is the angle it makes with
the magnetic meridian.
Surveying commonly comprises three distinct operations;
viz.:
1. The Field HeasniementSi or the process of determining by
direct measurement certain lines and angles.
2. The Oomptitation of the required parts from the measured
lines and angles.
3. The Flottmgi or representing on paper the measured and
computed parts in relative extent and position.
THE MEASUREMENT OF LINES.
§ 2. Instruments fob Measuring Lines.
The Chmter's Qhain is generally employed in measuring land.
It is 4 rods, or 66 feet, in length, and is divided into 100 links.
Hence, links may be written as hundredths of a chain.
The Engineer's Ohain is employed in surveying railroads,
canals, etc. It is 100 feet long, and is divided into 100 links.
A Tape Heasnre, divided into feet and inches, is employed
in measuring town-lots, cross-section work in railroad survey-
ing, etc.
In the United States Coast and Geodetic Survey, the meter
is the unit ; and, when great accuracy is required, rods placed
end to end, and brought to a horizontal position by means of
a spirit-level, are employed in measuring lines.
§ 3. Chaining.
Eleven tally-pins of iron or steel are used in chaining; also,
one or more slender poles shod with iron, and bearing a flag.
A forward chainman, or leader, and a hind chainman, or
follower, are required. A flag-staff having been placed at the
farther end of the line, or at some point in the line visible
CHAININO. 8
from the beginmng, the follower takes one end of the chain,
and a pin which he thrusts into the ground at the beginning
of the line. The leader moves forward in the direction of
the flag-staff, with the other end of the chain and the re*
maining ten pins, until the word "halt" from the follower
warns him that he has advanced nearly the length of the
chain.
At this signal he stops, and the follower, meanwhile hav-
ing placed his end of the chain at the beginning of the line,
directs the leader by the words "right" or "left" until the
chain ia exactly in line with the flag-staff. This being accom-
plished, and the chain stretched tightly in a horizontal posi-
tion, the follower sajrs, "down." The leader then puts in
a tally-pin exactly at the end of the chain, and answers,
"down"; after which the follower withdraws the pin at the
beginning of the line, and the chainmen move forward until
the follower nears the pin set by the leader. The follower
again says, "halt," and the operation just described is re-
peated. This process is continued until the end of the line
is reached.
If the tally-pins in the hands of the leader are exhausted
before the end of the line is reached, when he has placed the
last pin in the ground, he waits until the follower comes up
to him. The follower gives the leader the ten pins in his
hand, and records the fact that ten chains have been meas-
ured. The measuring then proceeds as before. If the distance
from the last pin to the end of the line is less than a chain,
the leader places his end of the chain at the end of the line,
and the follower stretches tightly such a part of the chain
as is necessary to reach to the last pin, and the number of
links is ceunted. The number of whole chains is indicated by
the number of pins in the hands of the follower, the last pin
remaining in the ground.
In measuring, the chain must be held in a horizontal posi-
tion. If the ground slopes, one end of the chain must be
raised until the horizontal position is attained. By means of
SUBVEYINa.
a plumb-line, or a slender staff, or, less accurately, hj drop-
ping a pin (heavy end downwards), the point vertically under
the raised end of the chain may be determined. If the slope
is considerable, half a chain or less may be used.
To oonstraot a perpendiotilar with a bhain :
1. When the point through which the perpendicular is to
pass is in the line :
Let AB (Fig. 1) repreBent the line, and P the point. Measure from
P to the right or left, PC- 40 links,
and place a stake at C. Let one end
of the chain be held at P, and the
end of the eightieth link at C; then,
taking the chain at the end of the
.... thirtieth link from P, draw it so that
Rg. I. ^® portions DO and DF are tightly
stretched, and place a stake at D.
DP will be the perpendicular required. (Why ?)
2. When the point is without the line :
Let AB (Fig. 1) be the line, and D the point Take any point in
the line, and stretch the chain between D and C; then, let the middle of
the part of the chain between (7 and D be held in place, and swing the
end at D around until it meets the line in P. DP will be the perpen-
dicular required. (Why ?)
§ 4. Obstacles to Chaining.
1. When a tree, building, or other obstacle is encountered
in measuring or extending a line, it may be passed by an off-
set in the following manner :
Let (Fig. 2) represent a building on the line AD. At S erect BE
/v^ ... perpendicular to
£ f ^^-nQ>-f f^ « AB, at S erect
I • r!^«___ '. ' JSF perpendicu-
^ ^ Fig. 2. '^ '^ 1" ^ ^-^J ** ^
erect FO^ BE
perpendicular to -EF; then, CD perpendicular to FC will be in the
required line, and AB + EF+ CD — AD, By constructing two other per-
pendiculars, B^E' and P'C", the accuracy of the work will be increased.
OBSTACLES TO CHAINING.
2. To measure across a body of water :
Let it be required to measure the liue ABCD (Fig. 3) crossing a river
between B and C. Measure j?^— 400 links; at E ereei-the perpendi-
cular EF'-' 600 links ; at B erect the perpendicular BG - 300 links.
Place a stake at Q, the intersection of AD and FG beyond the river.
Fig. S.
Then BC=^QO links. For, by similar triangles, EF: BG.iCE. CB.
But EF'^2BG- hence, CE^2CB, and CB rr- BE '^ 400 links. EG
and FG should be measured, in order to test the accuracy of the work.
EG ^FG-' 500 \inki.
Instead of the above distances, any convenient distan ces may be t aken.
For, if EF-^ 2BG, then CB - BE, and EG - FG - V^+W.
8. To measure a line the end of which is invisible from the
-beginning, and intermediate points unknown :
■~-^-
Fig. 4.
^""C"
^et AB (Fig. 4) represent the line. Set up a flag-staff at D, beyond
3 and visible from A. From B let fall 5(7 perpendicular to AD. Meas-
ure -AC and .BC. Then ^^^ — — •
AB^^ACf + JSS'.
To find intermediate points on AB :
At any point ^on ilC erect .EF perpendicular to AC, and determine
EG by the proportion AC CB : : AE . EG. G will be a point on AB,
The line AJ) is called a Baadom Liiie.
SUEVEYING.
THE MEASTJEEMENT OF AKGLES.
§ 5. The Vernier Compass*
The Vernier Oompass is shown on the following page.
The compass circle is divided into half-degrees, and is fig-
ured from 0® to 90® each way from the north and south points.
In the centre of the compass circle is the pivot which supports
the magnetic needle. The needle may be lifted from the pivot
by a spring and pressed against the glass covering of the
compass circle, when the instrument is not in use. The muin
plate moves around the compass circle through a small arc,
read by the vernier, for the purpose of allowing for the varia-
tion of the needle (§ 23). The sight standards at the extremi-
ties of the main plate have fine slits nearly their whole length,
with circular openings at intervals; on the edges of the north
standard tangent scales are placed; and on the outside of the
south standard two eye-pieces at the same distance from the
main plate as the zeros of the tangent scales, respectively.
The telescopic sight (a recent improvement by the Messrs.
Gurley), consists of a small telescope attached to the south
standard. The main plate is furnished with two spirit levels
at right angles, and turns horizontally upon the upper end of
the ball spindle, the lower end of which rests in a spherical
socket in the top of the tripod or JacoVs staff which sup-
ports the instrument. From the centre of the plate at the
top of the tripod a plumb-bob is suspended by which the
centre of the compass can be placed directly over a definite
point on the ground.
* The instruments described on this and the following pages are adjusted
by the maker. If .they should require readjustment, full directions will be
found in the manual furnished with the instruments.
The manual published by Messrs. W. & L. E. Gitelbt, Troy, N.Y., has
been freely used, by permission, in describing these instruments.
THE VERNIER COMPASS.
INSTRUMENTS AND THEIR USES.
§ 6. Uses of the Compass.
To take the bearing* of a line. Place the instrument bo that
the bob will be directly over one end of the line, and level by
pressing with the hands on the main plate until the bubbles
are brought to the middle of the spirit levels. Turn the south
end of the instrument toward you, and sight at the flag-staff
at the other end of the line. Read the bearing from the north
end of the needle. First, write N. or S. according as the
north end of the needle is nearer N. or S. of the compass cir-
cle ; secondly, write the number of degrees between the n^th
end of the needle and the nearest zero mark ; and thirdly,
write E. or W. according as the north end of the needle is
nearer E. or W. of the compass circle.
In Fig. 5 the bearing would be N. 45^ W.
In Fig. 6 the bearing would be S. 45* W.
In Fig. 7 the bearing would be S. 30* E.
In Fig. 8 the bearing would be N. 60* E.
If the needle coincides with the N. S. or E.W. line, the bear-
ing would be N., S., E., or
W., according as the north
end of the needle is over
N., S., E., or W.
As the compass circle is
divided into half-degrees,
the bearing may be deter-
mined pretty accurately to
quarter-degrees.
When' a fence or other
obstruction interferes with
placing the instrument
over the line, it may be
placed at one side, the
Rg. 7.
Rg. 8.
flag-staff being placed at an equal distance from the line.
* The magnetic bearing is meant nnless otherwise specified.
iO SITEVEYING.
Local DiBtarbanoes. Before a bearing is recorded, care should
be exercised that the chain, pins, and other instruments that
would affect the direction of the needle, are removed from the
vicinity of the compass. Even after the greatest care in this
respect is exercised, the direction of the needle is often affected
by iron pre, ferruginous rocks, etc.
Beverse Bearings. When the bearing of a line has been taken,
the instrument should be removed to the other end of the line
and the reverse bearing taken. The number of degrees should
be the same as before, but the letters should be reversed.
Tq take the bearing of a line one end of which oannot be seen
from the other. Bun a random line (§ 4, 3) ; then (Fig. 4),
tana45 = |g; '
whence, the angle CAB may be found. This angle combined
with the bearing of the random line will give the bearing
required.
Another method will be given in § 19.
To measure a horizontal angle by means of the needle. Place the
compass over the vertex of the angle, take the bearing of each
side separately, and combine these bearings.
To measure angles of elevation. Bring the south end of the
compass towards you, place the eye at the lower eye-piece,
and with the hand hold a card on the front side of the north
sight, so that its top edge will be at right angles to the divided
edge and coincide with the zero mark; then, sighting over
the top of the card, note upon a flag-staff the height cut by
the line of sight ; move the staff up the Elevation, and carry
the card along the edge of the sight until the line of sight
again cuts the same height on the staff; read off the degrees
of the tangent scale passed over by the card.
To measure angles of depression. Proceed in the same man-
ner as above, using the eye-piece and tangent scale on the
opposite sides of the sights, and reading from the top of the
sight.
INSTBUMEKTS AND THEIB USES.
11
§ 7. Vebniebs.
First form. Let AB (Fig. 9) represent a portion of a rod
for measuring heights (§ 32). The graduation to feet and hun-
dredths of a foot begins at the lower end, which rests on the
ground when the rod is in use. The line
extending nearly across the rod at the bot-
tom of the portion shown marks the begin-
ning of the fourth foofc. The face of the rod
is divided into four columns : in the first is
written the number of feet ; in the second,
the number of tenths ; and in the third, the
number of hundredths.
It is evident that, with the arrangement
just described, heights could be measured
only to hundredths of a foot. To enable us
to find the height more precisely, a contri-
vance called a Yemier is used. This is shown
at the right of the rod. It consists of a piece
of metal or wood, the graduated part of
which is -^^ of a foot in length ; and tins
is divided into ten equal parts. Hence, one
division of the vernier = -j^ of -^ = y^
of a foot; and one division of the vernier
exceeds one division of the rod by YHiy —
ihs = -rhrs ^^ «• ^oot.
The vernier slides along the face or side
of the rod.
To use the vernier, place the lower end
of the rod upon the ground, and move the vernier until its in-
dex or zero mark is opposite the point whose distance from
the ground is desired. In the figure, the height of the index
of the vernier is evidently 4.16 feet, increased by the distance
of the index above the next lower line (4.16) of the rod. We
shall now determine this distance.
12
BUBVBYING.
Observe which line of the vernier is exactly opposite a line
of the rod. In this case, the line of the vernier numbered 7 is
opposite a line of the rod. Then, since each division of the
vernier exceeds each division of the rod by xriW ^^ * fo<>^»
of a foot above the next lower line of the rod.
of a foot above the next lower line of the rod.
of a foot above the next lower line of the rod,
of a foot above the next lower line of the rod.
of a foot above the next lower line of the rod.
of a foot above the next lower line of the rod.
of a foot above the next lower line of the rod.
6 of the vernier is j^
6 of the vernier is yAtv
4 of the vernier is y^
3 of the vernier is ^jf^
2 of the vernier is j^^
1 of the vernier is Yfijjf
of the vernier is y^^
Hence, the required reading is 4.16 + .007 = 4.167 feet.
In general, the following rule is evident :
Move the vernier until its zero line is at the
required height ; read the height to the near-
est hundredth below the index, and write in
the thousandths* place the number of the divi-
sion line of the vernier which stands opposite
any line of the rod.
Second form. In this form (Fig. 10) the
graduated part of the vernier is -j-J^ of a foot
in length, and is divided into ten equal parts.
Hence, one division of the vernier = -^r of
tH ~ 1 0^0 ^^ ^ ^^^^ J ^^d one division of the
vernier is less than one division of the rod
^y xk-Tinnr = T(nnr of a foot.
The height of the index of the vernier
in Fig. 10 is 4.16 feet, increased by the dis-
tance of the index from the next lower line
(4.16) of the rod. We shall now determine
this distance.
We observe that the line of the vernier
numbered 7 stands exactly opposite a line
(3) of the rod. Hence,
INSTBUHEITTS AJETD THEIB USES.
18
. 6 of the vernier is j^^ of a foot above the next lower line of the rod.
5 of the vernier is yAht ^^ ^ ^^^ above the next lower line of the rod.
4 of the vernier is jjfg^ of a foot above the next lower line of the rod.
3 of the vernier is ^^ o^ a ^oot above the next lower line of the rod.
2 of the vernier is jjhn ^^ ^ ^^^^ above the next lower line of the rod.
1 of the vernier is yAtv ^^ * ^^^^ above the next lower line of the rod.
of the vernier is y/jtv ^^ ^ ^^^^ above the next lower line of the rod.
Hence, the required reading is 4.16 + .007 = 4.167 feet ;
and the rule is evidently the same as for the first form.
Oompass Tenien. Let LL* (Fig. 11) represent the limb of
the compass graduated to half-degrees, and VV* the vernier
divided into thirty equal spaces, equal to twenty-nine spaces
of the limb. Then one space of the vernier is less than one
space of the limb by 1', and the reading may be obtained to
single minutes.
In Fig. 11 the index or zero of the vernier stands between
32° and 32** 30', and the line of the vernier marked 9 coincides
with a line of the limb. Hence, the reading is 32° 9'.
When the index moves from the zero line of the limb in a
direction contrary to that in which the numbers of the limb
run, the number of minutes obtained as above must be sub-
tracted from 30' to obtain the minutes required.
If, however, the vernier be made double, that is, if it have
thirty spaces on each side of the zero line, it is always read
directly. The usual form of the double vernier, shown in
14
SUEVBYING.
Fig. 12, has only fifteen spaces on each side of the zero line.
When the vernier is turned to the right less than 15' past a
division line of the limb, read the lower figures on the left of
the zero line at any coincidence ; if moved more than 15' past
a division line of the limb, read the upper figures on the right
of the zero line at any coincidence ; and vice versa.
TTses of the Oompass Termer. The most important use of the
vernier of the vernier compass is in setting off the variation
of the needle (§ 23). If the variation of the needle at any
place is known, by means of the vernier screw the compass
circle may be turned through an arc equal to the variation.
If the observer stands at the south end of the instrument, the
vernier is turned to the right or left according as the varia-
tion is west or east. The compass will now give the bearings
of the lines with the true meridian.
In order to retrace the lines of an old survey, turn the eights
in the direction of a known line, and move the vernier until the
needle indicates the old bearing. The arc moved over by
the vernier will indicate the change of variation since the time
of the old survey. If no line is definitely known, the change
of variatixm from the time of the old survey will give the arc
to be set off.
INSTBUMEKTS AKD THEIB USES. 16
§ 8, The Surveyor's Transit.
This instrument is shown on page 17.
The compctaa circle is similar to that of the compass. The
vernier plate which carries the teleecape has two vemiera and
moves entirely around the gradvaied limb of the main plate.
The axis of the telescope carries a vertical circle which meas-
Tires vertical angles to single minutes by means of a vernier.
Under the telescope, and attached to it, is a spirit level by
means of which horizontal lines may be run, or the difference
of level between two stations found. The crass wires are two
fine fibres of spider's web, or fine platinum wires, which extend
across the tube of the telescope at right angles to each other ;
their intersection determines the optical axis or lirie of coUir
motion of the telescope. The transit is levelled by four level-
Ung screws which pass through a plate firmly fastened to the
ball spindle, and the lower ends of which rest in depressions
on the upper side of the tripod plate.
Two recent improvements (introduced by the Messrs. Gur-
ley) enable the surveyor to bring the transit quickly to an
approximately level position by the pressure of the hands,
after which the levelling screws are used ; also, to change the
position of the transit without changing the position of the
tripod legs, so as to bring the bob exactly over any point.
To lerd the transit by the levelling screws. Turn tiie instru-
ment until the spirit levels on the vernier plate are parallel
to the vertical planes passing through opposite pairs of level-
ling screws. Take hold of opposite screw heads with the
thumb and fore-finger of each hand, and tarn both thumbs
in or out as may be necessary to raise the lower side of the
parallel plate and lower the other until the desired correction
is made.
To use the tdesoope. Both the eye-piece and the object
glass may be moved in and out by a rack-and-pinion move-
ment. The eye-piece must be moved until the cross wires are
16 SUaVEYINO.
perfectly distinct; in which case, a slight movement of the
eye of the observer, from side to side, will .produce no appar-
ent change in the position of the threads upon the object.
The object glass must- be moved until the object is distinctly
visible; and this operation must be repeated whenever the
distance of the object is changed.
§ 9. Uses of the Transit.
The transit may be used for all the purposes indicated in
§ 6, but with much greater precision than the compass. The
principal icsey however, of the transit is in measuring horizontal
angles by means of the groduated limb and verniers.
To measnie a horixontal angle with the transit. Place the transit
over the vertex of the angle; level, and set the limb at zero.
-Turn the telescope in the direction of one of the sides of the
angle, clamp to the spindle ; unclamp the main plate, and turn
the telescope until it is in the direction of the other side of
the angle, and read the angle by the verniers. The object
of the two verniers on the vernier plate is to correct any mis-
take that might arise from the want of exact coincidence in
the centres of the verniers and the limb. The correct reading
may be obtained by adding to the reading of one vernier the
supplement of the reading of the other, and dividing by 2.
3y turning off a right angle by this method, perpeadiculars
may. be constructed with much greater facility than by the
chain.
§ 10. The Theodolite.
The telescope of the transit can perform a complete revo-
lution on its axis; whence the name transit. The theodolite
differs from the transit chiefly in that its telescope cannot be so
revolved. It is not much used in this country.
§ 11. The Kailroad Compass.
This instrument has all the features of the ordinary com-
pass, and has also a vernier plate and graduated limb for
measuring horizontal angles. _ ^
THE SURVEYOR'S TRANSIT.
INSTBUHENTS AND TEEIB USES.
19
§ 12. Plotting.
The principal plotting instruments are a ruler, pencil,
straight-line pen, hair-spring dividers, diagonal scale, a right
triangle of wood, and a circular protractor. A T-square will
also be found convenient.
A* h B* C
\
\
M
)
\
/
\
1
1
I
\
'
J
/
ID a 8 r
A
8 S 1
B
Fig. 13.
The Diagonal Scale. A portion of this scale is shown in Fig.
13. A£ is the unit. A£ and A'JB' are divided into ten
equal parts, and B is joined with A, the first division point to
the left of .B'; the first division point to the left of £ is joined
with the second to the left of -B', etc.
The part of the horizontal line numbered 1 intercepted be-
tween ££' and £h is evidently ^ of -^ = y^ of the unit ;
the part of the horizontal line numbered 2 intercepted between
££' and £h is yj^ of the unit, etc.
The method of using this scale, in laying off distances, will
be made plain by the following example :
Let it be required to lay off the distance 1.43.
Place one foot of the dividers at the intersection of the horizontal line
numbered 3 and the diagonal numbered 4, and place the other foot at
the intersection of the vertical line numbered 1 {CC) and the horizontal
line numbered 3 ; the distance between the feet of the dividers will be
the distance required. For, measuring along the horizontal line num-
bered 3,from C(y to BB' is 1 ; from BB' to J?A is U)3 ; and from Bh to
the diagonal numbered 4 is 4 ; and 1 + .03 + .4 — 1.43.
20 gtJBVEYING.
The Oircnlar Protractor. This instrument (Fig. 14) usually
consists of a semi-circular piece of bras3 or german silver, hav-
ing, its arc divided into degrees and its centre marked.
To lay off an angle with the protractor, place the centre
over the vertex of the angle, and make the diameter coincid,e
with the given side of the angle. Mark off the number of
degree^'in the given angle, and draw a line through this point
and the vertex.
Fig. 14.
Some protractors have an arm which carries a vernier, by
which angles may be constructed to single minutes.
To draw through a given point a line parallel to a given
line, make one of the sides of a triangle coincide with the
given line, and, placing a ruler against one of the other sides,
move the triangle along the ruler until the first side passes
through the given point ; then draw a line along this side.
To draw through a given point a line perpendicular to a
given line, make the hypotenuse of a right triangle coincide
with the given line, and, placing a ruler against one of the
other sides of the triangle, revolve the triangle about the ver-
tex of the right angle as a centre until its other perpendicular
side is against the ruler ; then move the triangle along the
ruler until the. hypotenuse passes through the given point,
and draw a line 6long the hypotenuse.
CHAPTEE II.
LAND SUBYEYINO.
§ 13. Definition.
Land STunreyiiig is the art of measuring, laying out, and
dividing land, and preparing a plot.
§ 14. Detebmination of Abeas.
Tlie unit of land measure is the
acre == 10 square chains
= 4 roods
= 160 square rods, perches, or poles.
Areas are referred to the horizontal plane, no allowance
being made for inequalities of surface.
For convenience of reference, the following rules for areas
are given :
Let A, B, and C7be the angles of a triangle, and a, &, and c
the opposite sides, respectively ; and let « = 1 (a + 4 + c).
Area of triangle ABO=^8{8 — a){9''b){9 — €) [a]
= J6c8in^ [b]
X g^sin-BsinO ^ r -i
~ sin(j5 + (7) ^ -•
= } base X altitude. [d]
Area of rectangle = base X altitude.
Area of trapezoid = 1 sum of parallel sides X altitude.
Pkobleu 1. To determine the area of a triangular field.
Measure the necessary parts with a Gunter's chain, or with a chain
and transit, and compute by formula [a], [b], [c], or [d].
22
SUEVEYING.
Problem 2. To find the area of a field having any ntunber of
straight sides.
(a) Divide the field into triangles by diagonals ; find the area of each
triangle and take the sum.
(b) Bun a diagonal, and perpendiculars from the opposite vertices to
this diagonal. The field is thus divided into right triangles, rectangles,
and trapezoids, the areas of which may be found and the sum taken.
Fig. IS.
Fig. i€.
To find the area of a field having an irregular
Problem 3.
bonndary line.
(a) Let AGBCD (Fig. 15) represent a field having a stream AEFQ
HKB as a boundary line. Bun the line AB, From E, F, G, H, and
K, prominent points on the bank of the stream, let fall perpendiculars
EE\ FF', etc., upon AB. Begarding AE, EF, etc., as straight lines, the
portion of the field cut ofif by AB is divided into right triangles, rect-
angles, and trapezoids, the necessary elements of which can be measured
and the areas computed. The sum of these areas added to the area of
ABCD will give the area required.
{b) When the irregular boundary line crosses the straight line joining
its extremities, as in Fig. 16, the areas of AEFHsktA RGB may be found
separately, as in the preceding case. Then the area required =» ABCD +
EGB-AEFH.
Problem 4. To determine the area of a field from two interior
stations.
Let ABCDi^ig. 17) represent a field, and Pand P' two stations within
it. Measure PP^ with great exactness. Measure the angles between PP
and the lines from P and P' to the corners of the field.
DETEEMINATION OP AREAS.
23
In the triangle PP'X>, PP' and the angles P'FD and PP'D are
known ; hence, PD may be found. In ^
like manner, PC may be found. Then,
in the triangle PDC, PD, PC, and the
angle DPC atb known; hence, the area
of PDC may be computed.
In like manner, the areas of all the
triangles about P and P' may be deter-
mined.
Area ABCD - PAD + PDC + PCS
+ PBA. Also,
Area ABCD - P'AD + P^DC^-P^CB
-\-P'BA.
Problem 5. To detenmne the area of a field from two exterior
statioxui. D
Let ABCD (Fig. 18) represent the field,
and P and P* the stations. Determine the
areas of the triangles PAD, PDC, PCB,
and PBA, as in the preceding problem.
Area ABCD = PAD + PDC + PBC-
PBA, Also,
Area ABCD - P^AD + P^DC + P'BA
^FBC.
Fig. 18.
Exercise I.
1. Eequired the area of a triangular field whose sides are
respectively 13, 14, and 15 chains.
2. Required the area of a triangular field whose sides are
respectively 20, 30, and 40 chains.
3. Eequired the area of a triangular field whose base is
12.60 chains, and altitude 6.40 chains.
4. Required the area of a triangular field which has two sides
4.50 and 3.70 chains, respectively, and the included angle 60®.
5. Required the area of a field in the form of a trapezium,
one of whose diagonals is 9 chains, and the two perpendiculars
upon this diagonal from the opposite vertices 4.50 and 3.25
chains.
24
SURVEYING*
Fig. 19.
6. Required the area of the field ABCDEF (Fig. 19), if
AE= 9.25 chains, FF^=^AO chains, BE= 13.75 chains, DZ)'
= 7 chains, DB = 10 chains, (7(7' =
4 chains, and ^^'=4.75 chains.
7. Required the area of the field
ABCDEF (Fig. 20), if
AF^= 4 chains, i?!F'= 6 chains,
^^'=6.50 chains, ^J7'= 9 chains,
^Z) = 14 chains, ^(7' = 10 chains,
AB^ = 6.50 chains, BB' = 7 chains,
CC =6.75 chains.
8. Required the area of the field AQBCD (Fig. 15), if the
diagonal -4(7=5, ^-B' (the perpen-
dicular from B\j^AQ) = \, DB' (the
perpendicular from D to ^C) = 1.60,
.£'^7'= 0.25, i^i^'= 0.25, {?{?' = 0.60,
HW=^^.h% J5rir'=0.54, ^Z7'=0.2,
.£"i^' = 0.50, i^'ff'=0.45, ff'^' =
0.45, ^'JT' = 0.60, and jr'5= 0.40.
9. Required the area of the field
AQBCD (Fig. 16), if AD=^% AC
= 5,^^=6, angle 1)^(7= 45^ angle ^^C=30^ AW =^
0.75, ^i^' = 2.25, ^^=2.53, ^(?' = 3.15, Z;^' = 0.60,
i^i^' = 0.40, and Q& ^Mh.
10. Determine the area of the field ABCD from two inte-
rior stations, P and P\ if PP'= 1.50 chains,
angle PFC = 89** 35', angle P'PB = 3** 35',
PP'B = 185** 30', P'PA = 113** 45',
PP'A = 309° 15', P'Pi> = 165** 40',
PP'D = 349** 45', P'PC = 303** 15'.
11. Determine the area of the field ABCD from two exte
rior stations P and P', if PP^ = 1.50 chains,
angle P'PP= 41** 10', angle PP'i>= 66** 45',
P'PA^ 55** 45', PP'(7= 95** 40',
P'PC= 77** 20', PP'P = 132** 15',
P'Pi> = 104** 45', PP'^ = 103** 0'.
EECTANGULAE BUEYEYING.
25
BECTANGULAR SURVEYING.
§ 15. Definitions.
An East and West Line is a line perpendicular to the mpig-
netic meridian.
The Latitude of a line is the distance between the east and
west lines through its extremities.
The Departoie of a line is the distance between the meridians
through its extremities.
Note. When a line extends north of the initial point the latitude is
called a northing ; when it extends south, a southing ; when it extends
east the departure is called an easting ; when it extends west, a westing.
The Meridian Distance of a point is its distance from a meridian.
Let AB (Fig. 21) represent a line, and NA8 the magnetic
meridian. Let BB^ be perpendicular to N8, ^
The bearing of the line AB is the angle
BAB\
The latitude of the line AB is AS,
The departure of the line AB is BB\
The meridian distance of the point B is
BB\
In the right triangle ABB\
AB' = ABxco8BAB\
and BB' = AB X sin BAB'.
Hence, latitude = distance X cos of bearing^
and departure = distance X sin of bearing.
The latitudes and departures correspond-
ing to any distance and bearing may be
found from the above formulas by means of
a table of natural sines and cosines, or from
" The Traverse Table." * Rg. 21.
♦See Table VII. of Wentworth & Hill's Five-Place Logwrithmic and
Trigonometric Tables.
26
SURVEYING.
§ 16. Field Notes, Computation, and Plotting.
The field notes are kept in a book provided for the purpose.
The page is ruled in three columns, in the first of which is
written the number of the station ; in the second, the bearing
of the side ; and in the third, the length of the side.
Example 1. To survey the field ABCD (Fig. 22).
Field Notes.
1
N. 20«> E.
8.66
2
S. 70° E.
6.00
3
S. 10«> E.
10.00
4
N. 70« W.
10.00
(a) To obtain the field notes.
Place the compass at A, the first sta-
tion, and take the bearing of AB (§ 6) ;
suppose it to be N. 20° E. Write the re-
sult in the second column of the field notes
opposite the number of the station. Meas-
ure AB = 8.66 chains, and write the result
in the third column of the field notes.
Place the compass at B, and, after test-
ing the bearing of AB (§ 6), take the
bearing of BQ measure BC, and write the
results in the field notes ; and so continue until the bearing and length
of each side have been recorded.
Fig. 22.
(b) To compute
the area.
I.
n.
m.
IV.
V.
VI.
VII.
Vlll.
EX.
X.
XI.
Side.
Bearing,
Diet
N.
S,
E.
w.
M.D.
D.M.D.
N.A,
S.A.
AB
N.20'E.
8.66
AB'
8.14
BB'
2.96
BB'
2.96
BB'
2.96
2ABB*
24.0914
BC
8. 7(rE.
5.00
....
B'C
1.71
CC"
4.70
...
CC
7.66
BB'+CC
10.62
....
2C'CBB'
18.1602
CD
S. 10* E.
10.00
CD'
9.85
DD"
1.74
...
DD'
9.40
CC+DD'
17.06
2D'DCC
168.0410
DA
N.TO^W.
10.00
D'A
3.42
....
....
DD'
9.40
DD'
9.40
2ADD'
32.1480
56.2424
186.2012
FIELD NOTES. 27
The Borvey may begin at any comer of the field ; bnt in computing
the area, the field notes shonld be arranged bo 1M.2012
that the most eastern or most western station 66.2424
will stand first For the sake of uniformity, we 2
shall always begin with the most western station, ^^ .
129.9688
04.98 aq. ehalni.
and keep the field on the right in passing around it. ^"^^ '
The field notes occupy the first three of the eleven columns in the
above tablet Columns IV., V., VI., and VII. contain the latitudes
and departures corresponding to the sides, and taken from the Traverse
Table. The lines represented by these numbers are indicated imme-
diately above each number. Column VIII. contains the meridian dis-
tances of the points J?, (7, i>, and ui, taken in order. Column IX. contains
the double meridian distances. Their composition is indicated by the
letters immediately above the numbers. Column X. contains the pro-
ducts of the double meridian distances by the northings in the same
line. The first number,
24.0944 - 2.96 x 8.14 - j&B' x ^JS' - 2 area of the triangle ABB* ;
32.1480 - 9.40 X 3.42 - Z>i)' x ^iX - 2 area of the triangle ADIX.
Column XI. contains the products of the double meridian distances by
the southings in the same line. The first number,
18.1602 - 10.62 X 1.71 - {BB' + CC) xffC
- 2 area of the trapezoid C'CBB';
168.0410 - 17.06 X 9.85 - {CC + DIX) xI/C
- 2 area of the trapezoid D'DCC'.
The sum of the north areas in column X.
- 56.2424- 2(^5^' + JLDiy).
The sum of the south areas in column XI.
- 186.2012 - 2(CCBB' + jyDCC).
But {CCBB -\-iyDCa)- (ABBf-^-ADD^)^ ABCD.
Hence, 2{C'CBBf + D^DCa) - 2{ABff + ADD^ - 2 ABCD ;
that is, 186.2012 - 56.2424 - 129.9588 - 2 ABCD,
Hence, area ABCD - } of 129.9588 - 64,9794 sq. ch. - 6.498 acres.
(c) To make the plot
The plot or map may be drawn to any desired scale. If a line one
inch in length in the plot represents a line one chain in length, the plot
is said to be drawn to a scale of one chain to an inch. In this case the
plot (Fig. 22) is drawn to a scale of eight chains to an inch.
Draw the line NA8 to represent the magnetic meridian, and lay off
the first northing Aff - 8.14 (§ 12). Draw the indefinite line BE per-
28 BUEVEYING.
pendicular to N8 and lay off B'B, the first easting » 2.96. Join A and
J?; then the line AB will represent the first side of the field. Through
B draw -BC" perpendicular to J?J?', and make 5C" = 1.71, the first
southing. Through C" draw C"(7 perpendicular to SC", and equal to
4.70, the second easting. Join jB and C, The line BC will represent
the second side of the field.
Proceed in like manner until ike field is completely represented. The
extremity of the last line I/A, measured &om Z>', should fall at A. This
will he a test of the accuracy of the plot.
By drawing the diagonal AC, and letting &11 upon it perpendiculars
from B and J9, the quadrilateral ABCD is divided into two triangles,
the bases and altitudes of which maybe measured and th« area com-
puted approximately.
Other methods of plotting will suggest themselves, bat the method
just explained is one of the best.
Balancing the Work
In the survey, we pass entirely around the field ; hence, we
move just as far north as south. Therefore, the sum of the
northings should equal the sum of the southings. In like
manner, the sum of the eastings should equal the sum of the
westings. In this way the accuracy of the field work may be
tested.
In Example 1, the sum of l^e northings is equal to the sum
of the southings, being 11.56 in each case ; and the sum of the
eastings is equal to the sum of the westings, being 9.40 in each
case. Hence, the work balances.
In actual practice the work seldom balances. When it does
not balance, corrections are generally applied to the latitudes
and departures, by the following rules :
The perimeter of the field : any one side
: : total error in latitude : correction ;
: : total error in departure : correction.
If special difficulty has been experienced in taking a par-
ticular bearing, or in measuring a particular line, the correc-
tions should be applied to the corresponding latitudes and
departures.
FIELD NOTES.
29
The amount of error allowable varies in the practice of dif-
ferent surveyors, and according to the nature of the ground.
An error of 1 link in 8 chains would not be considered too
great on smooth, level ground ; while, on rough ground, an
error of 1 link in 2 or 3 chains
might be allowed. If the error
is considerable, the field meas-
urements should be repeated.
Example 2. Let it be re-
quired to survey the field A£
CDUF(Fig,2S).
FnLD Notes.
1
N.TSOSO'W.
5.00
2
s. leoso'w.
5.00
3
N.28«30'W.
7.07
4
N.20«>00'E.
11.18
5
S. 43«>30'E.
5.00
6
S. 13°30'E.
10.00
818.0888
81.4955
101.5933
80.7967
8.0797 acret.
EzFLANATioir. The first station
in the field notes is D, but we re-
arrange the numbers in the tablet so
that A stands first. The northings
and southings balance, but the east-
ings exceed the westings by 1 link.
We apply the correction to Uie west-
ing 4.79 (the distance DE being in
doubt), making it 4.80, and rewrite
all the latitudes and departures in the next four columns, incorporating
the correction. In practice, the corrected numbers are written in red ink.
IJl ta b Q to K
1^ Is ts b Q to
1
ai i» s« !» » ?!
ij S q Si ii jj
8 8 8 8 8 8
^ 4 ^ » p ^
1
§ i § 1 i g
1
?•
s
g i b i :• i
3ft
S
i 5 : § i :
59
1
: i : g ^ h
?!
1
§ b ^ ; ; ;
55
5
i 5?: ^§gt=
Co
i : i g^^^iS
r^
«aiM M>] «k.b *
5
o g-^ 5^ g« S'^ g*;
a>s 3t>i So 8<5 StB
3C
t^ 0. q ta
^^ "i ^5 ®tt *^ ***
|»i to <3 Q
22
je
?-
1
: IS: i§B§:
so
30
SURYEYINa.
B*
The remainder of the computation does not require expla-
nation.
It will be seen that this method of computing areas is
perfectly general.
'v § 17. Supplying Omissions.
If, for any reason, the bearing
and length of any side do not ap-
pear in the field notes, the latitude
and departure of this side may be
found in the following manner :
Find the latitudes and departures
of the other sides as usual. The
difference between the northings
and southings will give the north-
ing or southing of the unknown
side, and the difference between
the eastings and westings will give
the easting or westing of the un-
known side.
If the length and bearing of the
unknown side are desired, they
may be found by solving the right
triangle, whose sides are the lati-
tude and departure found by the method just explained, and
whose hypotenuse is the length required.
Fig. 23.
§18. Irregular Boundaries.
If a field have irregular boundaries, its area may be found
by offsets, as explained in § 14, Prob. 3.
§ 19. Obstructions.
If the end of a line be not visible from its beginning, or if
the line be inaccessible, its length and bearing may be found
as follows :
OBSTRUCTIONS.
31
1. By means of a random line (§ 4, 3).
2. When it is impossible to run a random line, which is
frequently the case on account of the extent of the obstructioni
the following method may be used : ^
Let AJB (Fig. 24) represent an inacceasiblQ line
whose extremities A and B only are known, and
B invisible from A.
Set flag-stafib at convenient points, C and D.
Find the bearings and lengths of AQ CD^ and DB^
and then proceed to find the latitude and depar-
ture of AB, as in § 17.
Fig. 94.
Example. Suppose that we have the following notes (see
Fig. 24):
side.
Bearing.
Diet
fi.
S.
e.
w.
AC
CD
DB
S. 45«E.
E.
N.30«E.
3.00
3.50
4.83
4.18
2.12
2.12
3.50
2.42
4.18
2.12
8.04
Flff.25.
4 lg The northing of AB is 2.06, and the easting, 8.04 ; which
2 12 numbers may be entered in the tablet in the columns N. and E.,
r^ opposite the side AB.
If the bearing and length of AB are required, construct the
right triangle ^-BC(Fig. 25), making J.C- 8.04 and -BC- 2,06.
UxiBAC*
,BC^2j06^q256,
AC 8.04
Hence, the angle J?^C- 14*22'.
Also, AB - VZ^TW - V8.04« + 2.06« - 8.29.
Therefore, the bearing and length of AB are N. 76*» SS' E. 8.29.
82
SUEVEYIKG.
Exercise II.
IS'oTE. In examples 5 and 6 detours were made on account of inacces-
sible sides ($ 19, 2). The notes of the detours are written in braces.
1. 6. 8.
St9.
Bearings.
DIat.
1
S. 76«E.
6.00
2
S. 15«E.
4.00
3
S. 75«W.
6.93
4
N.45«>E.
5.00
5
N.45°W.
5.19i
2.
Sta.
Baaringa.
DIat.
1
N.45«>E.
10.00
2
S. 75° E.
11.55
3
S. 16° W.
18.21
4
N.45°W.
19.11
3.
Sta.
Bearfnga,
DIat
1
N.15°E.
3.00
2
N.75°E.
6.00
3
S. 160W.
6.00
4
N.75°W.
5.20
Sta.
Baaringa.
Dtat.
1
N.89°45'E.
4.94
2
S. 7°00'W.
2.30
3
S. 28°00'E.
1.52
4
S. 0°45'E.
2.57
5
N.84°45'W.
5.11
6
N. 2°30'W.
5.79
Sta.
Baaringa,
DIat.
1
S. 2°15'E.
9.68
N.51°45'W.
2.39
2]
S. 85°00'W.
6.47
S. 55°10'W.
1.62
3
N. 3045'E.
6.39
4
S. 66°46'E.
1.70
5
N.15°00'E.
4.98
6
s. 82045'E.
6.03
Sta.
Baaringa,
DIat
'{
S. 81°20'W.
4.28
N.76°30'W.
2.67
2
N. 5°00'E.
8.68
3
S. 87°30'E.
5.54
S. 7°00'E.
1.79
4"
S. 27°00'E.
1.94
S. 10°30'E.
5.35
•
N.76°45/W.
1.70
7.
Sta.
Baaringa.
DIat.
1
N. 6°15'W.
6.31
2
S. 81°50'W.
4.06
3
S. S^OO'E.
5.86
4
N.88°30'E.
4.12
Sta.
Baaringa.
DIat
1
N. 5°30'W.
6.08
2
S. 82°30'W.
6.51
3
S. S^OO'E.
5.33
4
E.
6.72
9.
Sta.
Baaringa,
DIat
1
N.20°00'E.
4.62J
2
N.73°00'E.
4.16i
3
S. 45°15'E.
6.18J
4
S. SSOSO'W.
8.00
5
Wtttint.
W»tint.
10.
sta.
Bearfnga.
DIat
1
S. 3°00'E.
4.23
2
S. 86°45'W.
4.78
3
S. 37°00'W.
2.00
4
N.81«00'W.
7.45
5
N.61°00'W.
2.17
6
N.32°00'E.
8.68
7
S. 75°50'E.
6.38
8
S. 14°45'W.
0.98
9
S. 79°15'E.
4.52
RECTANOULAB METHOD. 33
§ 20. Modification of the Rectanqulab Method.
The area of a field may be found by a modification of the
rectangular method, if its sides and interior angles are known.
Let A, B, C, -D, represent the inte-
rior angles of the field ABCD (Fig.
26). Let the side AB determine the
direction of reference.
The bearing of AB, with reference
to AB, is 0^
The bearing of BO, with reference
to AB, is the angle b = 180^- .B.
The bearing of CD, with reference a^
to AB, is the angle c = (7— 5. Rf • m.
The bearing of DA, with reference to AB, is the angle d=sA,
The area may now be computed by the rectangular method,
regarding AB as the magnetic meridian.
As the interior angles may be measured with considerable
accuracy by the transit, the latitudes and departures should
be obtained by using a table of natural sines and cosines.
Exeboibe III.
1. Find the area of the field ABCD, in which the angle
^ = 120^ J? = 60^ C'=15(r, and 2) = 30*; and the side
AB = 4 chains, J?(7= 4 chains, CD = 6.928 chains, and DA
= 8 chains.
2. Find the area of the farm ABCDE, in which the angle
A = 106^ 19', B = 99* 40', (7= 120* 20', D = 86** 8', and E^
127**33'; and the side ^5=^79.86 rods, 50^=121.13 rods,
CD = 90 rods, DE^ 100.65 rods, and EA = 100 rods.
§21. Gekebal Bemabks on DsTEBMiNiKa Abeas.
Operations depending upon the reading of the magnetic
needle must lack accuracy. Hence, when great accuracy is
required (which is seldom the case in land surveying), the
rectangular method (§§ 16-19) cannot be employed.
84 SURVEYING.
The best results are obtained by the methods explained in
§§ 14 and 20, the horizontal angles being measured with the
transit, and great care exercised in measuring the lines.
§ 22. The Variation op the Needle.
The Magnetic Deolinaticnii or variation of the needle, at any
place, is the angle which the magnetic meridian makes with
the true meridian, or north and south line. The variation is
said to be east or west, according as the north end of the
needle lies east or west of the true meridian. Western vari-
ation is indicated by the sign +, and eastern variation by the
sign -.
IrregTilar VariationB are sudden deflections of the needle,
which occur without apparent cause. They are sometimes
accompanied by auroral displays and thunder storms, and are
most frequent in years of greatest sun-spot activity.
Solar-Diurnal Variation. North of the equator, the north end
of the needle moves to the west, from 8 a.m. to 1.30 p.m., about
6' in winter and 11' in summer, and then returns gradually to
its normal position.
Seonlar Variation is a change in the same direction for about
a century and a half; then in the opposite direction for about
the same time.
The line of no variation, or the Agonic Lme, is a line joining
those places at which the magnetic meridian coincides with
the true meridian. In the United States, this line at present
(1882) passes through Michigan, Ohio, the Virginias, and the
Carolinas. It is moving gradually westward, so that the varia-
tion is increasing at places east of this line, and decreasing at
places west of this line. East of this line the variation is
westerly, and west of this line the variation is easterly.
The table on page 35, which has been prepared by permis-
sion from the report for 1879 of the United States Coast and
Geodetic Survey, shows the magnetic variation at different
places in the United States and Canada for several years;
also, the annual change for 1880.
TABIATION OF THE KEESLE.
35
CQCQO
P 80 B
p p I
hi >;'-P
go CD -
£-...* ■ p ,....• .
•<rtO»-'OOCDOb3t*^H-'b3tOtOCA3tOOOO«OCDOtO»i*-H-t— K-OtOU)WW^
^ IS3 en CO <I H- O W O >*^ "^ 05 O to 05 (Ul <^ P U) 50 00 00 Oi CD 00 l>3 J— j*i. QO 5
o, 'h-* i<i o to *>*!»• O en bo CO bo bi i^ bo bo co bo <© '-vr to to bi cD bi I*^ co bi to bo •
tOH-*co<:pcooooooooooooo^i^i-^-^^j-^-^-i-i-i^7-^*.T'^-»i-a-j^iS'
tO-»lCnOOOOOOl-»l-»l-»OCOCDOO^lOiOiCn>f»»OOOOfcOfcO>— H-H-H-OO^
•^rf^OdtOCOtOOOQOtO^pCnMMptNST^pppllsOCn
to bo bi to CO bi o bi *P bi I*^ bo U^h. b» bi co p P f^ 00 bo ^ »fk. h-* bi ^ 00 bi bi •
I
I
I I
»— ' I— »
CO*-*
III.
-vicnCAO
i.: il
ooooH-'>*w -^ t^ 01 05 to *T oa *T 00 3
»^b:>
too
I I I I 1 I I I I
CDOOOifcOOaj-'j^pOO
to H-» ^ "^ ^ »^ bo bo CO
OH-'Cooppo-^-<r
P P p p to >*!»• pi 74 1^ pi p to 00 *T c» p 3
s
I I
»— ' I— »
OitO
o o
I I I I I I I I ^ ^§
00 00 -^ to Oa p Jf». p to »-» J-» j-» to J-» 00 pi ^ 00 pi p 00 CO CO p p J-' 5
J**' H-» o p Q bi H-» bo bo bo J-* to ►-' ^ 'a bi o co co ijo bo ^ bo f-j bj to 2
OOCOCDPH-'>f:i.t*;*.>^tOOacD^OOoa>OOiCi3S^>— OCOH-'COCaO ?
[
OitO
I-»bo
00 00
<r <r p p 4»«. p 00 j-* {-•
bo bi -^r CO bo bo o bi b»
COtO-^COOOOOOOOOi
to to to CO to t^ p 00 p *T ;-! CO .Ol p p .►-' tO J
I-a p I^ <i CD <i CD I^ Vo bo bo *p CD bi I^ ►-^ c*o •
^T»i^OiH^OOiOOCOOiCnOiOiCn<^COCiCn !*
poo
bo to
<TO
Oi-qr Oi o f^
P IsO to O to CO CO jlW 00 pi -<r p p 00 •<! p p H- h-' h-* to 5
bo o 00 4^ »-' bo I^. bo CO bi I^ *p 'rf*" bi 4^ I^ to CO J-* CD to bo to o CO 00 2
00Ol00l-»00-^00t00i0i00i4^000lCnOC000OC0h-'t0»-'C*0toO =•
i-i.H-1 I I I I I H- H-H-^H-^»-|'
OiOOOOiOippOOM»H-»OOOOOf^COp1{*^P74pj-*ppP;»IJ--»J-'tOOO j
bi bi »-^ ino ►**• bo 'h-* *4»^ CO bo p ?*»»■ bi bo bi o »-^ to bo ^ o bo bi ^ bo b^
tOOOOOCDOiCO-^»-J»COtP»H-*tOtOH-»Ol^IOit— »OH-*4h.tOOH-COpOCO r
s
LL'. Ill I I I ^-*
Oi 00 ^ p p pi O CO j-» ►-» CO p 00 rf^ CO pi J^ p ex p to CD p
bi bi ' p *^ bi bo I-* bi p bo h-' bo <i bo bo I^ ^ o I-» I^ ►-' co
Oi4»»' O 00 Oi -^ to •-' O CD ^I 00 00 -vl O -^ O CO CO »-• CO to
»-J» (-» h-» »-J» »— ' «8
-qr»— ^ to CO 5
O »-» to CO CO CO 00 j^ to CO 4^ CO rfk. pi 00 00 CO »^ to CO p j^ CO
bi ^ bi to bi if!^ o to bi bi to *P bi o to bo 'a> co bi ^ o co <r
H-'tOtOCOh-' 5-
biI-»co<ibi S.
Ig.
36
STJRVEYINa.
§23. To Establish a True Meridian.
This may be done as follows :
1. By means of Burt's Solar Compass (§ 25).
2. By observations of Polaris,
The North Star or Polaris revolves about the pole at pres-
ent at the distance of about 1-^° ; hence, it is on the meridian
twice in 23 h. 56 m. (a siderial day), once above the pole
called the upper culmination, and 11 h. 58 m. later below the
pole called the lower culmination. It attains its greatest
eastern or western elongation, or greatest distance from the
meridian, 5 h. 59 m. after the culmination.
The following table gives the time of the upper culmination
of Polaris for 1882. The time is growing later at the rate of
about one minute in three years.
Month.
PlpstDay.
Elerenth Day.
Twenty-JtostDay.
H. M.
H. M,
H. M.
January . .
6 32 P.M.
6 53 P.M.
5 13 P.M.
February .
4 30 P.M.
3 60 p.m;
3 11 P.M.
March . . .
' 2 35 P.M.
' ' 1 56 P.M.
1 17 P.M.
April. . . .
12 29 P.M.
11 50 AM.
11 11 A.M.
May
10 31 A.M.
9 52 A.M.
9 13 A.M.
i June ....
8 30 A.M.
7 51 A.M.
7 12 A.M.
July ....
6 32 A.M.
5 53 A.M.
5 14 A.M.
I August. . .
4 31 A.M.
3 52 A.M.
3 13 A.M.
September .
2 29 A.M.
1 50 A.M.
1 11 A.M.
October . .
12 32 A.M.
11 52 P.M.
11 13 P.M.
November .
10 30 P.M.
9 50 P.M.
9 11 P.M.
December .
8 32 P.M.
7 52 P.M.
7 13 P.M.
The time of the upper culmination of Polaris maj be found
by means of the star Alioth, which is the star in the handle
of the Dipper (in the constellation of the Great Bear) next to
the four which form the bowl. It crosses the meridian about
the same time as Polaris. Suspend from a height of about 20
feet a plumb-line, placing the bob in a pail of water to lessen
^
TO ESTABLISH A TEUE MEEIDIAN.
.37
its vibrations. About 15 feet south of the plumb-line, upon a
horizontal board firmly supported at a convenient height, plac«
a compass sight fastened to a board a few inches square. At
night, when Alioth by estimation approaches the meridian,
place the compass sight in line with Polaris and the plumb-
line, and move it so , ,
as to keep it in this
line until the plumb- ^ ^p,,,
line also falls on Ali-
oth (Fig 27). Note
the time; then twen-
ty-two minutes later
Polaris will be on the
meridian.
If the lower cul- i ^^*
mination takes place ^
at nightjthe time may *
be found in a similar
manner.
Instead of the compass sight, any upright with a small open-
ing or slit may be used.
The plumb-line may be made visible by a light held near it.
(a) To locate the true meridian by the position of Polaris at its
1. By vsing the apparatus described in finding the time of
culmination. At the time of culmination bring Polaris, the
plumb-line, and the compass sight into line. The compass
sight and the plumb-bob will give two points in the true
meridian.
2. By means of the transit. Bring the telescope to bear on
Polaris at the time of culmination, holding a light near to
make the wires visible, if the observation is made at night.
The telescope will then lie in the plane of the meridian, which
may be marked by bringing the telescope to a horizontal
position.
Fig. 27.
88
SUEVEYINQ.
(b) To locate the meridian by the poBition of Polaris at greatest
elongatioiL
The Arimnth of a star is the angle which the meridian plane
makes with a vertical plane passing through the star and the
zenith of observer.
Let Z (Fig. 28) represent the zenith of the place, P the pole, and N
Polaris at its greatest elongation. Let ZP, ZN, and PN be
arcs of great circles, and let i\r be a right angle.
sin PiV= cob(90« - ZP)cos(9(y> - Z),
[Spher.Trig^|47.]
Bnt ZP« the complement of the latitude.
Hence, 90® — ZP» the latitude of the place.
Hence, sin PJV» cos latitude X sin Z.
Hence, sinZ»
sinPJy
cos latitude
Hence, Z (the azimuth of Polaris) can be found if the lati-
tude of the place and the greatest elongation of Polaris {PN)
are known.
The following table gives the latter element Jan. 1, 1882-93.
Geeatest ELOiraATioisr of Polaeis.
1882
1« W 12.5''
1886
in?' 56.6"
1890
P 16' 40.7"
1883
P 18' 53.5"
1887
P 17' 37.6"
1891
P 16^ 21.8"
1884
P 18' 34.5"
1888
P 17' 18.6"
1892
P10' 3"
1885
P 18' 15.5"
1889
P 16^ 69.7"
1893
P 15' 44.1"
The greatest elongation of Polaris, or the polar distance, is
given in the Nautical Almanac. The table gives this element
for Jan. 1. It may be found for other dates by interpolation.
To obtam a line in the direction of Polaris at greatest elongation.
1. By using the apparaiics for finding the time of culmina-
tion.. A few minutes before the time of greatest elongation
(5 h. 59 m. after culmination), place the compass sight in line
with the plumb-line and Polaris, and keep it in line with these
until the star begins to recede. At this moment the sight and
plumb-line are in the required line.
DIVIDINQ LAND.
39
2. By means of the transit, A few minutes before the time
of greatest elongation, bring the telescope to bear on the star,
and follow it, keeping the vertical wire over the star until it
begins to recede. The telescope will then be in the required
line.
To establish the meridian. Having the transit sighted in the
direction of the line just found, turn it through an angle equal
to the azimuth in the proper direction.
§ 24. Dividing Land.
A few problems in the Division of Land are solved in this
section, for the purpose of acquainting the student with some
of the methods employed. The surveyor must, for the most
part, depend on his general knowledge of Geometry and Trigo-
nometry, and his own ingenuity, for solutions of the problems
that arise in practice.
Problem 1. To divide a triangular field into two parts having a
given ratioi by a line through a given vertex. ^
Let ABC (Fig. 29) be the triangle, and A the
given vertex.
BD^
Divide BC&t 2), so that j~ equals the given
ratio, and join A and D. ABD and iiDCwill
be the parts required ; for
ABDiADCi.BD'.DC.
Problem 2. To oat off from a
tnangolar field a given area, by a
line parallel to the base.
Let ABC (Fig. 30) be the triangle,
and let DE be the division line re-
quired.
y/ABC.y/ADE-.AB.AD.
.AD'-AB-yl
ABC
Rg. 30.
40
SimVEYIKG.
FB03LEM 3. To divide any field into two parts having a given
ratio, by a line through & given point in
the perimeter.
Let ABODE (Fig. 2>1) represent the field,
P the given point, and PQ the required divi-
sion line.
The areas of the whole field and of the
required parts having been determined, run
the line PD from P to a corner D, dividing
the field, as near as possible, as required.
Determine the alrea PBCD.
The triangle PDQ represents the part
which must be added to PBCD to make the
Fig. 31. required division.
Area PDQ ^ixPDxDQxsin PDQ.
Hence, i)Q =
2 area PDQ
Pi)x sin PX>Q
Note. DQ^
2 area PDQ
perpendicular from P on DE"
Pon DEm&y be run and measured directly.
This perpendicular from
Pboblem 4. To divide a field into a given number of parts,
B BO that aooess to a pond of water is given
to each.
Let ABODE (Fig. 32) represent the field,
and Pthe pond. Let it be required to divide
the field into four parts. Find the area of
the field and of each part.
Let AP be one division line. Run PE,
and find the area APE. Take the differ-
ence between APE and the area of one of
the required parts ; this will give the area
of the triangle PQE, from which QE may be
found, as in Problem 3. Join P and Q ;
PAQ will be one of the required parts. In
like manner, PQE and PAS are determined
fourth part required.
whence, PSB must be the
\
EXAMPLES. 41
EXEECISE rV.
1. From the square ABCD, containing 6 A. 1 B. 24 p.,
part off 3 A. by a line J?jP parallel to AB,
2. From the rectangle ABCD, containing 8 A. 1 B. 24 p.,
part off 2 A. 1 E. 32 p. by a line EF parallel to AD — 1 cfa.
Then, from the remainder of the rectangle, part off 2 A. 8 b.
25 p., by a line (?JB' parallel to EB,
3. Part off 6 A. 3 b. 12 p. from a rectangle ABOD, con-
taining 15 A., by a line J?jP parallel to AB ; AD being 10 ch.
4. From a square A BCD, whose side is 9 ch., part off a
triangle which shall contain 2 A. 1 b. 36 p., by a line BE
drawn from B to the side AD.
5. From ABCD, representing a rectangle, whose length is
12.65 ch., and breadth 7.58 ch., part ofT a trapezoid which
shall contain 7 A. 3 B. 24 p., by a line BE drawn from B to
the side DC.
6. In the triangle ABC, AB= 12 ch., AC= 10 ch., and
BC= 8 ch. ; part off 1 a. 2 b. 16 p., by the line DE parallel
to AB.
7. In the triangle ABC, AB = 26 ch., AC= 20 ch., and
BC= 16 ch. ; part off 6 A. 1 b. 24 P., by the line DE par-
allel to AB.
8. It is required to divide the triangular field ABC smong
three persons whose claims are as the numbers 2, 3, and 5, so
that they may all have the use of a watering-place at C; AB
= 10 ch., AC= 6.85 ch., and CB = 6.10 ch.
9. Divide the five-sided field ABCHE among three per-
sons, X, Y, and Z, in proportion to their claims, X paying
$500, Y paying $750, and Z paying $1000, so that each may
have the use of an interior pond at P, the quality of the land
being equal throughout. Given AB = 8.64 ch., J5C= 8.27 ch.,
CS'= 8.06 ch., E:E=^ 6.82 ch., and EA = 9.90 ch. The per-
pendicular PD upon AB = 5.60 ch., PD' upon BC= 6.08 ch.,
PD" upon Cff=4.80 ch., PD'' upon £[E=6.U ch., and
PD"" upon EA = 5.40 ch. Assume Pff as the divisional
42 SUBVBYING.
fence between X *s and Y 's shares ; it is required to determine
the position of the fences PM and PN between X *s and Y 's
shares and Y 's and Z 's shares, respectively.
10. Divide the triangular field ABC, whose sides AB.AO,
and EC are 15, 12, and 10 ch., respectively, into three equal
parts, by fences EO and 2)-F parallel to BC.
11. Divide the triangular field ABC, whose sides AB, BC,
and -4 (7 are 22, 17, and 15 ch., respectively, among three per-
sons, A, B, and 0, by fences parallel to the base AB, so that
A may have 3 A., B, 4 a., and 0, the remainder.
25. United States Public Lands.
Burt's Solar Compass.
This instrument, which is exhibited on the following page,
may be used for most of the purposes of a compass or transit.
Its most important use, however, is to run north and south
lines in laying out the public lands.
A full description of the solar compass, with its principles,
adjustments, and uses, forms the subject of a considerable vol-
ume, which should be in the hands of the surveyor who uses
this instrument. The limits of our space will allow only a
brief reference to its principal features.
The main plate and standards resemble these parts of the
compass.
a is the latitude arc,
b is the declination arc,
h is an arm, on each end of which is a solar lens having its
focus on a silvered plate on the other end.
c is the hour arc.
n is the needle-box, which has an arc of about 36®.
To mn a north and south line with the solar compass. Set off
the declination of the sun on the declination arc. Set off the
latitude of the place (which may be determined by this instru-
ment) on the latitude arc. Set the instrument over the station,
'\
BURTS SOLAR COMPASa
LAYING OUT PUBLIC LANDS.
45
level, and turn the sights in a north and south direction, ap-
proximately, by the needle. Turn the solar lens toward the
sun, and bring the sun's image between the equatorial lines on
the silvered plate. Allowance being made for refraction, the
sights will then indicate a true north and south line.
The Solar AUachmerUfor Tranaiis.
This consists, essentially, of the solar apparatus of the solar
compass attached to the telescope of the ordinary transit.
Laying Out the Public Lands,
The public lands north of the Ohio River and west of the
Mississippi are generally laid out in townships approximately
six miles square.
A Frindpal Heridiaiii or true north and south line, is first
run by means of Burt's Solar Compass, and then an east and
west line, called a Base Line.
Parallels to the base line are run at intervals of six miles,
and north and south lines at
the same intervals. Thus, the
tract would be divided into
townships exactly six miles
square, if it were not for the
convergence of the meridians
on account of the curvature of "''"
the earth.
The north and south lines,
or meridians, are called Bange
Lines. The east and west lines,
or parallels, are called Town-
Bhip Lmes.
Let N8 (Fig. 33) represent a principal meridian, and WE
a base line ; and let the other lines represent meridians and
parallels at intervals of six miles.
The small squares. A, B, (7, etc., will represent townships.
B
Fig. 33.
46
SUBVETINQ.
6
7
18
19
SO
31
5
8
17
20
29
32
4
9
16
21
28
33
3
10
15
22
27
34
2
11
14
23
26
35
1
12
18
24
25
36
A would be designated thus: T. 3 N., R. 2 W. ; that is,
township three north, range two west ; which means that the
township is in the third tier north of the base line, and in the
second tier west of the principal meridian. JB and (7, respec-
tively, would be designated thus : T. 4 S., R. 3 W. ; and
T, 2 N., R. 2 E.
The townships are divided into sections approximately one
mile square, and the sections are di-
vided into quarter-sections. The town-
ship, section, and quarter-section corners
are permanently marked.
The sections are numbered, beginning
at the north-east comer, as in Fig. 34,
which represents a township divided
into sections. The quarter-sections are
^*** ^' designated, according to their position,
asN.E., N.W., S.E., andS.W.
Every fifth parallel is called a Standard Parallel or Oorreotion
Line.
Let iV3'(Fig. 35) represent a principal meridian ; WU a base
line ; rp, etc., meridians ; and
ms the fifth parallel. If Op
equals six miles, mr will be
less than six miles oij account
of the convergence of the
meridians. Surveyors are
instructed to make Op such
a distance that mr shall be
six miles; then mA, hk, etc.,
are taken similarly. At the
correction lines north of ms
the same operation is re-
peated.
The township and section
'** lines are surveyed in such an
order as to throw the errors on the north and outer townships
and sections.
/
w —
p
1
PLANE-TABLE SURVEYING. 47
If, in running a line, a navigable stream or a lake more than
one mile in length is encountered, it is meandered by marking
the intersection of the line with the bank and running lines
from this point along the bank to prominent points which are
marked, and the lengths and bearings of the connecting lines
recorded.
Six principed meridians have been established and con-
nected. In addition to these there are several independent
meridians in the Western States and Territories which will in
time be connected with each other and with the eastern
system.
§26. Planb-Tablb Survbyino.*
After the principal lines of a survey have been determined
and plotted, the details of the plot may be filled in by means
of the plane-table ; or, when a plot only of a tract of land is
desired, this instrument affords the most expeditious means of
obtaining it.
An approved form of the plane-table, as used in the United
States Coast and Geodetic Survey, is shown in the plate on
page 49.
The Table-top is a board of well-seasoned wood, panelled
with the grain at right angles to prevent warping, and sup-
ported at a convenient height by a Tripod and Levelliiig
Head.
The Alidade is a ruler of brass or steel supporting a telescope
or sight standards, whose line of sight is parallel to a plane
perpendicular to the lower side of the ruler, and embracing
its fiducial edge.
The Deolinatoire consists of a detached rectangular box con-
taining a magnetic needle which moves over an arc of about
10° on each side of the 0.
^In preparing this section the writer has consulted, hj permission, the treat-
we on the plane-table by Mr. £. Hergesheimer, contained in the report for 1880
ol the U.S. Coast and Geodetic Sarrej.
4b SURVEYING.
Two spirit levels at right angles are attached to the ruler or
to the declinatoire. In some instruments these are replaced
by a circular level, or by a detached spirit level.
The paper upon which the plot is to be made or completed
is fastened evenly to the board by clamps, the surplus paper
being loosely rolled under the sides of the board.
To place the table in position. This operation, which is some-
times called orienting the table, consists in placing the table
so that the lines of the plot shall be parallel to the correspond-
ing lines on the ground.
This may be accomplished by turning the table until the
needle of the declinatoire indicates the same bearing as at a
previous station, the edge of the declinatoire coinciding with
the same line on the paper at both stations.
If, however, the line connecting the station at which the
instrument is placed with another determined station is already
plotted, the table may be placed in position by placing it over
the station so that the plotted line is by estimation over and
in the direction of the line on the ground ; then making the
edge of the ruler coincide with the plotted line, and turning
the board until the line of sight bisects the signal at the other
end of the line on the ground.
To plot any point Let ab on the paper represent the line
AB on the ground ; it is required to plot c, representing C on
the ground.
1. By intersection.
Place the table in position at A (Fig. 36), plumbing a over A, and
Q making the fiducial edge of the
i+x^ ruler pass through a; turn the
/ ^'^v alidade about a until the line of
/ ^^^ sight bisects the signal at Q and
draw a line along the fiducial edge
of the ruler. Place the table in
position at B, plumbing b over J5,
and repeat the operation just de-
scribed, c will be the intersec-
p"«- 36' tion of the two lines thus drawn.
THE PLANE-TABLE.
PLANE-TABLE SURVEYINQ.
61
2. By rejection.
Place the table in position at A (Fig. 37), and draw a line in the di-
rection of (7, as in the former case; then remove the inatroment to C,
place it in position by the line drawn
-=sf
/
/
/
/
/
:\.
-B
c
from a, make the edge of the ruler
pass through i, and turn the alidade
about h until ^ is in the line of
sight. A line drawn along the
edge of the ruler will intersect the
line from a in c.
3. By radiaticn.
Place the table in position at A pj 3^
(Fig: 38), and draw a line from a
toward C, as in the former cases.
Measure AQ and lay off oc to the
same scale as ab.
To plot a Add A BCD
1. By radiation.
Set up the table at any point P,
and mark p on the paper over P. B
Draw indefinite lines from p to- ^ 33
ward A, B, C, Measure PA,
PB, , and lay off pa, pb, , to a suitable scale, and join a and b,
b and c, c and d, etc.
2. By progression.
Set up the table at A, and draw a line from a toward B. Measure
AB, and plot ab to a suitable scale. Set up the table in position at B,
and in like manner determine and plot be, etc.
3. By. intersection.
Plot one side as a base line. Plot the other corners by the method of
intersection, and join.
4. By resection.
Plot one side as a base line. Plot the other comers by the method of
resection, and join.
52
SUBVEYING.
The Three Point Problem,
Let A, B, C represent three points determined and plotted
zsa.bjC (Fig. 39) ; it is required to plot by the plane-table,
d representing a fourth undetermined point D.
Fig. 39.
Determine the point d by the method of resection ; first,
from A and B, then from A and C, If the three lines from
a, &, and c meet in a point, this will determine d. Ordinarily,
they will not meet in a point, but will form a triangle of error ^
ac he ab. Through a, 6, and ab\ a, c, and ac; and 5, c, and
be, respectively, pass circumferences of circles; these three
circles will intersect in the required point d.
CHAPTER III.
TBIANQULATION.*
§27. Intboductoby Rehabks.
Geographical positions upon tlie surface of the earth are
commonly determined by systems of triangles which connect a
carefully determined base line with the points to be located.
Let F (Fig. 40) represent a point whose position with refer-
ence to the base line AB is
required. Connect AB with
F by the series of triangles
ABC, ACL, ADE, and
DEF, so that a signal at Q
is visible from A and B, a
signal at D visible from A
and G, a signal at E visible
from A and D, and a signal at F visible from D and E, In
the triangle ABC, the side AB is known, and the angles at
A and B may be measured; hence, AO may be computed.
In the triangle ACD, AO \% known, and the angles at A and
C may be measured ; hence, AD may be computed. In like
manner DE and EF or DF may be determined. DF, or
some suitable line connected with DF, may be measured, and
this result compared with the computed value to test the
accuracy of the field measurements.
• In preparing thig chapter the writer has consulted, by permission, recent
reports of the United States Coast and Geodetic Sarvey.
64 SURVEYING.
Three orders of triangulation are recognized ; viz. :
Primaiy, in wUch the sides are from 20 to 150 miles in
length.
Seoondaryi in which the sides are from 5 to 40 miles in
length, and which connect the primary with the tertiary.
Tertiary, in which the sides are seldom over 5 miles in
length, and which bring the survey down to such dimensions
as to admit of the minor details being filled in by the compass
and plane-table.
§ 28. The Measueement of Base Lines.
Base lines should be measured with a degree of accuracy
corresponding to their importance.
Suitable ground must be selected and cleared of all obstruc-
tions. Each extremity of the line may be marked by cross
lines on the head of a copper tack driven into a stub which is
sunk to the surface of the ground. Poles are set up in line
about half a mile apart, the alignment being controlled by a
transit placed over one end of the line.
The preliminary measurement may be made with an iron
wire about one-eighth of an inch in diameter and 60" in
length. In measuring, the wire is brought into line by means
of a transit set up in line not more than one-fourth of a mile
in the rear. The end of each 60" is marked with pencil lines
on a wooden bench whose legs are thrust into the ground after
its position has been approximately determined. If the last
measurement exceeds or falls short of the extremity of the line,
the difierence may be measured with the 20" chain.
The final measurement is made with the base apparatus,
which consists of bars 6" long, which are supported upon
trestles when in use. These bars are placed end to end, and
brought to a horizontal position, if this can be quickly accom-
plished ; if not, the angle of inclination is taken by a sector,
or a vertical offset is measured with the aid of a transit, so
that the exact horizontal distance can be computed.
MEASUREMENT OF ANGLES. 65
A thermometer is attached to each bar, so that the tempera-
ture of the bar may be noted and a correction for temperature
applied.
The method of measuring lines varies according to the re-
quired degree of accuracy in any particular case, but the brief
description given above will give the student a general idea
of the methods employed.
§29. The Measurement of Angles.
Angles are measured by large theodolites, and the reading
taken by microscopes to single seconds.
In order to eliminate errors of observation, and errors aris-
ing from imperfect graduation of the circles, a large number
of readings is made and their mean taken. Two methods are
in use; viz., repetition and series.
The method of repetition consists, essentially, in measuring
the angles about a point singly, then taking two adjacent
angles as a single angle, then three, etc. ; thus '' closing the
horizon," or measuring the whole angular magnitude about a
point in several different ways.
The method of series consists, essentially, in taking the
readings of an angle with the circle or limb of the theodolite
in one position,' then turning the circle through an arc and
taking the readings of the same angle again, etc. ; thus read-
ing the angle from successive portions of the graduated circle.
On account of the curvature of the earth, the sum of the
three angles of a triangle upon its surface exceeds 180°. This
spherical excess, as it is called, becomes appreciable only when
the sides of the triangle are about 5 miles in length. To
determine the angles of the rectilinear triangle having the
same vertices, one-third of the spherical excess is deducted
from each spherical angle.
CHAPTER IV.
LEVELLING.
§30. Definitions, Cuevature, and Refraction.
A Level Bnrfaoe is a surface parallel with the surface of still
water.
A Level Line is a line in a level surface.
Levelling is the process of finding the difference of level of
two places, or the distance of one place above or below a level
line through another place.
The Line of Apparent Level of a place is a tangent to the level
line at that place. Hence, the line of apparent level is per-
pendicular to the plumb-line.
The Ooirection for Ourvatnre is the deviation of the line of
apparent level from the level line for any distance.
Let t (Fig. 41) represent the line of apparent level of the
place P, a the level line, d the diame-
ter of the earth ; then c represents the
correction for curvature. To compute
the correction for curvature:
Therefore,
c-\rd d
approximately, since c is very small
compared with c?, and < = a without
appreciable error.
Since d is constant (= 7920 miles, nearly), the correction for
curvature varies as the square of the distance.
Fig. 44.
THE LEVELLING BOD.
57
Example. What is the correction for curva-
ture for 1 mile?
By substituting in the formula deduced
above, a* 1' • o • ^
Hence, the correction for curvature for any
distance may be found in inches, approximate-
ly, by multiplying 8 by the square of the dis-
tance expressed in miles.
A correction for the refraction of the rays of
light is sometimes made by decreasing the cor-
rection for curvature by one-sixth of itself.
§ 31. The Y Level.
This instrument is shown on page 59.
The telescope is about 20 inches in length,
and rests on supports called y"«, from their
shape. The spirit level is underneath the tel-
escope, and attached to it. The levelling- head
and tiipod are similar to the same parts of the
transit.
§32. The Levelling Rod.
The rod shown in Fig. 42 is known as the
New York levelling rod. It is made of two
pieces of wood, sliding upon each other, and
held together in any position by a clamp. The
front surface of the rod is graduated to hun-
dredths of a foot up to 6} feet.
The target slides along the front of the rod,
and is held in place by two springs which press
upon the sides of the rod. It has a square
opening at the centre, through which the divi-
sion line of the rod opposite to the horizontal
line of the target may be seen. It carries a
vernier by which heights may be read to thou-
sandths of a foot (§ 7).
58
SUEVEYING.
If a greater height than 6^ feet is desired, the target is
made fast with its horizontal line at 6^ feet from the lower
end. The back part of the rod is then moved up until the
target is at the required height, and the reading is taken at
the side of the rod. When fully extended the rod is 12 feet
long,
§33. Difference of Level.
To find the difference of level between two places visible from an
intermediate place.
Let A and B (Fig. 43) represent the two places. Set the
Y level at a station equally distant, or nearly so, from A and
Fig. 43.
By but not necessarily on the line AB. Place the legs of the
tripod firmly in the ground, and level over each opposite pair
of levelling screws, successively. Let the rodman hold the
levelling rod vertically at A. Bring the telescope to bear up-
on the rod (§ 8), and by signal direct the rodman to move the
target until its horizontal line is in the line of apparent level
of the telescope. Let the rodman now record the height AA'
of the target. In like manner find BB\ The difference be-
tween AA' and BB^ will be the difference of level required.
If the instrument be equally distant from A and B^ or nearly
so, no correction for curvature will be necessary.
If the instrument be set up at one station, and the rod at
the other, the difference between the heights of the optical
axis of the telescope and the target, corrected for curvature
and refraction, will be the difference of level required.
THE Y LEVEL.
LEVELLING FOB SECTION.
61
To find the differanoe of level of two plaoes, one of whioh oannot
be seen from the other, and both invisible from the same plaoe; or,
when the two plaoes differ oonsiderablj in level
Let A and D (Fig. 44) represent the two places. Place the
level midway between A and some intermediate station B.
Fig. 44.
Find AA^ and BB\ as in the preceding case, and record the
former as a hach-sight and the latter as a fore-sight. Select
another intermediate station C, and in like manner find the
back-sight BB^^ and the fore-sight CC ; and so continue until
the place D is reached. The difference between the sum of the
fore-sights and the sum of the back-sights will be the difference
of level required.
§34. Levelling for Section.
The intersection of a vertical plane passing through two
places, with the surface of the earth, is called a section or pro-
file; and levelling for the purpose of determining a section is
called levelling for section.
A Datum Line, or datum level, is a horizontal line to which
all points in the section are referred.
A Benoh Hark is a permanent mark at or near the beginning
of a line or intermediate points on the line.
In levelling for section the field notes should record not
only the fore-sights and back-sights, but also the distances
62
SURVEYING.
:t|
■JU. "J
between tte stations at which the rod is set up, and, if a hori-
zontal plot is desired, the bearing of the line.
Let ABQ J (Fig. 45) represent a
section to be determined and plotted.
Measure the section and set up stakes at
prominent points A, -S, (7, D, etc. En-
ter in the second column of the field notes
the bearing of the parts of the section
AO and OJ, marking as a station the
point O where the direction changes from
"W. to N. 80° "W. In the third column
enter the distances between the stations.
Thus, ^J? = 4, ^(7=4, Ci> = 4.50
chains, etc. Place the level at some sta-
tion B from which the level of Ay (7, and
D may be taken. Set up the rod at A
and find AA! = 10 feet, and record it as
a back-sight. Measure B£* = 4.50, the
height of the axis of the telescope. Set
up the rod at (7 and find (X" = 2.25, and
in the same manner find JDD' = 8.25, and
record the last three numbers as fore-
sights, placing an * after 8.25 to denote
that it is the last back-sight of the series.
Set up the level at Uy and proceed in the
same manner to find the back-sight jDjD"
= 3, and the fore-sights EE' = ^, FF^
= 9, and GO^ = 2*. Set up the level at
7, and find the back-sight (?(?" = 9.76,
and the fore-sights SS^ = 7.50, IF = 5,
and JF = 1.75*. The field work is now
complete, and is recorded in the field
book as in the first five columns of the
following tablet.
The first column contains the number of the station at which
the level is placed.
JiL-
\2P— i
LEVELLING FOR SECTION.
63
The sixth column, headed "Rise," contains the differences
between the back-sights and fore-sights when the former are
the greater; and the seventh column, headed "Fall," contains
these differences when the latter are the greater.
The eighth column, headed " Reduced Levels," contains the
heights of the points B^ C, D, etc., above a datum line
through A,
The ninth column contains the names of the stations.
L
Bearing,
Diet
B.S,
F.S.
Rlee,
Fall.
Reduced
Levaie.
s.
' Remarka.
1
w.
0.00
4.00
0.00
+5.50
A
B
Instrument.
10.00
4.50
5.50
4.00
....
2.25
7.75
.... +7.75
C
4.50
....
8.25*
1.75*
+1.75
D
2
w.
9.37
3.00
4.00
• • • .
1.00
+ .75
E
Instrument.
5.25
■ . . •
9.00
....
6.00
-5.25
F
8.00
2.00*
1.00*
. . . .
+2.75
3
N.80°W.
5.63
9.75
7.50
2.25
• . . .
+5.00
H
5.00
5.00
4.76
• • . .
+7.50
I
Instrument.
6.00
1.75*
8.00*
+10.75
J
22.75
12.00
10.75
0.00
10.75
0.00
10.75
The numbers in the eighth column are found most eas- ^ ^«
ily by drawing a rough figure of the section, from which i2!oo
the combinations necessary to produce these numbers are iq^
made plain. Great care should be exercised in crossing
the datum line.
The difference of level oi A and Jm&j be found in
three ways : 1. By taking the difference of the sum of the
back-sights and the sum of the fore-sights having an * ; 2. By
taking the difference of the sums of the numbers having an *
in the next two columns; 3. It is the last number of the
column of reduced levels, provided the datum line passes
through A,
To plot the section line. Draw the datum line A J", Choose
any convenient scale (12 chains to the inch in this case), and
lay off AB'' =A = AB, J5"C^' = 4 = J5C; {7"i>"^= 4.50 = CD,
etc.
64
SURVEYING.
At the points B", C", D"\ etc., erect perpendiculars £"£',
C'C'f etc. Clioose any convenient scale (24 feet to the inch
in this case), and lay off -B"^= 5.50, the second number in
the column of reduced levels; C"0= 7.75, the third number in
this column, etc.
Through A, B, 0, etc., draw the required section line
ABO J.
The datum line is commonly taken so as to bring all the
stations above it. This is done by increasing equally all the
numbers in the column of reduced levels. Thus, in the above
example, if each of these numbers is increased by 6.26, the da-
tum line will pass through -F, the lowest station. If each of
these numbers is increased by 10, the datum level will lie 10
units below its present position.
The labor of computation and plotting is lessened by mak-
ing all the distances AB, BO, etc., equal; but this method has
some obvious disadvantages when section only is required.
If it is more convenient to set up the level at one side of
the line, it may be so placed, and the height BB^ found by
the rod in the usual manner.
§35. Substitutes for the Y Level.
For many purposes not requiring accuracy, the following
simple instruments in connection with a graduated rod will be
found sufficient.
Fig. 46.
^-
Fig. 47.
Fig. 48.
The Plumb Level (Fig. 46) consists of two pieces of wood
joined at right angles. A straight line is drawn on the up-
right perpendicular to the upper edge of the cross-head.
SUBSTITUTES FOR THE Y LEVEL.
65
The instrument is fastened to a support by a screw through
the centre of the cross-head. The upper edge of the cross-head
is brought to a level by making the line on the upright coin-
cide with a plumb-line.
A modified form is shown in Fig. 47. A carpenter's square
is supported by a post, the top of which is split or sawed so as
to receive the longer arm. The shorter arm is made vertical
by a plumb-line which brings the longer arm to a level.
The Water Level is shown in Fig. 48. The upright tubes are
of glass, cemented into a connecting tube of any suitable ma-
terial. The whole is nearly filled with water, and supported
at a convenient height. The surface of the water in the up-
rights determines the level.
By sighting along the upper surface of the block in which
the Spirit Level is mounted for the use of mechanics, a level
line may be obtained.
Exercise V.
1. Find the difference of level of two places from the fol-
lowing field notes : back-sights, 6.2, 3.2, and 4.2 ; foresights,
8, 9, and 8.9.
Find the difference of level and determine section from the
following field notes:
2.
Back-
SIghta.
Fora-Slghta.
Diatancaa.
0.750
0.100
3.450
11.200*
10.500
1.00
2.00
2.00
Back-
Sighta.
Fore-
Sighta.
Diatancea.
2.461
6.895
1.35
1.468
5.321
0.75
....
12.000*
1.00
3.812
8.264
1.00
2.200
5.600
1.00
2.111
9.322
1.50
....
8.000
1.00
3.212
7.444
2.00
10.000
1.00
1.211
4.321
1.25
11.750*
1.00
66
SURVEYING.
4.
5.
6.
Baok-
Sights.
Fore-
sights.
Distances.
Back-
sights.
Fore-
sights.
Distances.
Back-
sights.
Fore-
Sights.
Distances.
3.50
5.65
4.60
1.84
0.16
2.00
0.34
3.16
1.00
4.10
10.85
3.20
1.02
8.70
1.00
5.86
5.61
1.00
5.04
9.25
3.80
2.83
7.91
1.00
4.19
4.24
1.00
3.84
12.91
3.60
5.20
10.63
2.00
5.44
1.20
1.00
4.12
7.65
2.80
4.13
8.71
1.92
4.96
3.20
1.00
10.49
3.92
3.00
4.04
8.02
2.08
4.73
1.32
1.00
12.96
3.03
6.00
2.92
3.16
7.92
6.04
2.00
2.00
6.10
5.33
2.00
3.96
1.00
1.00
2.17
6.87
1.00
5.91
1.83
1.00
2.13
7.00
1.80
3.70
0.90
1.00
7.
8.
Back-
sights.
Fore-
Sights.
Distances.
Back-
sights.
Fore-
Sights.
Distances.
13.71
7.88
5.19
4.47
4.53
1.00
9.40
16.30
7.96
....
9.22
1.00
3.87
11.71
2.27
5.07
1.00
2.63
12.41
3.08
0.24*
1.00
14.62
0.95
5.08
6.36
1.87*
1.00
17.00
1.45
3.40
6.14
1.47*
1.00
10.66
15.40
6.59
6.62
2.10
1.00
2.87
17.00
4.01
2.24*
0.16
3.40
10.32
2.18
10.42
4.32*
0.84
§36. TOPOGBAPHICAL LEVELLING.
The principal object of topographical surveying is to show
the contour of the ground. This operation, called topographi-
cal levelling, is performed by representing on paper the
curved lines in which parallel horizontal planes at uniform
distances apart would meet the surface.
It is evident that all points in the intersection of a horizontal
TOPOGEAPHIOAL LEVELLING.
67
plane with the surface of the ground are at the same level.
Hence, it is only necessary to find points at the same level,
and join these to determine a line of intersection.
The method commonly employed will be understood by a
reference to Fig. 49. The ground ABCD is divided into
equal squares, and a numbered
stake driven at each intersec-
tion. By means of a level and
levelling rod the heights of the
other stations above m and D,
the lowest stations, are deter-
mined. A plot of the ground
with the intersecting lines is
then drawn, and the height of
each station written as in the
figure.
Suppose that the horizontal
planes are 2 feet apart ; if the
first passes through m and i>, the second will pass through p,
which is 2 feet above m ; and since n is 3 feet above m, the
second plane will cut the line mn in a point 8 determined by
the proportion mn : ms : : 3 : 2. In like manner the points ^, g,
and r are determined.
The irregular line tap qr represents the intersection of
the second horizontal plane with the surface of the ground.
In like manner the intersections of the planes, respectively,
4, 6, and 8 feet above m, are traced.
It is evident that the more rapid the change in level the
nearer these lines will approach each other.
CHAPTER V.
RAILROAD SURVEYING.
§37. Gekebal Eemabks.
When the general route of a railroad has been determined,
a middle surface line is run with the transit. A profile of this
line is determined, as in § 34. The levelling stations are com-
monly 1 chain (100 feet) apart. Places of different level are
connected by a gradient line, which intersects the perpendic-
ulars to the datum line at the levelling stations in points
determined by simple proportion. Hence, the distance of each
levelling station, above or below the level or gradient line
which represents the position of the road bed, is known.
§38. Ceoss Section Woek.
-C--
AC r
Fig. 50.
Excavations. If the road bed lies below the surface, an exca-
vation is made.
Let ACDB (Fig. 50) represent a cross section of an excava-
tion, /a point in the middle surface line, /' the corresponding
point in the road bed, and CD the width of the excavation at
the bottom. The slopes at the sides are commonly made so
BAILBOAJ) CUBVE8.
that AA' = iA'a and ££'=-iDB\ ff and CD being
known, the points A^ B^ C\ and U are readily determined by
a level and tape measure.
If from the area of the trapezoid ABBfA! the areas of the
triangles AA^C and BB^D be deducted, the remainder will be
the area of the cross section.
In like manner the cross section at the next station may be
determined. These two cross sections will be the bases of a
frustum of a quadrangular pyramid whose volume will be the
amount of the excavation, approximately.
Embankments. If the road bed lies above the Bur£BU)e, an
embankment is made, the cross section of which is like that of
the excavation, but inverted.
Fig. 51.
Fig. 51 represents the cross section of an embankment
which is lettered so as to show its relation to Fig. 50.
§39. Railroad Curves.
When it is necessary to change the direction of a railroad,
it is done gradually by a
curve, usually the arc of
a circle.
Let AB and AO (Fig.
52) represent two lines to
be thus connected. Take
AD-=AE=t. The in-
tersection of the perpen-
diculars DC and EC de-
termines the centre (7, and '^*' ***
the radius of curvature DC=r, The length of the radius
70
SURVEYING.
of curvature will depend on the angle A and the tangent AJD.
For, ini the right triangle ABO, -
. . tani)-4e=^, ortaniJ. = j:.
Hence, r = < tan^-4.
To Lay out the Curve.
First Method. Let Dm (Fig. 53) represent a portion of the
tangent. It is required to find mP, the
perpendicular to the tangent meeting the
curve at P.
mP=Dn = CD-Ch,
But CD = r,
and
Fig. 53.
Hence,
0^ = ^01^- Pn
f.
Second Methodt It is required to find
mP (Fig. 54) in the direction of the
centre.
mP^mC-PC.
Fig. 54.
Third Method.
Fig. 55.
But mC = y/D(f+Dm^^y/7+^.
Hence,
mP = V?+?— r.
Place transits at D and E (Fig. 55). Direct
the telescope of the former
to E, and of the latter to A,
Turn each toward the curve
the same number of degrees,
and mark P, the point of
intersection of the lines of
sight. P will be a point in
the circle to which AD and
^j^are tangents at D and J?, respectively.
BAILBOAD CUBVES.
71
Fonrth Method. Measure
AI} = AE(Yig.56), Com-
pute r. Select any distance
I (the length of the chain
will be found convenient).
Place the transit at D and
direct toward A. Turn oflf
an angle ADP determined
by the equation sin ADP
the angle A and the tangents
A
= -~, and measure DP
2r
Rg. 56.
l\ make the angle ADP ^2 ADP
the angle ADP' = ^ADP B,ndi PP'
and PP = l] make
= I; etc. P, JP, P", etc., will be in the circumference of the
required arc.
AN 8 WEBS.
EZSBOIBE I.
1. 8 a. 64 p.
2. 29 A. 7| p.
8. 4 a. 5Ap.
4. 115^ p.
5. 8 a. 78 p.
6. 13 A. 6^ p.
7. 11 A. 157 p.
8. 7.51925.
EzEBOISS II.
9.
10.
11.
18.0785.
2 A. 58i p.
4 A. 35 p.
1. 2 A. 27 p.
2. 20 a. 12 p.
8. 2 a. 54 p.
4. 2 A. 152 p.
5. 8 a. 54 p.
6. 5 a. 48 p.
7. 2 A. 78 p.
8.
9.
10.
8 a. 122 p.
6 A. 2 p.
9 A. 41 p.
EZBSOISE ni.
1. 2 A. 12^ P. 2.
EXSBOISE IV.
i A. 92 p.
1.
2.
8.
4.
5.
AIJ=S.7bch.
^^=8.50 ch.;
EG = 8.42ch.
^^=4.55 ch.
AI!=5.&0cii.
CJ: =4.455 ch.
^D = 2.275 ch.;
5J7=1.82ch.
AD = 4.61 cL;
^J?= 8.608 ch.
Tlie distances on A£ are
2, 3, and 5 ch.
9. J?Jf(onJL4)=2.5087ch.
^i^(on^^) =6.489 ck
10. L6tI!0>DF,
AE = 12.24:1 ch.
AG= 9.798 ch.
AI)= 8.659 ch.
AF^ 6.928 ch.
11. Let I)G>EF,
00" = 14.862 ch.
CD = 18.118 ch.
GF= 11.404 ch.
LCJ?= 10.062 ch.
then
then
74 ANSWEBS.
EXEBCISE V.
1. 13.3 ft. 6. Difference = 42.52 ft.
2. Difference = 27.292 ft. 6. Difference = 19.14 ft.
3. Difference = 31.9 ft. T. Difference = 16.26 ft.
4. Difference = 9.21 ft. & . Difference = 23.87 ft.
66
SURVEYING.
4.
5.
6.
Back-
Sights.
Fore-
sights.
Distances,
Back-
Sights.
Fore-
Sights.
Distances.
Back-
sights.
Fore-
Sights.
Distances.
3.50
5.65
4.60
1.84
0.16
2.00
0.34
3.16
1.00
4.10
10.85
3.20
1.02
8.70
1.00
5.86
5.61
1.00
5.04
9.25
3.80
2.83
7.91
1.00
4.19
4.24
1.00
3.84
12.91
3.60
5.20
10.63
2.00
5.44
1.20
1.00
4.12
7.65
2.80
4.13
8.71
1.92
4.96
3.20
1.00
10.49
3.92
3.00
4.04
8.02
2.08
4.73
1.32
1.00
12.96
3.03
6.00
2.92
3.16
7.92
6.04
2.00
2.00
6.10
5.33
2.00
3.96
1.00
1.00
2.17
6.87
1.00
5.91
1.83
1.00
2.13
7.00
1.80
3.70
0.90
1.00
7.
8.
Back-
Sights.
Fore-
Sights.
Distances.
Back-
sights.
Fore-
Sights.
Distances.
13.71
7.88
5.19
4.47
4.53
1.00
9.40
16.30
7.96
9.22
1.00
3.87
11.71
2.27
....
5.07
1.00
2.63
12.41
3.08
....
0.24*
1.00
14.62
0.95
5.08
6.36
1.87*
1.00
17.00
1.45
3.40
6.14
1.47*
1.00
10.66
15.40
6.59
6.62
2.10
1.00
2.87
17.00
4.01
....
2.24*
0.16
3.40
10.32
2.18
10.42
4.32*
0.84
§ 36. TOPOGBAPHICAL LEVELLING.
The principal object of topographical surveying is to show
the contour of the ground. This operation, called topographi-
cal levelling, is performed by representing on paper the
curved lines in which parallel horizontal planes at uniform
distances apart would meet the surface.
It is evident that all points in the intersection of a horizontal
TOPOGEAPHIOAL LEVELLING.
67
plane with the surface of the ground are at the same level.
Hence, it is only necessary to find points at the same level,
and join these to determine a line of intersection.
The method commonly employed will be understood by a
reference to Fig. 49. The ground ABCD is divided into
equal squares, and a numbered
stake driven at each intersec-
tion. By means of a level and
levelling rod the heights of the
other stations above w and i>,
the lowest stations, are deter-
mined. A plot of the ground
with the intersecting lines is
then drawn, and the height of
each station written as in the
figure.
Suppose that the horizontal
planes are 2 feet apart ; if the
first passes through m and i>, the second will pass through p,
which is 2 feet above m ; and since n is 3 feet above w, the
second plane will cut the line mn in a point 8 determined by
the proportion mn : ms : : 3 : 2. In like manner the points ^, g,
and r are determined.
The irregular line tsp qr represents the intersection of
the second horizontal plane with the surface of the ground.
In like manner the intersections of the planes, respectively,
4, 6, and 8 feet above m, are traced.
It is evident that the more rapid the change in level the
nearer these lines will approach each other.
66
SUEVEYING.
4.
5.
6.
Back-
Sights.
Fore-
sights.
Distances.
Bacl(-
Sights.
Fore-
Sights.
Distances,
Back-
sights.
Fore-
Sights.
Distances.
3.50
5.65
4.60
1.84
0.16
2.00
0.34
3.16
1.00
4.10
10.85
3.20
1.02
8.70
1.00
5.86
5.61
1.00
5.04
9.25
3.80
2.83
7.91
1.00
4.19
4.24
1.00
3.84
12.91
3.60
5.20
10.63
2.00
5.44
1.20
1.00
4.12
7.65
2.80-
4.13
8.71
1.92
4.96
3.20
1.00
10.49
3.92
3.00
4.04
8.02
2.08
4.73
1.32
1.00
12.96
3.03
6.00
2.92
3.16
7.92
6.04
2.00
2.00
6.10
5.33
2.00
3.96
1.00
1.00
2.17
6.87
1.00
5.91
1.83
1.00
2.13
7.00
1.80
3.70
0.90
1.00
7.
8.
Back-
sights.
Fore-
Sights.
Distances.
Back-
sights.
Fore-
Sights.
Distances,
13.71
7.88
5.19
4.47
4.53
1.00
9.40
16.30
7.96
9.22
1.00
3.87
11.71
2.27
....
5.07
1.00
2.63
12.41
3.08
....
0.24*
1.00
14.62
0.95
5.08
6.36
1.87*
1.00
17.00
1.45
3.40
6.14
1.47*
1.00
10.66
15.40
6.59
6.62
2.10
1.00
2.87
17.00
4.01
2.24*
0.16
3.40
10.32
2.18
10.42
4.32*
0.84
§ 36, TOPOGEAPHICAL LEVELLING.
The principal object of topographical surveying is to show
the contour of the ground. This operation, called topographi-
cal levelling, is performed by representing on paper the
curved lines in which parallel horizontal planes at uniform
distances apart would meet the surface.
It is evident that all points in the intersection of a horizontal
TOPOGEAPHICAL LEVELLING.
67
plane with the surface of the ground are at the same level.
Hence, it is only necessary to find points at the same level,
and join these to determine a line of intersection.
The method commonly employed will be understood by a
reference to Fig. 49. The ground ABCD is divided into
equal squares, and a numbered
stake driven at each intersec-
tion. By means of a level and
levelling rod the heights of the q
other stations above m and D,
the lowest stations, are deter-
mined. A plot of the ground
with the intersecting lines is
then drawn, and the height of
each station written as in the
figure.
Suppose that the horizontal
planes are 2 feet apart ; if the
first passes through m and J9, the second will pass through p,
which is 2 feet above m ; and since w is 3 feet above m, the
second plane will cut the line mn in a point 8 determined by
the proportion mn : m« : : 8 : 2. In like manner the points t, q,
and r are determined.
The irregular line tsp qr represents the intersection of
the second horizontal plane with the surface of the ground.
In like manner the intersections of the planes, respectively,
4, 6, and 8 feet above m, are traced.
It is evident that the more rapid the change in level the
nearer these lines will approach each other.
CHAPTER V.
RAILROAD SURVEYING.
§37. General Kemabks.
When the general route of a railroad has been determined,
a middle surface line is run with the transit. A profile of this
line is determined, as in § 34. The levelling stations are com-
monly 1 chain (100 feet) apart. Places of different level are
connected by a gradient line, which intersects the perpendic-
ulars to the datum line at the levelling stations in points
determined by simple proportion. Hence, the distance of each
levelling station, above or below the level or gradient line
which represents the position of the road bed, is known.
38. Cross Section Work.
Excavations. If the road bed lies below the surface, an exca-
vation is made.
Let ACDB (Fig. 50) represent a cross section of an excava-
tion, /a point in the middle surface line, /' the corresponding
point in the road bed, and CD the width of the excavation at
the bottom. The slopes at the sides are commonly made so
SAILBOAD CUBVES.
that AA':=iA'C, and 55' = |i>J5'. //' and CD being
known, the points A, B, 0\ and i>' are readily determined by
a level and tape measure.
If from the area of the trapezoid ABB' A' the areas of the
triangles -4-4' C and BB^JD be deducted, the remainder will be
the area of the cross section.
In like manner the cross section at the next station may be
determined. These two cross sections will be the bases of a
frustum of a quadrangular pyramid whose volume will be the
amount of the excavation, approximately.
Embankments. If the road bed lies above the surfiace, an
embankment is made, the cross section of which is like that of
the excavation, but inverted.
Fig. 51.
Fig. 51 represents the cross section of an embankment
which is lettered so as to show its relation to Fig. 50.
§39. Railroad Curves.
When it is necessary to change the direction of a railroad,
it is done gradually by a
curve, usually the arc of
a circle.
Let AB and AO (Fig.
52) represent two lines to
be thus connected. Take
AD = AE=^t. The in- /
tersection of the perpen-
diculars DC and JEC de-
termines the centre (7, and
the radius of curvature DC=r.
Fig. 52.
The length of the radius
70
SURVEYING.
of curvature will depend on liie angle A and the tangent AD,
For, in the right triangle ADC, -
iwDAO^^, or tani^ = ^.
AD t
Hence, r = ^ tan ^-4.
Fint Method.
To Lay ovi the Curve.
Let Dm (Fig. 63) represent a portion of the
tangent. It is required to find mP, the
perpendicular to the tangent meeting the
curve at P.
mP=Dn = CD—Cn.
CD = r
Cn = -\^CP-Pfi^
Fig. 53.
Hence,
mP= r — Vr* — )?.
Second Method. It is required to find
mP (Fig. 54) in the direction of the
centre.
mP=mC-PC.
Fig. 54.
But mC = ^D(f+Dm^ = \^^i^+?.
Hence,
mP = V?+¥-r.
Third Method. Place transits at D and U (Fig. 65). Direct
the telescope of the former
to U, and of the latter to A.
Turn each toward the curve
the same number of degrees,
and mark P, the point of
intersection of the lines of
sight. P will be a point in
the circle to which AD and
ASeare tangents at D and J?, respectively.
Fig. 55.
KAILEOAD CUKVES. 71
FonrUi Method. Measure the angle A and the tangents
AD = AE (Fig. 56). Com- ^
pute r. Select any distance ^y^ \
I (the length of the chain n ^^^^^^'^^---JZ \
will be found convenient). ^y:^^^"" ^^^"^X
Place the transit at D and ^y/' ^
direct toward A, Turn off ' \f
an angle ADP determined V
by the equation sin-4-DP Rg. 56.
= --, and measure DP=l\ make the angle ADP = 2 ADP
2r
and PP = l\ make the angle ^i)JP" = 3 ^i)P and PP^
= 1; etc. P, Pf P\ etc., will be in the circumference of the
required arc.
yui LOOABITHMS.
19. If the given mantissa cannot be found in the table, find in the
table the two adjacent mantissas between which the given mantissa
lies, and the four figures corresponding to the smaller of these two
mantissas will be the first four significant figures of the required
number. If more than four figures are desired, they may be found by
interpolation, as in the following examples :
Find the number corresponding to the logarithm 1.48762.
Here the two ac^acent mantissas of the table, between which the given
majDtissa 48762 lies, are found to be (page 6) 48756 .and 48770. The corre-
sponding nombers are 8078 and 8074. The smaller of these, 8078, contains
the first four significant figures of the required number.
The difference between the two adjacent mantissas is 14, and the difference
between the corresponding numbers is 1.
The difference between the smaller of the two adjacent mantissas, 48756,
and the given mantissa, 48762, is 6. Therefore, the number to be annexed to
8078 is ^ of 1 = 0.428, and the fifth significant figure of the required number
is 4.
Hence, the required number is 80.784.
Find the number corresponding to the logarithm 7.82326 — 10.
The two adjacent mantissas between which 82826 lies are (page 13) 82821
and 82828. The number corresponding to the mantissa 82321 is 6656.
The difference between the two adjacent ma)itissas is 7, and the difference
between the corresponding numbers is 1.
The difference between the smaller mantissa, 82321, and the given mantissa,
82826, is 5. Therefore, the number to be annexed to 6656 is ^ of 1 = 0.7, and
the fifth significant figure of the required number is 7.
Hence, the required number is 0.0066567.
In using a five-place table the numbers corresponding to mantissas
may be carried to five significant figures, and in the first part of the
table to six figures.*
20. The logarithm of the reciprocal of a number is called the
Oologarithin of the number.
K A denote any number, then
colog-4 = log— = log 1 — log -4 (§ 3) = —log -4.
.a.
%
Hence, the cologarithm of a number is equal to the logarithm of
the number with the minus sign prefixed, which sign affects the entire
logarithm, both characteristic and mantissa.
* In most tables of logarithms proportional parts are given as an aid to
interpolation; but, after a little practice, the operation can be performed
nearly as rapidly without them. Their omission allows a page with larger-
faced type and more open spacing, and consequently less trying to the eyes.
INTRODUOTION.
IX
In order to avoid a n^atiye mantissa in the oologarithm, it is
customary to sabstitate for -~ log A its equivalent
(10-log^)-.10.
Hence, the cologarithm of a number is found by subtracting the
logarithm of the number from 10, and then annexing — 10 to the
remainder.
The best way to perform the subtraction is to begin on the left and
subtract each figure of log^ fh>m 9 until we reach the last significant
figure, which must be subtracted from 10.
K log.^ is greater in absolute value than 10 and less than 20, then
in order to avoid a negative mantissa, it is necessary to write •* log^
in the form
(20-.log^)-.20.
So that, in this case, colog^ is found by subtracting log^ from 20,
and then annexing — 20 to the remainder.
Fmd the cologarithm of 4007.
10 -10
Pages. log 4007= 8.60282
colog 4007 = 6.89718-10
Fmd the cologarithm of 103992000000.
20 -20
Page 2. log 108992000000 = 11 .01700
colog 108992000000= 8.98800-20
If the characteristic of log^ is negative, then the subtrahend, ^10
or — 20, will vanish in finding the value of colog ul.
Find the cologarithm of 0.004007.
10-10
log 0.004007= 7.60282-10
colog 0.004007= 2.39718
With practice, the cologarithm of a number can be taken from the
table as rapidly as the logarithm itself.
By using cologarithms the inconvenience of subtracting the loga-
rithm of a divisor is avoided. For dividing by a number is
equivalent to multiplying by its reciprocal. Hence, instead of
subtracting the logarithm of a divisor its cologarithm may be added.
LOGABITHMS.
Computation by Logarithms.
ZL (1) Find the value of a?, if a; = 72214 x 0.08203.
Page 14. log 72214 = 4.86862
Page 16. log 0.08203 =8.91397-10
By §2. logx =3.77269
Page 11. z = 6923.63
(2) Find the value of a, if a? = 5250 -i- 23487.
Page 10. log 6260 = 3.72016
Page 4. colog 23487 =6.62917-10
Page 4. log X = 9.34933 - 10 = log 0.22363
.-. X =0.22363
(3) Find the value ofx,\fx=. ^'^^ ^ ^^^^ ^ ^^^
^^ ' 899.1x0.00337x23435
Page 16. log 7.66 =0.87862
Page 9. log 4667 = 3.66904
Page 11. log 667 =2. 76368
Page 17. colog 899.1 = 7.04619 - 10
Page 6. colog 0.00337 =2.47237
Page 4. colog 23436 = 6.63013 — 10
Page 6. : logx =2.44983 = log 281.78
.-. X =281.73
(4) Find the cube of 376.
Page 7. log 376 =2.67619
Multiply by 3 (§ 4) , 3
Page 10. log 376= = 7.72667 = log 63168600
.-. 376' =63168600
(5) Find the square of 0.003278.
Page 6. log 0.003278= 7.61561-10
Page 2. log 0.003278« = 16.03122 - 20 = log 0.000010746
.-. 0.003278«= 0.000010746
(6) Find the square root of 8322.
Page 16. log 8322 = 3.92023
Divide by 2 (§ 6) , 2 ) 3.92023
log V8322 = 1.96012 = log 91.226
.-. V8322 =91.226
If the given number is a proper fraction, its logarithm will have as
a subtrahend 10 or a multiple of 10. In this case, before dividing the
logarithm by the index of the root, both the subtrahend and the num-
INTBODUCTION.
ber preceding the mantissa should be increased by such a number as
will make the subtrahend, when divided by the index of the root,
10 or a multiple of 10.
(7) Find the square root of 0.000043641.
Pages, log 0.000048641 = 5.63989-10
10 -10
Divide by 2 (§ By, 2 )15.63989 -"JO
Page 13. log V0.00004"364l = 7.81995- 10 = log 0.0066062
.-. V0.000043641= 0.0066062
(8) Find the sixth root of 0.076553.
Page 15. log 0.076568 = 8.88397-10
50 -50
Divide by 6 (§ 6>j 6 )68.88397-60
Page 13 log v^ O.076668 = 9.81400 - 10 = log 0.66168
.-. \^0.076663 = 0.66168
TABLE II.
22. This table (page 20) contains the value of the number tt, its
most useful combinations, and their logarithms.
Find the length of an arc of 47** 32' 57" in a unit circle.
47° 32' 67" = 171177"
log 171177 = 5.23344
log— =4.68557-10
a"
log arc 47® 32' 57" = 9.91901 - 10 = log 0.82994
. \ length of arc = 0.82994
Find the angle if the length of its arc in a unit circle = 0.54936.
log 0.54936 = 9.73986 - 10
cologi-=loga" =5.31443
log angle = 5.05429 = log 113316
.-. angle = 113316"= 31^28' 36"
23. The relations between arcs and angles given in Table II. are
readily deduced from the circular measure of an angle.
In Circular Measure an angle is defined by the equation
angle = -^,
radius
in which the word arc denotes the length of the arc corresponding to
the angle, when both arc and radius are expressed in terms of the
same linear unit.
XU LOGARITHMS.
Sinoe the arc and radios for a given angle in different circles vary
in tlie same ratio, the value of the angle given by this equation is
independent of the value of the radius. If the radius is unity, the
equation defining the angle becomes
angle = arc.
That is, in circular measure an angle is measured by the length of its
arc in a unit circle. Therefore,
If the arc = circumference, the angle = 2ir.
If the arc = semicircumference, the angle = «-.
If the arc =» quadrant, the angle ^iw.
If the arc = radius (= 1), the angle = 1 ;
that is, in circular measure the angular unit is the angle whose arc is
equal in length to the radius of the circle.
Since 180^ in common measure equals v units in circular measure,
therefore
1° in common measure = r^ units in circular measure ;
180®
1 unit in circular measure = in common measure.
TT
By means of these two equations, the value of an angle expressed
in one measure may be changed to its value in the other measure.
Thus, the angle whose arc is equal to the radius is an angle of
180°
1 unit in circular measure, and is equal to , or 57° 17' 45'', very
nearly.
TABLB III.
24. This table (pp. 21-49) contains the logarithms of the trigo-
nometric functions of angles. In order to avoid negative character-
istics, the characteristic of every logarithm is printed 10 too lai^e.
Therefore, — 10 is to be annexed to each logarithm.
On pages 28-49 the characteristic remains the same throughout
each column, and is printed at the top and the bottom of the column.
But on page 30 the characteristic changes one unit in value at the
places marked with bars. Above these bars the proper characteristic
is printed at the top, and below them at the bottom, of the column.
25. On pages 28-49 the log sin, log tan, log cot, and log cos, of 1°
to 89°, are given to every minute. Conversely, this part of the table
gives the value of the angle to the nearest minute when log sin, log tan,
log cot, or log cos is known, provided log sin or log cos lies between
8.23822 and 9.99992, and log tan or log cot lies between 8.23829 and
11.76171.
IHTBODXrCTION. XIU
If the exact value of the given li^arithm of a ftinction is not found
in the table, the value nearest to it is to be taken, unless interpolation
is employed as explained in § 26.
If the angle is less than 45^, the number of degrees is printed at
the top of the page, and the number of minute3 in the column to the
left of the columns containing the logarithm. If the angle is greater
than 45^, the number of degrees is printed at the bottom of the page,
and the number of minutes in the column to the right of the columns
containing the logarithms.
K the angle is less than 45°, the names of its fhnctions are printed
at the top of the page ; if greater than 45°, at the bottom of the
page. Thus,
Page 88. log sin 21« 37' = 9.66681 — 10.
Page 45. log cot BB^ 68' = 10.12473 - 10 = 0.18478.
Page 87. logcos69°H'= 9.64969-10.
Page 49. log tan 46<' 69' = 10.01491 - 10 = 0.01491.
Page 48. If log cos == 9.87468 — 10, angle = 41<' 28'.
Page 84. If log cot = 9.39368 — 10, angle := 76'' 6'.
Iflog Shi = 9.47760 — 10, the nearest log sin In the table is 9.47774 — 10
(page 86), and the angle corresponding to this value is 17*^ 29'.
If logtan = 0.76620= 10.76620 — 10, the nearest log tan in the table is
10.76490 — 10 (page 82), and the angle corresponding to this value is 80® 16'.
26. If it is desired to obtain the logarithms of the functions of
angles that contain seconds, or to obtain the value of the angle in
degrees, minutes, and seconds, firom the logarithms of its functions,
interpolation must be employed. Here it must be remembered that.
The difference between two consecutive angles in the table is 60".
Log sin and log tan increase as the angle increases; log cos and
log cot diminish as the angle increases.
Find log tan 70° 46' 8".
Page 87. log tan 70*» 46' = 0.46731.
The diiTerenee hetween the mantissas of log tan 70® 46' and log tan 70® 47'
is41, and^of 41 = 6.
As the ftmction is increasing, the 6 most be added to the figure In the fifth
place of the mantissa 46731 ; and
Therefore log tan 70® 46' 8" = 0.46786.
Fmd log cos 47** 35' 4".
Page 48. log cos 47® 86' = 9.82899 - 10.
The diiTerenee between this mantissa and the mantissas of the next log cos
is 14, and ^ of 14 = 1.
As the function is decreasing, the 1 mnst be subtracted firom the figure in
the fifth place of the mantissa 82899 ; and
Therefore log cos 47® 86' 4" = 9.82898 - 10.
XIV LOGARITHMS.
Find the angle for which log sin = 9.45359 — 10.
Page 35. The mantissa of the nearest smaller log sin in the table is 45334.
The angle corresponding to this value is 16° 30'.
The difference between 45334 and the given mantissa, 45359, is 25.
The difference between 45334 and the next following mantissa, 45377, is
43, and 11 of 60" = 35".
As the ftmction is increasing, the 35" must be added to 16° 30' ; and the
required angle is 16° 30' 35".
Find the angle for which log cot = 0.73478.
Page 32. The mantissa of the nearest smaller log cot in the table is 73415.
The angle corresponding to this value is 10° 27'.
The difference between 73415 and the given mantissa is 63.
The difference between 73415 and next following mantissa is 71, and
^of60" = 53".
As the fhnction is decreasing, the 53" must be subtracted firom 10° 27' ;
and the required angle is 10° 26' 7".
27. If log sec or log esc of an angle is desired, it may be found
from the table by the formulas,
sec A =s 7 ; hence, log sec A = colog cos A.
cos A
CSC A = -: — 7 ; hence, log esc A = colog sin A,
sm-d.
Page 31. log sec 8° 28' = colog cos 8° 28' = 0.00476.
Page 42. log esc 59° 36' 44" = colog sin 59° 36' 44" = 0.06418.
2& K a given angle is between 0° and 1®, or between 89** and 90** ;
or, conversely, if a given log sin or log cos does inot lie between the
limits 8.23822 and 9.99992 in the table; or, if a given log tan or
log cot does not lie between the limits 8.23829 and 11.76171 in the
table ; then pages 21-24 of Table III. must be used.
On page 21, log sin of angles between 0° and 0° 3', or log cos of the
complementar}'^ angles between 89® 57' and 90**, are given to every
second; for the angles between 0** and 0*'3', log tan = log sin, and
log cos = 0.00000 ; for the angles between 89** 57' and 90**, log cot =
log cos, and log sin = 0.00000.
On pages 22-24, log sin, log tan, and log cos of angles between 0**
and 1**, or log cos, log cot, and log sin of the complementary angles
between 89** and 90**, are given to every 10".
Whenever log tan or log cot is not given, they may be found by the
formulas,
log tan = colog cot. log cot = colog tan.
Conversely, if a given log tan or log cot is not contained in the
table, then the colog must be found; this will be the log cot or
log tan, as the case may be, and will be contained in the table.
INTBODUCTION.
XV
On pages 25-27 the logarithms of the functions of angles between
1° and 2°, or between 88° and 90°, are given in the manner employed
on pages 22-24. These pages should be used if the angle lies between
these limits, and if not only degrees and minutes, but degrees, min-
utes^ and multiples of 10" are given or required.
When the angle is between 0° and 2°, or 88° and 90°, and a greater
degree of accuracy is desired than that given by the table, interpor
lation may be employed ; but for these angles interpolation does not
always give true results, and it is better to use Table IV.
Find log tan 0° 2' 47", and log cos 89° 37' 20".
Page 21. Iogtan0°2'47" =logshiO° 2' 47" = 6.90829-10.
Page 23. log cos 89° 87' 20" = 7.81911 - 10.
Find log cot 0° 2' 15".
10 -10
Page 21. log tan 0° 2' 16" = 6.81591-10
Therefore, log cot 0<> 2' 15" = 8.18409
Find log tan 89° 38' 30".
10 -10
Page 23. log cot 89° 88' 80" = 7.79617-10
Therefore, log tan 89° 88' 80"= 2.20383
Find the angle for which log tan = 6.92090 — 10.
Page 21. The nearest log tan is 6.92110 — 10.
The corresponding angle for which Is 0° 2' 62".
Find the angle for which log cos = 7.70240 -- 10.
iPage 22. The nearest log cos is 7.70261 — 10.
The corresponding angle for which is 89° 42' 40".
Find the angle for which log cot = 2.37368.
This log cot is not contained in the table.
The cologcot = 7.62632 — 10 = log tan.
The log tan in the table nearest to this is (page 22) 7.62510—10, and the
angle corresponding to this value of log tan is 0° 14' 30".
29. K an angle x is between 90° and 360°, it follows, from formu-
las established in Trigonometry, that,
between 90° and 180°,
log sin X = log sin (180° — x) ,
log cos X = log cos (180° — aj)„,
log tan X = log tan ( 1 80° — x)^,
log cot X = log cot (180° — a;)n ;
between 180° and 270°,
log sin X = log sin (x — 180°)^,
log cos X = log cos (x — 180°)„,
log tan X = log tan (x — 180°) ,
log cot X = log cot {x — 180°) ;
ZYl LOOABITHMS.
between 270° and SeO*",
log sin a; S3 log sin (860° — a?),,
log cos » =s logcos (360° — oj) ,
logtana? =: logtan (360° — a;).,
log cot » = log cot (360* — x)^.
The letter n is placed (according to custom) after the logarithms
of those ftmctions which are n^^tive in value.
The above formulas show, without farther explanation, how to find
by means of Table III. the logarithms of the functions of any angle
between 90° and 360°.
Thus, logsinlS7»46'22" = log8in 42* 14' 38" = 9.82766 -10.
log cos 187® 46' 22" = log, cos 42« 14' 88" = 9.86940, - 10.
logtan 137" 46' 22" = log, tan 42« 14' 88" = 9.96816, - 10.
log cot 187*» 46' 22" = log, cot 42«> 14' 88" = 0.04186,.
logsin 209*» 82' 60" = log, sin 29» 82' 60'' = 9.69297, - 10.
logcos880O27'10"slogcos 29'' 82' 60" = 9.98949 -10.
Conversely, to a given logarithm of a trigonometric function there
correspond between 0° and 360° four angles, one angle in each quad-
rant, and so related that if x denote the acute angle, the other three
angles are 180°— a;, 180°+fl?9 and 860°— a?.
If besides the given logarithm it is known whether the function is
positive or n^^tive, the ambiguity is confined to two quadrants,
therefore to two angles.
Thns, if the log tan = 9.47461 - 10, the angles are IS'' 86' 17" in Quadrant I.
and 196'' 86' 17" in Quadrant III. ; but if the log tan = 9.47461, - 10, the angles
are 168*' 28' 48" in Quadrant II. and 848o 28' 43" in Quadrant IV.
To remove all ambiguity, further conditions are required, or a
knowledge of the special circumstances connected with the problem in
question.
TABLE IV.
30. This table (page 50) must be used when great accuracy is
desired in working with angles between 0° and 2°, or between 88°
and 90°.
The values of S and T are such that when the angle a is expressed
in seconds,
S = log sina — log a",
T = log tan a — log a".
Hence follow the formulas given on page 50.
The values of S and T are printed with the characteristic 10 too
large, and in using them —10 must always be annexed.
INTEODirOTIOK.
XVU
Find log sin 0*^58' 17".
0« 68' 17" = 8497"
log 8497 =8.64370
8=4.68666-10
log sin 0"* 68' 17" = 8.22926 - 10
Find log tan 0** 52' 47.5".
0^62' 47.6" = 8167.6"
log8167.6 = 8.60072
T = 4.68661 -10
logtan 0» 52' 47.6" = 8.18688 - 10
Find log COS 88** 26' 41. 2".
900-88* 26' 41.2" = 1° 88' 18.8"
= 6698.8"
log 6698.8 = 8.74809
8 = 4.68662-10
log cos 88» 26' 41.2" = 8.48861 - 10
Find logtan 89** 54' 37.362".
90* — 89*> 64' 87.862" = 0« 6' 22.688"
= 822.688"
log 822.638 = 2.60871
T = 4.68668 — 10
log cot 89«» 64' 87.862" = 7. 19429 - 10
log tan 89*' 64' 87.862" = 2.80671
Find the angle, if log sin =s 6.72306—10.
6.72806-10
: 4.68667 — 10
8 =
8abtract,
2.08749 = log 109.015
109.016" =0*»1' 49.016"
Find the angle for which log cot = 1.67604.
: 8.32896 - 10
: 4.68664 - 10
cologcot =
T =
Subtract,
8.68882 = log 4848.8
4848.8" = V 12' 28.8"
Find the angle for which logtan = 1.55407.
cologtan = 8.44598 — 10
T = 4.68669 - 10
Subtract, 8.76024 = log 6767.6
6767.6" = 1° 36' 67.6",
and 90' - 1° 35' 67.6" = 88*» 24' 2.4".
Therefore, the angle reqaired is 88® 24' 2.4".
TABLE V.
3L Table Y. (pp. 51-53), contains the natural sines, cosines,
tangents, and cotangents of angles from 0° to 90^, at intervals of 10^
If greater accuracy is desired it maj' be obtained by intcipolation.
TABLE VI.
32. This table (p. 54), containing the circumferences and areas
of circles, does not require explanation.
Note. In preparing the preceding explanations, firee use has been made
of the Logarithmic Tables by F. G. Gaoss, from which, also, Tables II. and
YI. have been taken.
xvm
LOGARITHMS.
TABLE VII.
33. Tliis table (pp. 55-60) gives the latitude and departure to
three places of decimals for distances from 1 to 10, corresponding
to bearings from C" to 90"^ at intervals of 15'.
If the bearing does not exceed 45^ it is found in the lefi-hand
column, and the designations of the columns under ^^ Distance" are
taken from the top of the page ; but if the bearing exceeds 45^, it is
found in the n^^-hand column, and the designations of the columns
under '' Distance" are taken from the bottom of the page.
The method of using the table will be made plain by the following
examples : —
(1) Let it be required to find the latitude and departure of the
course N. 85** 15' E. 6 chains.
On p. 60, left-hand column, look for 85^ 15' ; opposite this bearing, in the
vertical colonm headed '' Distance 6," are found 4.900 and 3.463 under the
headings ** Latitude*' and ^'Departure" respectively. Hence, latitude or
northing = 4.900 chains, and departure or easting = 3.463 chains.
(2) Let it be required to find the latitude and departure of the
course S. 87** W. 2 chains.
As the bearing exceeds 45*^, we look in the right-hand column of p. 55, and
opposite 87^ in the column marked *' Distance 2 " we find (taking the designa-
tions of the columns from the bottom of the page) latitude = . 105 chains,
and departure = 1.997 chains. Hence, latitude or southing = .105 chains, and
departure or westing = 1.997 chains.
(3) Let it be required to find the latitude and departure of the
course N. 15** 45' W. 27.36 chains.
In this case we find the required numbers for each figure of the distance
separately, arranging the work as in the following table. In practice, only
the last columns under '* Latitude " and *' Departure " are written.
Distance.
Latitude.
Departure.
20 =2X10
7
0.3 =3-T-10
0.06 = 6^100
1.925X10 =19.25
6.737
2.887 -r- 10 = 0.289
5.775-5-100= 0.058
0.543X10 =5.48
1.90
0.814 -r- 10 =0.081
1.628 -7- 100 = 0.016
27.36
26.334
7.427
Hence, latitude = 26.334 chains, and departure = 7.427 chains.
itABLE I.
THK
COMMON OB BRIGGS LOGARITHMS
OF THB
NATXJEAL NUMBEES
From 1 to 10000.
>
1-100
V
log
V
log
V
log
V
log
V
log
1
2
3
4
6
6
7
8
9
10
11
12
13
14
15
16
17
Id
19
20
0.00000
0.30103
0.47 712
0.60206
0.69897
0.77815
0.84 510
0.90309
0.95424
1.00000
1.04139
1.07918
1. 11 394
1.14613
1.17609
1.20412
1.23 045
1.25 527
1.27 875
1.30103
21
22
23
24
26
26
27
28
29
SO
31
32
33
34
36
36
37
38
39
40
1.32222
1.34242
1.36173
1.38021
1.39 794
1.41497
1.43136
1.44 716
1.46240
1.47 712
1.49136
1.50 515
1. 51 851
1.53148
1.54407
1. 55 630
1.56820
1.57978
1.59106
1.60 206
41
42
48
44
46
46
47
48
49
60
61
62
63
54
55
56
57
58
69
60
1.61278
1.62 325
1.63 347
1.64345
1.65 321
1.66276
1.67 210
1. 68 124
1.69020
1.69897
1.70 757
1.71600
1.72 428
1. 73 239
1.74036
1.74819
1. 75 587
1.76343
1.77085
1.77815
61
62
63
64
66
66
67
68
69
70
71
72
73
74
76
76
77
78
79
80
1.78533
1.79239
1.79934
1.80618
1.81291
1.81954
1.82607
1.83 251
1.83885
1.84510
1.85126
1. 85 733
1.86332
1.86923
1. 87 506
1.88081
1.88649
1.89 209
1.89 763
1.90309
81
82
83
84
86
86
87
88
89
90
91
92
93
94
95
96-
97
98
99
100
1.90849
1.91381
1.91908
1.92428
1.92942
1.93 450
1.93 952
1.94448
1.94939
1.95 424
1.95 904
1.96379
1.96848
1.97 313
1. 97 772
1.98 227
1.98677
1.99123
1.99564
2.00000
H
log
S
log
S
bg
K
log
S
log
1-100
100-160
N
1 2
8
4
5
6
7
8
•
100
00000
00043 00087
00130
00173
00217
00260
00303
00346
00389 1
101
00432
00475 00 518
00561
00604
00647 00689 00732 00775 00817 |
102
00860
00903 00945
00988 01030
01072
01115
01157
01199
01242 1
103
01284
01326 01368 01410
01452
01494
01536
01578
01620
01662 1
104
01703
01745 01787 01828 01870
01912
01953
01.995
02036 02078 1
106
02119 02160 02202
02 243
02 284
02325
02366
02407 02449
02490 1
106
02531
02 572 02612
02 653
02 694
02 735
02 776 02816 02857 02898 |
107
02938 02979 03019 03060 03100
03141
03181
03 222
03 262
03302 1
108
03342
03383 03423
03 463
03 503
03 543
03 583
03623
03 663
03 703 1
109
03 743
03 782 03 822
03 862
03 902
03941
03981
04021
04060
04100 1
110
04139 04179 0f2ia 04258 04 297
04336 04376 04415 04454 04493 |
111
04 532 04 571 04610 04650 04689
04 727 04 766 04805
04844
04883 1
112
04922
04961 04999
05038
05 077
05115
05154
05192
05 231
05 269 1
113
05308 05346 05385 05 423 05461
05 500 05 538 05 576 05 614 05 652 |
lU
05 690 05 729 05 767 05 805
05 843
05 881
05 918
05 956
05994 06032 1
116
06070 06108 06145 06183 06 221
06258 06296 06333 06371 06408 |
116
06446 06483*06521
06 558
06 595
06633
06670 06707 06744 06781 |
117
06819 06856 06893
06930
06 %7
07004
07041
07078 07115 07151 |
118
07188 07225 07262 07298 07335
07372
07408
07445
07482
07 518 1
119
07555 07 591 07628 07664 07 700
07737 07773 07809 07846 07882 |
120'
07918
07954 07990 08027
08063
08099 08135 08171
08207
08243 1
121
08279 08314 08350 08386 08422
08458 08493 08529 08565 08600 |
122
08636
08672 08 707
08 743
08 778
08814
08849 08884
06920
08955 1
123
08991
09026 09061
09096 09132
09167 09202 09237 09272 09307 |
124
09342 09377 09412 09447
09482
09517 09 552 09587 09621
09656
126
09691
09726 09760 09795
09830
09864
09899
09934
09968
10003
126
10037
10072 10106
10140
10175
10209
10243
10278
10312
10346
127
10380 10415 10449
10483
10 517
10551
10585
10619
10 653
10687
128
10 721
10755 10789
10823
10857
10890
10924
10958
10992
11025
129
11059
11093 11126
11160
11193
11227
11261
11294
11327
11361
180
11394
11428 11461
11494
11528
11561
11594
11628
11661
11694
131
11727
11760 11793
11826
11860
11893
11926
11959
11992
12024
132
12057
12090 12123
12156
12189
12 222
12 254
12 287
12 320
12352
133
12 385
12418 12450
12483
12 516
12 548
12 581
12 613
12 646
12678
134
12 710
12 743 12 775
12 808
12840
12 872
12 905
12 937
12 969
13 001
136
13 033
13066 13098
13130
13162
13194
13 226
13 258
13 290
13 322
136
13 354
13 386 13 418
13 450
13 481
13 513
13 545
13 577
13609
13 640
137
13672
13 704 13 735
13 767
13 799
13 830
13 867.
13 893
13925
13956
138
13988
14019 14051
14082
14114
14145
14176
14208
14239
14270
139
14301
14333 14364
14395
14426
14457
14489
14 520
14551
14 582
140
14613
14644 14675
14 706
14 737
14 768
14 799
14829
1486P
14891
141
14922
14953 14983
15 014
15 045
15 076
15106
15137
15168
15198
142
15 229
15 259 15 290
15 320
15 351
15 381
15 412
15 442
15 473
15 503
143
15 534
15 564 15 594
15 625
15 655
15 685
15 715
15 746
15 776
15 806
144
15836
15866 15 897
15 927
15 957
15 987
16017
16047
16077
16107
146
16137
16167 16197
16227
16256
16286
16316
16346
16376
16406
146
16435
16465 16495
16524
16 554
16584
16613
16643
16673
16 702
147
16 732
16 761 16 791
16 820
16850
16879
16909
16938
16967
16997
148
17026
17056 17085
17114
17143
17173
17202
17 231
17260
17289
149
17319
17348 17377
17406
17435
17464
17493
17 522
17 551
17580
150
17609
17638 17667
17 696
17 725
17 754
17 782
17811
17840
17869
N
1 2
8
4
5
6
7
8
100-160
160-200 8
N
12
8
4
5 6 7 8 9
160
17609 17638 17667
17696
17 725
17 754 17 782 17811 17840 17869
161
17898 17926 17955
17984
18013
18041 18070 18099 18127 18156
152
18184 18213 18 241
18270
18 298
18327 18355 18384 18412 18441
163
18469 18498 18 526
18 554
18 583
18611 18639 18667 18696 18 724
164
18 752 18 780 18806
18837
18865
18893 18921 18949 18977 19005
166
19033 19061 19089
19 117
19145
19173 19201 19229 19 257 19285
166
19312 19340 19368
193%
19424
19451 19479 19 507 19535 19 562
167
19590 19618 19645
19673
19700
19 728 19 756 19 783 19811 19838
168
19866 19893 19921
19948
19976
20003 20030 20058 20085 20112
16»
20140 20167 20194
20222
20249
20276 20303 20330 20358 20385
160
20412 20439 20466
20493
20 520
20548 20575 20602 20629 20656
161
20683 20 710 20 737
20 763
20 790
20817 20844 20871 20898 20925
162
20952 20978 21005
21032
21059
21085 21112 21139 21165 21192
163
21219 21245 21272
21299
21325
21352 21378 21405 21431 21458
164
21484 21511 21537
21564
21590
21617 21643 21669 21696 21722
166
21748 2177i 21801
21827
21854
21880 21906 21932 21958 21985
166
22011 22037 22063
22089
22115
22141 22167 22194 22 220 22 246
167
22 272 22298 22324
22 350 22 376
22401 22427 22453 22 479 22 505
168
22 531 22 557 22 583
22 608
22634
22660 22686 22712 22737 22763
169
22 789 22 814 22 840
22 866
22891
22917 22943 22968 22994 23019
170
23045 23070 23096
23121
23147
23172 23198 23 223 23249 23 274
171
23300 23325 23350
23376
23 401
23426 23452 23477 23502 23528
172
23553 23578 23603
23 629
23 654
23679 23 704 23 729 23 754 23779
173
2380i 23830 2385i
23880 23905
23930 23955 23980 24005 24030
174
24055 24080 24105
24130
24155
24180 24 204 24229 24254 24279
176
24304 24329 24353
24378
24403
24428 24452 24 477 24 502 24 527
176
24551 24576 24 601
24625
24650
24 674 24699 24 724 24 748 24 773
177
24 797 24822 24846
24871
24895
24920 24 9H 24969 24993 25 018
178
25042 25066 25091
25115
25139
25164 25188 25 212 25 237 25 261
179
25 285 25 310 25 334
25 358
25 382
25406 25 431 25 455 25 479 25 503
180
25 527 25 551 25 575
25 600
25 624
25 648 25 672 25 6% 25 720 25 744
181
25 768 25 792 25 816
25 840
25 864
25 888 25912 25935 25 959 25 983
182
26007 26031 26055
26079
26102
26126 26150 26174 26198 26 221
183
26245 26269 26293
26316
26340
26364 26387 26411 26435 26458
184
26482 26505 26529
26 553
26576
26600 26623 26647 26670 26694
186
26 717 26 741 26764
26 788
26811
26834 26858 26881 26905 26928
186.
26*951 26975 26998
27021
27045
27068 27091 27114 27138 27161
187
27184 27 207 27231
27254
27277
27300 27323 27346 27370 27393
188
27416 27439 27462
27485
27508
27531 27554 27577 27600 27 623
189
27646 27669 27692
27 715
27 738
27 761 27 784 27807 27830 27852
190
27875 27898 27921
27944
27967
27989 28012 28035 28058 28081
191
28103 28126 28149
28171
28194
28217 28240 28262 28285 28307
192
28330 28353 28375
28398
28421
28443 28466 28488 28 511 28 533
193
28 556 28 578 28601
28 623
28646
28668 28691 28 713 28 735 28 758
194
28 780 28803 28825
28847
28870
28 892 28914 28937 28959 28981
195
29003 29026 29048
29070
29092
29115 29137 29159 29181 29203
196
29226 29248 29270
29292
29314
29 336 29 358 29380 29403 29425
197
29447 29469 29491
29 513
29535
29557 29 579 29601 29623 29645
198
29667 29688 29 710
29 732
29 754
29776 29798 29820 29842 29863
199
29885 29907 29929
29951
29973
29994 30016 30038 30060 30081
200
30103 30125 30146
30168
30190
30211 30233 30255 30276 30298
X
12
3
4
5 6 7 8 9
160-200
200-260
N
12 8 4
5 6 7 8 9
200
30103 30125 30146 30168 30190
30211 30233 30255 30276 30296
201
30320 30341 30363 30384 30406
30428 30449 30471 30492 30 514
302
30535 30557 30578 30600 30621
30643 30664 30685 30707 30728
203
30 750 30771 30 792 30814 30835
30856 30878 30899 30920 30942
204
30963 30984 31006 31027 31048
31 069 31 091 31 112 31 133 31 154
206
31175 31197 31218 31239 31260
31281 31302 31323 31345 31366
206
31387 31408 31429 31450 31471
31492 31513 31534 31555 31576
207
31597 31618 31639 31660 31681
31702 31723 31744 31765 31785
208
31806 31827 31848 31869 31890
31911 31931 31952 31973 31994
209
32015 32035 32056 32077 32098
32118 32139 32160 32181 32 201
210
32222 32243 32263 32284 32305
32325 32346 32366 32387 32406
211
32428 32449 32469 32490 32 510
32 531 32 552 32 572 32 593 32 613
212
32 634 32654 32675 32 695 32 715
32 736 32 756 32 777 32 797 32818
213
32838 32 858 32 879 32899 32919
32940 32960 32980 33001 33021
2U
33041 33 062 33082 33102 33122
33143 33163 33183 33 203 33 224
215
33244 33264 33284 33304 33325
33345 33365 33385 33405 33 42S
216
33445 33465 33486 33506 33526
33 546 33566 33 586 33606 33626
217
33 646 33666 33686 33 706 33 726
33.746 33 766 33 786 33 806 33 826
218
33 846 33 866 33 885 33 905 33 925
33945 339iS 33985 34005 34025
219
34044 34064 34084 34104 34124
34143 34163 34183 34203 34 223
220
34 242 34 262 34 282 34301 34321
34341 34361 34380 34400 34420
221
34439 34459 34479 34498 34 518
34 537 34 557 34 577 34 596 34616
222
34635 34655 34674 34694 34 713
34 733 34 753 34 772 34 792 34811
223
34830 34850 34869 34889 34908
34928 34947 34967 34986 35 005
224
35025 35044 35064 35083 35102
35122 35141 35160 35180 35199
226
35 218 35238 35257 35 276 35 295
35 315 35 334 35 353 35 372 35 392
226
35 411 35 430 35 449 35 468 35 488
35 507 35 526 35 545 35 564 35 583
227
35 603 35 622 35 641 35 660 35 679
35 698 35 717 35 736 35 755 35 774
228
35 793 35 813 35 832 35851 35 870
35889 35908 35 927 35 946 35965
229
35 984 36003 36021 36040 36059
36078 36097 36116 36135 36154
230
36173 36192 36211 36229 36248
36267 36286 36305 36324 36342
231
36361 36380 36399 36418 36436
36455 36474 36493 36511 36 530
232
36 549 36 568 36586 36605 36624
36642 36661 36680 36698 36 717
233
36 736 36 754 36773 36 791 36810
36829 36847 36866 36884 36903
234
36922 36940 36959 36977 36996
37014 37033 37051 37070 37088
236
37107 37125 37144 37162 37181
37199 37218 37236 37 254 37 273
286
37291 37310 37328 37346 37365
37383 37401 37420 37438 37457
237
37475 37493 37 511 37 530 37 548
37 566 37 585 37603 37621 37639
238
37658 37676 37694 37 712 37 731
37 749 37 767 37 785 37 803 37822
238
37840 37858 37876 37 894 37912
37931 37949 37 %7 37985 38 003
240
38021 38039 38057 38075 38093
38112 38130 38148 38166 38184
241
38202 38220 38 238 38256 38274
38292 38310 38328 38346 38364
342
38382 38399 38417 38435 38453
38471 38489 38507 38525 38 543
243
38561 38578 38 596 38614 38632
38650 38668 38686 38 703 38 721
244
38739 38 757 38775 38 792 38810
38828 38846 38863 38881 38899
246
38917 38934 38952 38970 38987
39005 39023 39041 39058 39076
246
39094 39111 39129 39146 39164
39182 39199 39217 39235 39252
247
39270 39 287 39305 39322 39340
39358 39375 39393 39410 39428
248
39445 39463 39480 39498 39 515
39 533 39 550 39 568 39 585 39602
249
39620 39637 39655 39672 39690
39 707 39 724 39 742 39 759 39 777
250
39794 39811 39829 39846 39863
39881 39898 39915 39933 39950
N
12 8 4
5 6 7 8 9
200-260
260-300 «
N
O 1 2 8 4
5 6 7 8 9
2S0
3979^ 39811 39829 39846 39863
39881 39898 39915 39933 39950
351
39967 3998S 4O0Q2 40019 40037
40054 40071 40088 40106 40123
25S
40140 40157 40175 40192 40209
40226 40243 40261 40278 40295
263
40312 40329 40346 40364 40381
40398 40415 40432 40449 40466
264
40483 40500 40518 40535 40552
40569 40586 40603 40620 40637
266
40654 40671 40688 40705 40 722
40739 40756 40773 40790 40807
266
40824 40841 40858 40875 40892
40909 40926 40943 40960 40976
267
40993 41010 41027 41044 41061
41 078 41 095 41 111 41 128 41 145
368
41162 41179 41196 41212 41229
41246 41263 41280 41296 41313
260
41330 41347 41363 41380 41397
41 414 41 430 41 447 41 464 41 481
260
41497 41514 41531 41547 41564
41581 41597 41614 41631 41647
361
41664 41681 41697 41714 41731
41 747 41 764 41 780 41 797 41 814
263
41830 41847 41863 41880 41896
41913 41929 41946 41963 41979
368
41996 42012 42029 42045 42062
42078 42095 42111 42127 42144
364
42160 42177 42193 42 210 42 226
42 243 42 259 42 275 42 292 42 308
266
42325 42341 42357 42374 42 390
42406 42423 42439 42455 42472
866
42 488 42 504 42 521 42 537 42 553
42570 42586 42602 42619 42635
867
42651 42667 42684 42 700 42 716
42 732 42 749 42 765 42 781 42 797
368
42813 42830 42846 42862 42878
42894 42911 42927 42943 42959
360
42975 42991 43008 43024 43040
43056 43072 43088 43104 43120
270
43136 43152 43169 43185 43 201
43217 43233 43249 43265 43281
371
43297 43313 43329 43345 43361
43377 43393 43409 43425 43441
378
43457 43473 43489 43 505 43 521
43 537 43 553 43 569 43 584 43600
878
43616 43632 43648 43664 43680
43696 43 712 43 727 43 743 43 759
874
43 775 43 791 43807 43823 43838
43854 43870 43886 43902 43917
376
43933 43949 43%5 43981 43996
44012 44028 44044 44059 44075
376
44091 44107 44122 44138 44154
44170 44185 44201 44 217 44232
877
44248 44264 44279 44 295 44311
44326 44342 44358 44373 44389
378
44404 44420 44436 44451 44467
44483 44498 44514 44 529 44 545
378
44560 44 576 44 592 44607 44623
44638 44654 44669 44685 44 700
S80
44 716 44 731 44 747 44 762 44 778
44793 44809 44824 44840 44855
381
44871 44886 44902 44917 44932
44948 44963 44979 44994 45010
888
45025 45040 45056 45071 45086
45 102 45 117 45 133 45 148 45 163
383
45 179 45 194 45 209 45 225 45 240
45255 45 271 45 286 45 301 45 317
384
45 332 45 347 45 362 45 378 45 393
45408 45423 45439 45454 45469
386
45484 45 500 45 515 45 530 45 545
45 561 45 576 45 591 45606 45 621
386
45 637 45 652 45667 45 682 45697
45 712 45 728 45 743 45 758 45 773
387
45 788 45803 45818 45834 45849
45864 45879 45894 45909 45 924
888
45 939 45954 45969 45984 46000
46015 46030 46045 46060 46075
389
46090 46105 46120 46135 46150
46165 46180 46195 46210 46225
200
46240 46255 46270 46285 46300
46315 46330 46345 46359 46374
891
46389 46404 46419 46434 46449
46464 46479 46494 46509 46523
392
46538 46553 46568 46583 46598
46613 46627 46642 46657 46672
303
46687 46702 46716 46 731 46746
46761 46776 46790 46805 46820
394
46835 46850 46864 46879 46894
46909 46923 46938 46953 46967
396
46982 46997 47012 47026 47 041
47056 47070 47085 47100 47114
396
47129 47144 47159 47173 47188
47202 47217 47232 47246 47 261
397
47276 47290 47305 47319 47334
47349 47363 47378 47392 47407
398
47422 47436 47451 47465 47 480
47494 47 509 47 524 47 538 47 553
899
47 567 47582 47 596 47611 47625
47640 47654 47669 47683 47698
800
47 712 47 727 47741 47 756 47 770
47 784 47 799 47813 47828 47842
N
O 1 2 3 4
5 6 7 8 9
260-2
100
6
300-360
N
O
1 2
8
4
5
« 7 8 1
300
47 712
47 727 47 741
47 756 47 770
47 784 47799 47813 47 828 47 »4^ 1
301
47 857
47 871 47885
47900
47914
47929 47943 47958 47972 47986 |
302
48001
48015 48029 48044 48058
48073
48087 48101 48116 48130 1
303
48144
48159 48173
48187
48202
48216 48230 48244 48259 48273 |
304
48287 48302 48316 48330 48344
48359
48373 48387 48401 48416
306
48430 48444 48458 48473
48487
48501
48515 48530 48544 48 558
806
48 572
48586 48601
48615
48629
48643
48657 48671 48686 48 700
307
48 714 48 728 48 742 48 756 48 770
48 785
48 799 48813 48827 48 841
308
48855
48869 48883
48897
48911
48926 48940 48954 48968 48982 1
309
48996 49010 49024
49038
49052
49066 49080 49094 49108 49122 |
810
49136 49150 49164 49178 49192
49206
49220 49234 49248 49262 1
811
49276 49290 49304
49318
49332
49346 49360 49374 49388 49402 |
812
49415
49429 49443
49457
49471
49485
49499 49513 49527 49 541 1
813
49554
49 568 49582
49 5%
49610
49624 49638 49651 49665 49679 |
814
49693
49 707 49721
49 734
49 748
49 762
49776 49790 49803 49817 1
316
49 831 49845 49859
49872
49886
49900 49914 49927 49941 49955 I
816
49969
49982 49996
50010
50024
50037
50051 50065 50079 50092
817
50106
50120 50133
50147
50161
50174
50188 50202 50215 50229
818
50243
50256 50270
50284
50297
50311
50325 50338 50352 50365
819
50379
50393 50406
50420
50433
50447
50461 50474 50488 50 501
820
50 515
50529 50542
50556
50 569
50583
50 5% 50610 50623 50637
321
50651
50664 50678
50691
50 705
50 718
50 732 50 745 50 759 50 772
322
50 786
50 799 50813
50826
50840
50853
50866 50880 50893 50907
328
50920
50934 50947
50 %1
50974
50987
51001 51014 51028 51041
824
51055
51068 51081
51095
51108
51121
51135 51148 51162 51175
326
51188
51202 51215
51228
51242
51255
51268 51282 51295 51308
826
51322
51335 51348
51362
51375
51388
51402 51415 51428 51441
827
51455 51468 51481
51495
51508
51521
51534 51548 51561 51574
328
51587
51601 51614
51627
51640
51654
51667 51680 51693 51706
329
51720
51733 51746
51759
51772
51786
51799 51812 51825 51838
380
51851
51865 51878
51891
51904
51917
51930 51943 51957 51970
331
51983
51996 52009
52022
52035
52048
52061 52 075 52088 52101
332
52114
52127 52140
52153
52166
52179
52192 52 205 52218 52 231
333
52 244
52 257 52 270
52 284
52 297
52310
52323 52336 52 349 52362
334
52 375
52 388 52401
52414
52427
52 440
52453 52466 52479 52492
336
52 504
52 517 52 530
52 543
52 556
52 569
52 582 52 595 52608 52 621
336
52 634
52 647 52 660
52 673
52 686
52699
52 711 52 724 52 737 52 750
337
52 763
52 776 52 789 52 802 52 815
52 827
52 840 52 853 52 866 52 879
338
52 892
52905 52917
52930
52 943
52 956
52 %9 52 982 52994 53 007
339
53 020
53 033 53 046
53 058
53 071
53084
53 097 53110 53122 53135
840
53148
53161 53173
53186
53199
53 212 53 224 53 237 53 250 53 263 |
341
53 275
53 288 53 301
53 314
53 326
53339
53352 53 364 53 377 53 390
342
53403
53415 53 428
53 441
53 453
53466
53479 53 491 53 504 53 517
343
53 529
53 542 53 555 53 567 53 580
53593
53 605 53 618 53631 53 643
344
53 656
53 668 53 681
53 694
53 706
53 719 53 732 53 744 53 757 53 769 1
346
53 782
53 794 53 807
53 820
53 832
53 845 53 857 53 870 53882 53 895 |
346
53 908
53 920 53933
53 945
53 958
53 970
53 983 53 995 54 008 54 020
347
54033
54045 54058
54070
54 083
54095
54108 54120 54133 54145
348
54158
54170 54183
54195
54 208
54 220
54 233 54 245 54258 54 270
349
54 283
54 295 54307
54320
54332
54 345
54357 54370 54382 54 394
350
54407
54419 54432
54444
54456
54469
54481 54 494 54 506 54 518
N
1 2
3
4
5
6 7 8 9
300-360
360-4
too
7
N
O
1
2
8
4
5
6 7
S
9
850
54407
54419
54432
54 4H
54456
54469
54481 54494 54 506
54518
361
54 531
54 543
54 555
54568
54 580
54593 54605 54617
54630
54642
362
54654
54667
54679
54691
54 704
54 716
54 728 54 741
54 753
54 765
363
54 777
54 790
54802
54814
54827
54839 54851 54864
54 876 54888 |
354
54900
54913
54925
54937
54949
54962
54974 54986
54998
55 011 1
366
55 023
55035
55 047
55 060
55 072
55084
55096 55108
55121
55133 1
866
55145
55157
55169
55182
55194
55206 55218 55230 55242 55255 |
367
55267 55 279
55 291
55 303
55 315
55 328 55 340 55352
55364
55 376
368
55 388
55400
55 413
55 425
55 437
55 4*9
55461 55473
55485
55 497
369
55 509
55 522
55 534
55 546
55 558
55 570
55 582 55 594
55606
55 618
aeo
55630
55 642
55 654
55666
55 678
55 691
55 703 55 715
55 727
55 739
361
55 751
55 763 55 775
55 787
55 799
55 811
55823 55 835
55 847
55 859
362
55 871
55 883
55 895
55907
55 919
55931
55 943 55955
55 967
55 979
363
55 991
56003
56015
56027
56038
56050
56062 56074
56086
56098
364
56110
56122
56134
56146
56158
56170
56182 56194 56205
56217
366
56229
56241
56253
56265
56 277
56289
56301 56312
56324
56336
366
56348
56360
56372
56384
563%
56407
56419 56431
56443
56455
867
56467
56478
56490
56 502
56 514
56526
56538 56549
56561
56573
368
56585
56597
56608
56620
56632
56644
56656 56667 56679
56691
369
56703
56 714
56 726
56 738
56 750
56761
56773 56785
56 797
56808
370
56820
56832
56 844
56855
56867
56879 56891 56902
56914
56926
371
56937
56949
56961
56972
56984
56996
57008 57019 57031
57043
372
57054
57066
57078
57089
57101
57113
57124 57136
57148
57159
373
57171
57183
57194
57206
57217
57229
57241 57252
57 264
57276
374
57287
57299
57 310
57322
57334
57345
57357 57368
57380
57392
376
57403 57415
57426
57438
57449
57461
57473 57484
574%
57 507
376
57 519
57 530
57 542
57 553
57 565
57 576
57 588 57600
57611
57623
377
57634
57646
57657
57669
57680
57692
57 703 57 715
57 726 57 738 |
378
57 749
57 761
57 772
57 784
57 795
57807
57818 57830
57841
57852
379
57864
57875
57887
57898
57910
57921
57933 57 9H
57955
57 %7
380
57978
57990 58001
58013
58024
58035
58047 58058
58070
58081
381
58092
58104
58115
58127
58138
58149
58161 58172
58184
58195
882
58206
58218
58229
58 240 58252
58263
58274 58286 58297
58309
: 383
58320
58331
58343
58354
58365
58377
58388 58399
58410
58422
384
58433
58444
58456
58467
58478
58490
58501 58512
58524
58535
386
58 546
58 557
58 569
58 580
58 591
58602
58614 58625
58636
58647
386
58659
58670
58681
58692
58 704
58 715
58 726 58 737
58749
58 760
387
58 771
58 782
58 794
58805
58 816
58 827
58838 58850
58861
58 872
388
58883
58894
58906
58917
58928
58939
58950 58961
58973
58984
389
58995
59006
59017
59028
59040
59051
59062 59073
59084
59095
300
59106
59 118 59 129
59140
59151
59162
59173 59184
59195
59 207
391
59218
59229
59240
59251
59262
59273
59284 59295
59306
59318
392
59329
59340
59351
59362
59373
59384
59395 59406
59417
59428
398
59439
59450
59 461
59472
59483
59494
59 506 59517
59 528
59539
394
59550 59561
59 572
59 583
59594
59605
59616 59627
59638
59649
396
59660
59671
59682
59693
59 704
59 715
59 726 59 737
59 748
59 759
396
59 770
59 780
59 791
59 802
59 813
59 824
59835 59846
59857
59868
397
59879
59890
59901
59912
59 923
59934
59945 59956
59966
59977
398
59988
59999
60 010
60021
60 032
60043
60054 60065
60076
60086
399
60097
60106
60119
60130
60141
60152
60163 60173
60184
60195
400
60 206
60217
60 228
60239
60 249
60260
60271 60 282
60293
60304
N
1
2
3
4
5
6 7
8
9
360--^
LOO
400-460
N
12 3
4
5 6 7 8
400
60206 60217 60228 60239
60249
60260 60271 60282 60293
60 304
401
60314 60325 60336 60347
60358
60369 60379 60390 60401
60412
402
60423 60433 604H 60455
60466
60477 60487 60498 60509
60 520
408
60531 60 541 60552 60 563
60574
60584 60595 60606 60617
60627
404
60638 60649 60660 60670 60681
60692 60703 60713 60724 60735 |
406
60746 60756 60767 60778 60788
60799 60810 60821 60831
60842
406
60853 60863 60874 60885
60895
60906 60917 60927 60938
60949
407
60959 60970 60981 60991
61002
61013 61023 61034 61045
61055
408
61066 61077 61087 61098
61109
61 119 61 130 61 140 61 151
61162
409
61 172 61 183 61 194 61 204 61 215
61225 61236 61247 61257
61268
410
61278 61289 61300 61310
61321
61331 61342 61352 61363
61374
411
61384 61395 61405 61416
61426
61437 61448 61458 61469
61479
412
61490 61500 61511 61521
61532
61542 61553 61563 61574
61584
413
61595 61606 61616 61627
61637
61648 61658 61669 61679
61690
414
61 700 61 711 61 721 61 731
61742
61752 61763 61773 61784
61794
416
61805 61815 61826 61836
61847
61857 61868 61878 61888
61899
416
61909 61920 61930 61941
61951
61962 61972 61982 61993
62 003
417
62014 62024 62034 62045
62 055
62066 62076 62086 62097
62107
418
62118 62128 62138 62149
62159
62170 62180 62190 62 201
62 211
419
62 221 62 232 62 242 62 252
62 263
62273 62284 62294 62304
62 315
420
62 325 62 335 62 346 62 356
62 366
62377 62387 62397 62408
62418
421
62428 62 439 62449 62459
62469
62480 62490 62500 62511
62 521
422
62 531 62 542 62 552 62 562
62 572
62 583 62 593 62603 62 613
62 624
423
62634 62644 62655 62665
62 675
62685 62696 62706 62 716
62 726
424
62 737 62 747 62 757 62 767
62 778
62 788 62 798 62808 62818
62 829
425
62 839 62 849 62859 62 870
62 880
62890 62900 62910 62921
62931
426
62 941 62951 62 %1 62972
62 982
62992 63002 63012 63022
63 033
427
63043 63053 63063 63073
63 083
63094 63104 63114 63124
63134
428
63 144 63 155 63 165 63 175 63 185
63195 63 205 63 215 63 225
63 236
429
63 246 63256 63266 63276 63286
63296 63306 63317 63 327
63 337
430
63 347 63357 63 367 63 377 63 387
63397 63407 63417 63428
63 438
431
63448 63458 63468 63478 63488
63498 63 508 63 518 63 528
63 538
482
63 548 63 558 63 568 63 579 63 589
63599 63609 63619 63629 63639 |
433
63649 63659 63669 63679 63689
63699 63 709 63 719 63 729
63 739
434
63 749 63 759 63 769 63 779 63 789
63799 63809 63819 63829
63 839
435
63 849 63 859 63 869 63 879
63 889
63899 63909 63919 63929
63 939
436
63949 63959 63969 63979
63 988
63 998 64008 64018 64028
64038
437
64048 64058 64068 64078
64088
64098 64108 64118 64128
64137
438
64147 64157 64167 64177
64187
64197 64207 64 217 64 227 64 237 |
439
64 246 64256 64 266 64 276
64 286
64 296 64306 64 316 64 326
64 335 1
440
64345 64355 64365 64375
64385
64395 64404 64414 64424 64434 |
441
644H 64454 64464 64473
64483
64493 64 503 64 513 64 523
64 532
442
64 542 64 552 64 562 64 572
64 582
64591 64601 64611 64621
64631
443
64640 64650 64660 64670
64 680
64689 64699 64 709 64 719
64 729
444
64 738 64 748 64 758 64 768
64 777
64 787 64 797 64807 64816
64 826
445
64836 64846 64856 64865
64875
64885 64895 64904 64914
64924
446
64933 64943 64953 64963
64972
64982 64992 65002 65011
65 021
447
65 031 65 040 65 050 65 060
65 070
65079 65089 65099 65108
65118
448
65128 65137 65147 65157
65167
65 176 65 186 65 1% 65 205
65 215
449
65 225 65 234 65 244 65 254
65 263
65 273 65 283 65 292 65 302
65 312
450
65 321 65 331 65 341 65 350
65 360
65369 65 379 65 389 65 398
65 408
N
12 3
4
5 6 7 8
400-460
460-600
N
12 3
4
5 6
7
8
• 1
450
65 321 65 331 65 341 65 350
65 360
65 369 65 379
65389 65398 65408 |
451
65 418 65 427 65 437 65 447
65 456
65 466 65 475
65 485
65495
65 504
452
65 514 65 523 65 533 65 543
65 552
65 562 65 571
65 581
65 591
65 600
453
65 610 65 619 65 629 65 639
65 648
65 658 65 667
65 677
65 686
65 696
464
65 706 65 715 65 72i 65 734
65 744
65 753 65 763
65 772
65 782
65 792
456
65 801 65 811 65 820 65 830
65 839
65849 65858
65 868
65 877
65 887
466
65 896 65 906 65 916 65 925
65 935
65 9H 65 954
65 963
65 973
65 982
457
65 992 66001 66011 66020 66030
66039 66049
66058
66068
66077
458
66087 66096 66106 66115 66124
66134 66143
66153
66162
66172
459
66181 66191 66200 66210
66219
66229 66238
66247
66257
66 266
460
66276 66285 6629i 66304
66314
66323 66332
66342
66351
66361
461
66370 66380 66389 66398
66408
66417 66427
66436
66445
66455
462
66464 66474 66483 66492
66 502
66511 66521
66 530
66 539
66 549
463
66 558 66 567 66577 66586
66 596
66605 66614
66624
66633
66642
464
66652 66661 66671 66680
66689
66699 66 708
66 717
66 727
66 736
465
66 745 6675i 66764 66 773
66 783
66 792 66801
66811
66820
66829
466
66839 66848 66857 66867
66876
66885 66894
66904
66913
66922
467
66932 66941 66950 66960
66 %9
66978 66987
66997
67006
67015
468
67025 67034 67043 67052
67062
67071 67080
67089
67099
67108
469
67117 67127 67136 67145
67154
67164 67173
67182
67191
67201
470
67 210 67219 67228 67 237
67 247
67256 67 265
67274
67 284
67293
471
67302 67311 67321 67330
67 339
67348 67357 67367 67376 67385 I
472
67394 67403 67413 67422
67431
67440 67449
67459
67468
67477
473
67486 67495 67 504 67 514
67 523
67 532 67 541
67 550
67 560
67 569
474
67 578 67 587 67 596 67605
67 614
67624 67633
67642
67651
67660
476
67669 67679 67688 67697
67 706
67 715 67724 67 733
67 742
67 752
476
67 761 67 770 67 779 67 788
67 797
67806 67815
67 825
67834
67843
477
67852 67861 67870 67879
67 888
67897 67906 67916 67925
67934
478
67943 67952 67961 67970 67979
67988 67997
68006
68015
68024
479
68034 68043 68052 68061
68070
68079 68088
68097
68106
68115
480
68124 68133 68142 68151
68160
68169 68178 68187
681%
68205
481
68215 68224 68233 68242
68 251
68260 68269
68 278
68287
68 2%
48«
68305 68314 68323 68332
68341
68350 68359 68368 68377 68386 |
483
68395 68404 68413 68422
68431
68440 68449
68458
68467
68476
484
68485 68494 68502 68511
68 520
68529 68 538
68 547
68 556
68565
486
68574 68583 68592 68601
68610
68619 68628 68637 68646 68655 |
486
68664 68673 68681 68690
68 699
68 708 68 717
68 726
68 735
68 744
487
68753 68762 68771 68780
68 789
68 797 68806
68815
68824
68 833
488
68842 68851 68860 68869
68878
68886 68895
68904
68913
68922
489
68931 68940 68949 68958
68966
68975 68984
68993
69002
69011
490
69020 69028 69037 69046
69055
69064 69073
69082
69090 69099 |
491
69108 69117 69126 69135 69144
69152 69161
69170
69179
69188
492
69197 69 205 69214 69223
69 232
69 241 69249 69 258
69267
69 276
493
69285 69294 69302 69311
69320
69 329 69338
69346
69355
69364
494
69 373 69381 69390 69399
69408
69417 69425
69434
69443
69452
495
69461 69469 69478 69487
69496
69 504 69 513
69 522
69 531
69539
496
69 548 69 557 69 566 69 574
69 583
69 592 69601
69609
69618
69627
497
69636 69644 69653 69662
69 671
69679 69688
69697
69 705
69 714
498
69 723 69 732 69 740 69 749
69 758
69 767 69 775
69 784
69 793
69801
499
69810 69819 69827 69836 69845
69 854 69862
69871
69 880
69 888
500
69897 69906 69914 69923
69932
69940 69949
69958 69966
69975
N
12 3
4
5 6
7
8
9
460-600
10
500-660
IT
12 3
4
6
6
7
8
9
600
69897
69906 69914 69923
69932
69940
69949
69958
69966
69975
601
69984
69992 70001 70010
70018
70027
70036
70044
70053
70062
602
70070
70079 70088 70096 70105
70114
70122
70131
70140
70148
603
70157
70165 70174 70183
70191
70 200
70209
70217
70226
70234
604
70 243
70252 70260 70269
70278
70286 70295
70303
70312
70321
606
70329
70338 70346 70355
70364
70372
70381
70389
70398
70406
606
70415
70424 70432 70441
70449
70458
70467
70475
70484
70492
607
70501
70 509 70 518 70526
70 535
70544
70 552
70 561
70 569
70578
608
70586 70 595 70603 70612
70621
70629
70638
70 646
70655
70663
«09
70672
70680 70689 70697
70 706
70 714
70 723
70 731
70 740
70 749
510
70 757
70 766 70 774 70 783
70 791
70800
70808
70817
70 825
70 834
611
70842
70851 70859 70868
70876
70885
70 893
70902
70910
70919
612
70927
70935 70944 70952
70961
70969
70978
70986 70995
71003
613
71012
71020 71029 71037
71046
71054
71063
71071
71079
71088
614
71096 71105 71113 71122
71130
71139
71147
71155
71164
71172
616
71181
71189 71198 71206
71214
71223
71231
71240
71248
71257
616
71265
71273 71282 71290
71299
71307
71315
71324
71332
71341
617
71349
71357 71366 71374
71383
71391
71399
71408
71416 71425 1
618
71433
71441 71450 71458
71466
71475
71483
71492
71500
71508
610
71517
71525 71533 71542
71550
71559
71567
71575
71584
71592
520
71600
71609 71617 71625
71634
71642
71650
71659
71667
71675
621
71684
71692 71700 71709
71717
71725
71734
71742
71750
71759
622
71767
71775 71784 71792
71800
71809
71817
71825
71834
71842
623
71850
71858 71867 71875
71883
71892
71900
71908
71917 71925 1
624
71933
71941 71950 71958
71966
71975
71983
71991
71999
72008
626
72016
72024 72 032 72 041
72049
72057
72066
72 074
72082
72090
626
72099
72107 72115 72123
72132
72140
72148
72156 72 165
72173
627
72181
72189 72198 72 206
72 214
72 222
72 230
72 239
72 247 72 255 1
628
72 263
72 272 72280 72 288
72 296
72 304
72313
72321
72329 72337 |
629
72 346
72 354 72 362 72 370
72 378
72 387
72 395
72403
72411
72419
530
72428
72 436 72444 72452
72460
72 469
72477
72 485
72493
72 501
631
72 509
72 518 72 526 72 534
72 542
72 550
72 558
72 567 72 575
72 583
632
72 591
72 599 72 607 72 616
72 624
72 632
72640 72648 72656 72665 |
633
72673
72 681 72 689 72 697
72 705
72 713
72 722
72 730
72 738
72 746
634
72 754
72 762 72 770 72 779
72 787
72 795
72 803
72 811
72 819
72 827
636
72 835
72 843 72 852 72 860
72 868
72876
72 884
72 892
72900
72 908
636
72 916 72 925 72 933 72941
72 949
72957
72 965
72 973
72981
72989
637
72 997
73 006 73 014 73 022
73 030
73 038
73 046
73 054
73 062
73 070
638
73 078
73 086 73 094 73102
73111
73119
73127
73135
73143
73151
639
73159
73167 73175 73183
73191
73199
73 207
73 215
73 223
73 231
540
73 239
73 247 73 255 73 263
73 272
73 280
73 288
73 296
73 304
73 312
541
73 320
73 328 73 336 73 344
73 352
73 360
73 368
73 376
73 384
73 392
642
73 400
73 408 73416 73 424
73 432
73 440
73 448
73 456
73 464
73 472
643
73 480
73 488 73 4% 73 504
73 512
73 520
73 528
73 536
73 544
73 552
644
73 560
73 568 73 576 73 584
73 592
73 600
73 608
73 616
73 624
73 632
546
73 640
73 648 73 656 73 664
73 672
73 679
73 687
73 695
73 703
73 711
646
73 719
73 727 73 735 73 743
73 751
73 759
73 767
73 775
73 783
73 791
647
73 799
73 807 73 815 73 823
73 830
73 838
73 846
73 854
73 862
73 870
548
73 878
73 886 73 894 73 902
73 910
73 918
73 926
73 933
73 941
73 949
549
73 957
73 965 73 973 73 981
73 989
73 997
74005
74 013
74020
74028
550
74 036
74044 74052 74060
74068
74076
74084
74092
74099
74107
N
12 3
4
5
6
7
8
9
600-660
660-600
11
,_..
12 3
4
5
6 7 8
550
74036 74044 74052 74060 74068
74076
74084 74092 74099
74107
551
74115 74123 74131 74139
74147
74155
74162 74170 74178
74186
552
74194 74202 74 210 74 218
74225
74 233
74241 74249 74257 74265 1
663
74273 74 280 74 288 74 2%
74304
74312
74320 74327 74335
74 343
664
74351 74359 74367 74374
74382
74390
74398 74406 74414
74421
666
74429 74437 74445 74453
74461
74468
74476 74484 74492
74500
656
74507 74 515 74 523 74531
74 539
74547
74554 74562 74 570
74578
667
74586 74593 74601 74609
74617
74624
74632 74640 74648
74656
668
74663 74 671 74679 74687
74 695
74 702
74 710 74 718 74 726
74 733
669
74 741 74 749 74 757 74 764
74 772
74 780
74 788 74 7% 74803
74 811
560
74819 74827 74834 74842 74850
74858
74865 74873 74881
74 889
661
748% 74904 74912 74920
74927
74935
74943 74950 74958
74966
562
74974 74981 74989 74997 75 005 |
75012
75 020 75 028 75 035
75 043
663
75 051 75 059 75 066 75 074
75 082
75089 75 097 75105 75113
75120
664
75128 75136 75143 75151
75159
75166
75174 75182 75189
75197
666
75205 75213 75220 75 228 75236
75 243
75 251 75 259 75 266
75 274
666
75 282 75 289 75 297 75 305
75 312
75 320
75 328 75335 75 343
75351
667
75 358 75 366 75 374 75 381
75 389
75 397
75404 75 412 75 420
75 427
668
75 435 75 442 75 450 75 458
75 465
75473
75 481 75 488 75 4%
75 504
669
75 511 75 519 75 526 75 534
75 542
75 549 75 557 75 565 75 572 75 580 |
670
75 587 75 595 75 603 75 610
75 618
75 626
75 633 75 641 75 648
75 656
671
75 664 75 671 75 679 75 686
75 694
75 702
75 709 75 717 75 724
75 732
672
75 740 75 747 75 755 75 762
75 770
75 778
75 785 75 793 75 800
75 808
673
75 815 75 823 75 831 75 838
75 846
75 853
75861 75868 75876
75 884
674
75 891 75 899 75 906 75 914
75 921
75 929
75 937 75 9H 75 952
75 959
676
75 %7 75974 75 982 75 989
75 997
76005
76012 76020 76027
76035
676
76042 76050 76057 76065
76072
76080 76087 76095 76103
76 110
677
76118 76125 76133 76140
76148
76155
76163 76170 76178
76185
678
76193 76200 76208 76215
76223
76230
76238 76245 76253
76 260
679
76268 76 275 76283 76290
76 298
76305
76313 76320 76328
76335
580
76343 76350 76358 76365
76373
76380
76388 76395 76403
76410
681
76418 76425 76433 76440
76448
76455
76462 76470 76477
76485
682
76492 76500 76507 76 515
76 522
76 530
76537 76545 76552
76559
683
76 567 76574 76582 76589
76597
76604
76612 76619 76626
76634
684
76641 76649 76656 76664
76671
76678
76686 76693 76 701
76 708
686
76716 76723 76730 76738
76 745
76 753
76 760 76768 76 775
76 782
686
76 790 76797 76805 76812
76819
76827
76834 76842 76849
76856
687
76864 76871 76879 76886
76893
76901
76908 76916 76923
76930
588
76938 76945 76953 76960
76 %7
76975
76982 76989 76997
77004
689
77012 77019 77026 77034
77041
77048
77056 77063 77070
77078
500
77085 77093 77100 77107
77115
77122
77129 77137 77144
77151
691
77159 77166 77173 77181
77188
77195
77203 77210 77217
77225
692
77232 77 240 77 247 77 254
77262
77 269
77 276 77283 77291
77298
698
77 305 77313 77320 77 327
77335
77342
77349 77357 77364
77371
694
77379 77386 77393 77401
77408
77415
77422 77430 77 437
77444
696
77452 77459 77466 77474
77481
77488
77495 77 503 77 510
77517
696
77 525 77 532 77 539 77 546
77 554
77 561
77 568 77 576 77 583
77 590
697
77 597 77605 77612 77619
77627
77634
77641 77 648 77656
77 663
698
77670 77677 77685 77692
77 699
77 706
77 714 77 721 77 728
77 735
699
77 743 77 750 77757 77 764 77 772
77 779
77786 77 793 77801
77808
600
77815 77 822 77830 77 837
77 844
77851
77859 77866 77873
77 880
N
12 3
4
5
6 7 8
9
660 -e
;oo
12
600-660
N
12
8
4
8
6
7
8 9
600
77 815 77822 77830
77837
77 844
77851
77859
77866
77 873 77880
601
77887 7789i 77902 77909 77916
77924
77931
77938
77945 77952
602
77960 77967 77974
77981
77988
77996
78003
78010
78017 78025
603
78032 78039 78046
78053
78061
78068 78075
78082
78089 78097
604
78104 78111 78118
78125
78132
78140
78147
78154
78161 78168
605
78176 78183 78190
78197
78204
78211
78219
78226
78233 78240
606
78247 78 254 78 262
78269
78 276
78283
78 290 78297 78305 78312 1
607
78319 78326 78 333
78 340
78347
78355 78362 78369 78376 78383 |
608
78390 78398 78405
78412
78419
78426
78433
78440 78447 78455 |
609
78462 78469 78476
78483
78490
78497
78 504
78512
78 519 78526
610
78533 78 540 78547
78554
78 561
78 569
78 576
78 583
78590 78597
611
78604 78611 78618
78 625
78633
78640
78647
78654
78661 78668
612
78675 78682 78689
78 696
78 704
78 711
78 718 78 725
78732 78 739
613
78746 78753 78760
78 767
78 774
78 781
78 789
78 796
78803 78810
614
78817 78824 78 831
78838
78845
78852
78 859
78866
78873 78880
619
78888 7889i 78902
78909
78916
78923
78930
78937
78944 78951
616
78958 78965 78972
78979
78986
78993
79000
79007
79014 79021
617
79029 79036 79043 790i0 79057
79064
79071
79078 79085 79092 1
618
79099 79106 79113
79120
79127
79134
79141
79148 79155 79162 |
619
79169 79176 79183
79190 79197
79204
79211
79218
79225 79232
620
79239 79246 79253
79260
79267
79274
79281
79288
79295 79302
621
79309 79316 79323
79330
79337
79344
79351
79358
79365 79372
622
79379 79386 79393
79400
79407
79414
79421
79428 79435 79442 |
623
79449 79456 79463
79470
79477
79484
79491
79498
79 505 79 511
624
79 518 79 525 79532
79 539
79546
79 553
79560
79 567
79574 79581
625
79588 7959i 79602 79609 79616
79623
79630
79637
79644 79650
626
79657 79664 79671
79678
79685
79692
79699
79706
79 713 79 720
627
79727 79 734 79 741
79 748
79 754
79761
79 768
79 775
79782 79789
628
79796 79803 79810
79817
79824
79831
79837
79844
79851-79858
629
79865 79872 79879
79886
79893
79900
79906
79913
79920 79927
630
79934 79941 79948 79955 79962
79 %9
79 975
79982
79989 79996
631
80003 80010 80017
80024
80030
80037
80044
80051
80058 80065
632
80072 80079 80085
80092
80099
80106
80113
80120
80127 80134
633
80140 80147 80154
80161
80168
80175
80182
80188
80195 80 202
634
80209 80216 80223
80229
80236
80243
80 250 80257
80264 80271
635
80277 80284 80291
80298 80305
80312
80318
80325
80332 80339
636
80346 80353 80359
80366
80373
80380
80387
80393
80400 80407
637
80414 80421 80428
80434
80441
80448 80455
80462
80468 80475
638
80482 80489 80496
80502
80 509
80516
80 523
80530
80536 80 543
639
80 550 80557 80564
80 570
80577
80584
80591
80598
80604 80611
640
80618 8062i 80632
80638
80645
80652
80659
80665
80672 80679
641
80686 80693 80699
80 706
80 713
80 720
80 726
80 733
80 740 80 747
642
80 754 80760 80 767
80 774
80 781
80 787
80 794
80801
80808 80 814
643
80821 80828 80 835
80841
80848
80 855
80862
80 868
80875 80 882
644
80889 80895 80902
80909
80916
80922
80929
80936 80943 80949 |
645
80956 80963 80969
80976
80983
80990
80996
81003
81010 81017
646
81023 81030 81037
81043
81050
81057
81064
81070
81077 81084
647
81090 81097 81104
81111
81117
81124
81131
81137
81144 81151
648
81158 81164 81171
81178
81184
81191
81198
81204
81211 81218
649
81224 81231 81238 81245
81251
81258 81265 81271 81278 81285 |
660
81291 81298 81305
81311
81318
81325
81331
81338 81345 81351 1
N
12
3
4
S
6
7
8 9 1
600-660
660-700
13
N
18 8 4
5 6 7 8
650
81291 81298 81J0i 81311 81318
81325 81331 81338 81345 81351
661
81358 81365 81371 81378 8138i
81391 81398 81405 81411 81418
663
8142i 81431 81438 81445 81451
81458 81465 81471 81478 81485
653
81491 81498 81505 81511 81518
81525 81531 81538 81544 81551
664
81558 81564 81571 81578 81584
81591 81598 81604 81611 81617
666
81624 81631 81637 81644 81651
81657 81664 81671 81677 81684
666
81 690 81 697 81 704 81 710 81 717
81723 81730 81737 81743 81750
667
81 757 81 763 81 770 81 776 81 783
81790 81796 81803 81809 81816
668
81823 81829 81836 81842 81849
81856 81862 81869 81875 81882
669
81889 81895 81902 81908 81915
81921 81928 81935 81941 81948
660
81954 81961 81968 81974 81981
81987 81994 82000 82007 82014
661
82020 82027 82033 82040 82046
82053 82060 82066 82073 82079
662
82086 82092 82099 82105 82112
82119 82125 82132 82138 82145
663
82151 82158 82164 82171 82178
82184 82191 82197 82204 82210
664
82217 82223 82230 82236 82243
82249 82256 82263 82269 82276
666
82282 82289 82295 82302 82308
82315 82321 82328 82334 82341
666
82347 82354 82360 82367 82373
82380 82387 82393 82400 82406
667
82413 82419 82426 82 432 82439
82 445 82452 82458 82465 82 471
668
82478 82484 82491 82497 82504
82 510 82 517 82 523 82 530 82 536
669
82543 82549 82556 82562 82569
82575 82582 82588 82595 82601
670
82607 82614 82620 82627 82633
82640 82646 82653 82659 82666
671
82672 82679 82685 82692 82698
82 705 82 711 82 718 82 724 82 730
672
82737 82743 82750 82756 82763
82 769 82776 82782 82789 82795
678
82802 82808 82814 82821 82827
82834 82840 82847 82853 82860
674
82866 82872 82879 82885 82892
82898 82905 82911 82918 82924
676
82930 82937 82943 82950 82956
82963 82969 82975 82982 82988
676
82995 83001 83006 83014 83020
83027 83033 83040 83046 83052
677
83059 83065 83072 83078 83085
83091 83097 83104 83 110 83117
678
83123 83129 83136 83142 83149
83155 83161 83168 83174 83181
679
83187 83193 83 200 83 206 83 213
83219 83225 83232 83238 83245
680
83251 83257 83264 83270 83276
83283 83289 83296 83302 83308
681
83315 83321 83327 83334 83340
83347 83353 83359 83366 83372
682
83378 83385 83391 83398 83404
83410 83417 83423 83429 83436
683
83442 83448 83455 83461 83467
83474 83480 83487 83493 83499
684
83 506 83 512 83 518 83 525 83 531
83537 83544 83550 83556 83563
686
83569 83575 83582 83588 83594
83601 83607 83613 83620 83626
686
83632 83639 83645 83651 83658
83664 83670 83677 83683 83689
687
83 696 83 702 83 708 83 715 83 721
83 727 83 734 83 740 83 746 83 753
688
83 759 83 765 83 771 83 778 83 784
83 790 83 797 83803 83809 83816
689
83822 83828 83835 83841 83847
83853 83860 83866 83872 83879
600
83885 83891 83897 83904 83910
83916 83923 83929 83935 83942
691
83948 83954 83960 83967 83973
83979 83985 83992 83998 84004
692
84011 84017 84023 84029 84036
84042 84048 84055 84061 84067
693
84073 84080 84086 84092 84098
84105 84111 84117 84123 84130
694
84136 84142 84148 84155 84161
84167 84173 84180 84186 84192
695
84198 84 205 84 211 84 217 84 223
84230 84 236 84242 84 248 84 255
696
84 261 84 267 84 273 84 280 84 286
84 292 84298 84305 84311 84317
697
84323 84 330 84 336 84 342 84 348
84 354 84 361 84367 84373 84379
698
84386 84 392 84398 84404 84410
84417 84423 84429 84435 84442
699
84448 84454 84460 84466 84473
84479 84485 84491 84497 84 504
700
84 510 84 516 84 522 84 528 84 535
84 541 84547 84 553 84 559 84 566
K
O 1 2 8 4
5 6 7 8
660-700
14
700-760
N-
12 8 4
5 6 7 8
0.
700
84 510 84516 84 522 84 528 84 535
84 541 84 547 84553 84559
84566
701
84572 84 578 84 584 84 590 84 597
84603 84609 84615 84621
84628
702
84634 84640 84646 84652 84658
84665 84671 84677 84683
84689
70S
84696 84 702 84 708 84714 84 720
84 726 84 733 84 739 84 745 84 751 |
704
84 757 84 763 84 770 84 776 84 782
84 788 84 794 84800 84807
84813 1
705
84819 84825 84831 84837 84844
84850 84856 84862 84868 84874 |
706
84880 84887 84893 84899 84905
84911 84917 84924 84930
84936 1
707
84942 84948 84954 84960 84967
84973 84979 84955 84991 84997 |
708
85003 85009 85016 85022 85028
85 034 85 040 85046 85 052
85058 1
700
85065 85071 85077 85083 85089
85095 85 101 85107 85114
85120 1
710
85 126 85 132 85 138 85 144 85 150
85156 85163 85169 85175 85 181 |
711
85187 85193 85199 85 205 85 211
85 217 85 224 85 230 85 236
85 242
712
85 248 85 254 85 260 85 266 85 272
85 278 85 285 85 291 85297
85 303
713
85 309 85 315 85 321 85 327 85 333
85 339 85 345 85 352 85 358
85 364
714
85 370 85376 85 382 85 388 85 394
85400 85406 85412 85418
85 425
715
85 431 85 437 85 443 85 449 85 455
85461 85467 85473 85479
85 485
716
85 491 85497 85 503 85 509 85 516
85522 85 528 85534 85 540
85 546
717
85 552 85 558 85 564 85 570 85 576
85 582 85 588 85 594 85600
85 606
718
85 612 85 618 85 625 85 631 85 637
85 643 85 649 85 655 85 661
85 667
719
85 673 85 679 85 685 85 691 85 697
85 703 85 709 85 715 85 721
85 727
720
85 733 85 739 85 745 85 751 85 757
85 763 85 769 85 775 85 781
85 788
721
85 794 85800 85806 85 812 85 818
85 824 85 830 85 836 85 842
85 848
722
85 854 85 860 85 866 85 872 85 878
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86147
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86183 86189 86195 86201
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86213 86219 86225 86231 86237
86243 86249 86255 86261
86267
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86273 86279 86285 86291 86297
86303 86308 86314 86320
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780
86332 86338 86344 86350 86356
86362 86368 86374 86380
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86864 86870 86876 86882 86888
86894 86900 86 906 86911
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86923 86929 86935 86941 86 947
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86976
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87011 87017 87023 87029 87 035 [
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87070 87075 87081 87087
87093 1
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87128 87134 87140 87146
87151 1
744
87157 87163 87169 87175 87181
87186 87192 87198 87 204 87210 1
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87216 87 221 87 227 87233 87 239
87245 87251 87256 87262 87268 f
746
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87 390 87 396 87402 87408 87413
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87622 87628 87633 87639 876*5
87651 87656 87662 87668 87674
763
87679 87685 87691 87697 87 703
87 708 87 714 87 720 87726 87731
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87 766 87772 87 777 87 783 87 789
756
87795 87800 87806 87812 87818
87823 87829 87835 87841 87846
766
87852 87858 87864 87869 87875
87881 87887 87892 87898 87904
767
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87938 87 9H 87950 87955 87961
768
87967 87973 87978 87984 87990
87996 88001 88007 88013 88018
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88024 88030 88036 88041 88047
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761
88138 88144 88150 88156 88161
88167 88173 88178 88184 88190
762
88195 88201 88207 88213 88218
88224 88230 88235 88241 88247
768
88252 88258 88264 88270 88275
88281 88287 88292 88298 88304
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88309 88315 88321 88326 88332
88338 88343 88349 88355 88360
766
88366 88372 88377 88383 88389
88395 88400 88406 88412 88417
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88423 88429 88434 88440 88446
88451 88457 88463 88468 88474
767
88480 88485 88491 88497 88502
88508 88513 88519 88525 88530
768
88536 88542 88547 88553 88559
88564 88570 88576 88581 88587
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770
88649 88655 88660 88666 88672
88677 88683 88689 88694 88 700
771
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88846 88852 88857 88863 88868
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88874 88880 88885 88891 88897
88902 88908 88913 88919 88925
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777
89042 89048 89053 89059 89064
89070 89076 89081 89087 89092
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89154 89159 89165 89170 89176
89182 89187 89193 89198 89204
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89293 89298 89304 89310 89315
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89321 89326 89332 89337 89343
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89570 89575 89581 89586 89592
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89653 89658 89664 89669 89675
89680 89686 89691 89697 89702
789
89708 89713 89719 89 724 89 730
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89763 89768 89774 89 779 89785
89 790 89 796 89801 89807 89812
791
89818 89823 89829 89834 89840
89845 89851 89856 89862 89867
792
89873 89878 89883 89889 89894
89900 89905 89911 89916 89922
793
89927 89933 89938 89944 89949
89955 89960 89966 89971 89977
794
89982 89988 89993 89998 90004
90009 90 015 90020 90026 90031
796
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90064 90069 90075 90080 90086
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90200 90206 90211 90217 90222
90227 90233 90238 90244 90249
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90255 90260 90266 90271 90276
90282 90287 90293 90298 90304
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90309 90314 90320 90325 90331
90336 90342 90347 90352 90358
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850
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93877 93882 93887 93892 93897
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870
93952 93957 93962 93967 93 972
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94126 94131 94136 94141 94146
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94176 94181 94186 94191 94196
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94201 94206 94 211 94216 94221
94226 94 231 94 236 94 240 94245
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N
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5 6 7 8 9
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95 497 95 501 95 506 95 511 95 516
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95593 95 598 95602 95607 95612
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912
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96142 96147 96152 96156 96161
96166 96171 96175 96180 96185
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96213 96218 96 223 96227 96232
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96237 96242 96246 96251 96256
96261 96265 96270 96275 %280
918
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937
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97243 97 248 97253 97 257 97262
939
97267 97271 97 276 97 280 97 285
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940
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97 749 97 754 97 759 97 763 97 768
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97 772 97 777 97 782 97 786 97 791
97795 97800 97804 97809 97813
N
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98114 98118 98123 98127 98132
968
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969
98182 98186 98191 98195 98200
96204 98209 98214 98218 98223
960
98227 98232 98236 98241 98245
98250 98254 98259 98263 98268
961
98272 98277 98281 98286 98290
98 295 98299 98304 98308 98313
962
98318 98322 98327 98331 98336
98340 98345 98349 98354 98358
963
98363 98367 98372 98376 98381
98385 98390 98394 98399 98403
964
98408 96412 98417 98421 98426
98430 98435 98439 98444 98448
965
98453 98457 98462 98466 98471
98475 98480 98484 98489 98493
966
98498 98502 98507 98511 98516
98520 96525 98529 98534 98538
967
96543 96547 98552 98556 98 561
98565 98 570 98 574 98579 98 583
968
98588 98592 98597 98601 98605
98610 98614 98619 98623 98628
968
98632 98637 98641 98646 98650
98655 98659 98664 98668 98673
970
98677 98682 98686 98691 98695
98700 98704 98709 98713 98717
971
98722 98726 96731 98735 98740
98744 98749 98753 98758 98762
972
98767 98771 98776 98780 98784
98789 98 793 98798 98802 98807
978
98811 98816 98820 98825 98829
98834 98838 98843 98847 98851
974
98856 96860 98865 98869 98874
98878 98883 98887 98892 98896
976
98900 98905 98909 98914 96918
98923 98927 98932 98936 98941
976
98945 98949 98954 98958 98963
98967 98972 98976 98981 98985
977
98989 98994 98998 99 003 99007
99012 99016 99021 99025 99029
978
99034 99038 99043 99047 99052
99056 99061 99065 99069 99074
979
99078 99083 99087 99092 99096
99100 99105 99109 99114 99118
980
99123 99127 99131 99136 99140
99145 99149 99154 99158 99162
981
99167 99171 99176 99180 99185
99189 99193 99198 99202 99207
982
99211 99216 99220 99 224 99 229
99233 99238 99242 99247 99251
983
99255 99260 99264 99269 99273
99277 99282 99286 99291 99295
984
99300 99304 99308 99313 99317
99322 99326 99330 99335 99339
986
99344 99348 99352 99357 99361
99366 99370 99374 99379 99383
986
99388 99392 99396 99401 99405
99410 99414 99419 99423 99427
887
99432 99436 99441 99445 99449
99454 99458 99463 99467 99471
988
99476 99480 99484 99489 99493
99498 99502 99506 99511 99515
989
99 520 99 524 99528 99 533 99 537
99 542 99 546 99 550 99555 99 559
990
99564 99568 99572 99577 99 581
99 585 99 590 99 594 99 599 99603
991
99607 99612 99616 99 621 99 625
99629 99634 99638 99642 99647
992
99651 99656 99660 99664 99669
99673 99677 99682 99686 99691
993
99695 99699 99704 99708 99712
99 717 99 721 99 726 99 730 99 734
994
99739 99743 99747 99752 99756
99760 99765 99769 99774 99778
996
99782 99787 99791 99795 99800
99804 99808 99813 99817 99822
996
99826 99830 99835 99839 99843
99848 99852 99856 99861 99865
997
99870 99874 99878 99883 99887
99891 99896 99900 99904 99909
998
99913 99917 99922 99926 99930
99 935 99939 99944 99948 99952
999
99957 99%1 99965 99970 99974
99978 99983 99987 99991 999%
1000
00000 00004 00009 00013 00017
00022 00026 00030 00035 00039
N
12 3 4
5 6 7 8 9
960-1000
20 TABLE H-LOGAEITHMS OF CONSTAMS.
Circumference of the Circle in degrees , . = 360
Circumference of the Circle in minutes = 21 600
Circumference of the Circle in seconds = 1 296 000
If the radius r = 1, half the Circumference of the Circle is
» = 3. 14 159265 358979323 846264338328
log
2.55 630250
4.33445 375
6.11260500
0.49 714987
Also:
2k = 6.
47r=12.
^= 1.
3
= 0.
-i-= 0.
«-= 0.
TT
1=1.
IT
i.= 0.
47r
28318531
56637061
57079633
04 719 755
18879020
78539816
52 359878
31830989
15915 494
95492 966
27323954
23 873 241
log
0.79817987
1.09920986
0.19611988
0.02 002 862
0.62 208 861
9.89508988-10
9.71899 862-10
9.50285 013-10
9.20182013-10
9.97 997138-10
0.10491012
9.37 791139-10
7r« =
1 _
v^ =
1 ^
/e
<lh-
9.86960440
0. 10 132 118
1.77245 385
0.56418958
0.97 720502
1.12 837917
1.46459189
0.68278406
2.14 502940
0.62035 049
0.80599 598
0.99429975
9.00570025-10
0.24 857494
9.75142 506-10
9.98998 569-10
0.05 245 506
0.16 571662
9.83 428338-10
0.33143 325
9.79 263 713-10
9.90 633 287-10
Arc a, whose length is equal to the radius r, is
- 180
in degrees
.a' .
in minutes a' .
in seconds a" .
10800
648 000
= 57.29577951*;.
= 3 437.74677'..
= 206264.806"..
Arc 2 a, whose length is equal to twice the radius,2r, is
860
in degrees 2a° = —
TT
in minutes 2a' =
in seconds 2a" =
21600
TT
1296000
= 114. 59 155 903«
= 6875.49354'..
....= 412 529.612"..
If the radius r = 1, the length of the arc is :
1 ^jr_
a^ "*' 180
1 _ IT
a' "" 10 800
1 ^ TT
a" * * * * 648 000
for 1 degree . . .
for 1 minute...
for 1 second...
1 TT
for J degree — — = rrr
^ ^ 2a° 360
for i minute. . . .-^.. . . = ^ . . .
^ 2a' 21 600
1 TT
for i second — =
^ 2a" 1296000
.= 0.01745 329..
.= 0.00029089..
.= 0.00000485..
. = 0.00872 665..
.= 0.00014 544..
.= 0.00000242..
Sin 1" in the unit circle = 0. 00000485. . .
log
1.75 812 263
3.53 627388
5.31442 513
2.05 915 263
3.83 730388
5.61545 513
8.24187 737-10
6.46372 612-10
4.68 557487-10
7.94084 737-10
6.16269612-10
4.38454487 — 10
4.68557487 — 10
21
TABLE III.
THE LOGARITHMS
TELIGONOMETRIC FUJ^CTIONS:
Erom (Tto 0** 3', or 89** 57' to 90^ for every seoond;
From (f* to 2"", or 88** to 9(r, for everj ten seoondBi
Rwinrt
NOTB. To
log sin
88^ fOT e
aU the log
ytrjmiiinte.
3 appended
I
000
ftrithms —
10 is to iM
0°
log tan »* log ill
bg 001 -10. 00
ff
0»
1'
' 2'
ft
ff
0'
V
2'
99
o
6.46 373
6.76 476
00
30
6. 16 270
6.63 982
6.86167
80
1
4.68 557
6.47 090
6. 76 836
69
31
6.17 694
6.64 462
6.86 455
29
2
4.98 660
6.47 797
6. 77 193
68
32
6. 19 072
6.64 936
6. 86 742
28
3
5. 16 270
6.48 492
6. 77 548
67
33
6.20 409
6.65 406
6.87 027
27
4
5.28 763
6.49175
6.77 900
66
34
6. 21 705
6.65 870
6. 87 310
26
6
5.38 454
6.49 849
6. 78 248
66
36
6.22 964
6.66 330
6. 87 591
26
6
5. 46 373
6. 50 512
6.78 595
64
36
6.24 188
6.66 785
6. 87 870
24
7
5. 53 067
6. 51 165
6. 78 938
63
37
6. 25 378
6.67 235
6. 88 147
23
8
5.58 866
6.51808
6. 79 278
62
38
6. 26 536
6.67 680
6.88 423
22
5.63 982
6.52 442
6.79 616
61
39
6.27 664
6.68 121
6.88 697
21
10
5.68 557
6. 53 067
6. 79 952
50
40
6.28 763
6.68 557
6.88 969
20
11
5. 72 697
6.53 683
6.80 285
40
41
6.29 836
6.68 990
6. 89 240
19
12
5. 76 476
6. 54 291
6.80 615
48
42
6.30 882
6. 69 418
6. 89 509
18
13
5. 79 952
6.54 890
6.80 943
47
43
6.31904
6.69 841
6. 89 776
17
U
5. 83 170
6.55 481
6.81268
46
44
6.32 903
6. 70 261
6.90 042
16
16
5.86167
6.56064
6. 81 591
46
46
6.33 879
6. 70 676
6.90 306
16
16
5.88 969
6. 56 639
6.81911
44
46
6.34 833
6.71088
6.90 568
14
17
5.91602
6. 57 207
6. 82 230
43
47
6. 35 767
6. 71 496
6.90 829
13
18
5.94 085
6.57 767
6.82 545
42
48
6.36 682
6.71900
6.91088
12
19
5.96 433
6.58 320
6.82 859
41
49
6.37 577
6. 72 300
6.91346
11
20
5.98 660
6.58 866
6. 83 170
40
50
6.38454
6. 72 697
6.91602
10
21
6.00 779
6.59 406
6. 83 479
39
61
6.39 315
6.73 090
6.91857
9
22
6.02 800
6. 59 939
6.83 786
38
62
6. 40 158
6. 73 479
6.92 110
8
23
6.04 730
6.60 465
6.84 091
37
63
6.40 985
6.73 865
6. 92 362
7
24
6. 06 579
6.60 985
6.84 394
36
64
6.41797
6. 74 248
6.92 612
6
25
6.08 351
6. 61 499
6.84 694
36
66
6.42 594
6. 74 627
6.92 861
6
26
6. 10 055
6. 62 007
6.84 993
34
66
6.43 376
6.75 003
6.93 109
4
27
6. 11 694
6. 62 509
6. 85 289
33
67
6.44 145
6. 75 376
6.93 355
3
28
6. 13 273
6.63 006
6. 85 584
32
68
6.44 900
6. 75 746
6.93 599
2
20
6. 14 797
6.63 4%
6. 85 876
31
69
6.45 643
6. 76 112
6.93 843
1
80
6. 16 270
6.63 982
6. 86 167
30
00
6.46 373
6. 76 476
6.94 085
//
69'
58'
67'
ff
ft
50'
58'
57'
ff
log cot = log OOB
log sin = 10. 00 000
89°
log cos
22
0°
f ff
logdn
log tan
log 001
ff 9
9 ff
log tin
kgtaa
logOM
99 9
_
10.00000
06O
10
7.46373
7.46373
10.00000
050
10
5.68 557
5.68 557
10.00000
50
10
7.47090
7.47091
10.00000
60
20
5.98660
5.98660
10.00000
40
20
7.47 797
7.47 797
10.00000
40
30
6.16270
6.16270
10.00000
30
30
7.48491
7.48492
10.00000
80
40
6.28 763
6.28 763
10.00000
20
40
7.49175
7.49176
10.00000
20
60
6.38454
6.38454
10.00000
10
60
7.49849
7.49849
10.00000
10
1
6.46373
6.46373
10.00000
058
11
7.50 512
7.50 512
10.00000
48
10
6.53 067
6.53 067
10.00000
50
10
7. 51 165
7. 51 165
10.00000
60
20
6.58866
6.58866
10.00000
40
20
7.51808
7.51809
10.00000
40
30
6.63 982
6.63 982
10.00000
30
30
7.52442
7.52443
10.00000
80
40
6.68 557
6.68557
10.00000
20
40
7.53 067
7.53067
10.00000
20
60
6.72 697
6.72697
10.00000
10
50
7.53683
7.53 683
10.00000
10
2
6.76476
6.76476
10.00000
058
12
7.54 291
7.54 291
10.00000
048
10
6.79952
6.79952
10.00000
50
10
7.54890
7.54 890
10.00000
60
20
6.83170
6. 83 170
10.00000
40
20
7.55 481
7. 55 481
10.00000
40
30
6. 86 167
6. 86 167
10.00000
30
30
7.56064
7.56064
10.00000
80
40
6.88%9
6.88%9
10.00000
20
40
7.56639
7.56639
10.00000
20
50
6.91602
6.91602
10.00000
10
50
7.57 206
7.57207
10.00000
10
8
6.94085
6.94085
10.00000
57
18
7. 57 767
7.57 767
10.00000
47
10
6.96433
6.96433
10.00000
50
10
7.58320
7.58320
10.00000
60
20
6.98660
6.98661
10.00000
40
20
7.58866
7.58 867
10.00000
40
30
7.00779
7.00779
10.00000
30
30
7.59406
7.59406
10.00000
30
40
7.02800
7.02 800
10.00000
20
40
7.59939
7.59939
10.00000
20
60
7.04 730
7.04 730
10.00000
10
50
7.60465
7.60466
10.00000
10
4
7.06 579
7.06579
10.00000
056
14
7.60985
7.60986
10.00000
046
10
7.08351
7.08 352
10.00000
50
10
7. 61 499
7. 61 500
10.00000
50
20
7.10055
7.10055
10.00000
40
20
7.62 007
7.62 008
10.00000
40
30
7. 11 694
7.11694
10.00000
30
30
7.62 509
7.62 510
10.00000
30
40
7. 13 273
7. 13 273
10.00000
20
*40
7.63 006
7.63 006
10.00000
20
60
7.14 797
7. 14 797
10.00000
10
50
7.63 496
7.63497
10.00000
10
5
7.16270
7.16270
10.00000
055
15
7.63 982
7.63 982
10.00000
045
10
7.17694
7.17 694
10.00000
50
10
7.64 461
7.64462
10.00000
60
20
7.19072
7.19073
10.00000
40
20
7.64 936
7.64937
10.00000
40
30
7.20409
7.20409
10.00000
30
30
7.65 406
7.65 406
10.00000
30
40
7. 21 705
7. 21 705
10.00000
20
40
7.65 870
7. 65 871
10.00000
20
60
7.22964
7.22 964
10.00000
10
60
7.66330
7.66330
10.00000
10
6
7. 24 188
7.24188
10.00000
54
16
7.66 784
7.66 785
10.00000
044
10
7. 25 378
7.25 378
10.00000
60
10
7. 67 235
7.67 235
10.00000
50
20
7.26536
7.26536
10.00000
40
20
7.67 680
7.67680
10.00000
40
30
7.27664
7.27664
10.00000
30
30
7.68121
7. 68 121
10.00000
30
40
7. 28 763
7.28 764
10.00000
20
40
7.68 557
7.68 558
9.99999
20
50
7.29836
7.29836
10.00000
10
50
7.68989
7.68990
9.99999
10
7
7.30882
7.30882
10.00000
058
17
7.69417
7.69418
9.99999
48
10
7.31904
7.31904
10.00000
60
10
7.69 841
7.69842
9.99999
60
20
7.32 903
7.32903
10.00000
40
20
7. 70 261
7.70 261
9.99999
40
30
7. 33 879
7.33 879
10.00000
30
30
7.70 676
7.70677
9.99999
30
40
7.34833
7.34 833
10.00000
20
40
7.71088
7.71088
9.99999
20
50
7.35 767
7.35 767
10.00000
10
50
7.71496
7. 71 496
9.99999
10
8
7.36682
7.36682
10.00000
52
18
7.71900
7.71900
9.99999
42
10
7.37 577
7. 37 577
10.00000
60
10
7.72 300
7.72301
9.99999
60
20
7.38454
7.38455
10.00000
40
20
7.72 697
7.72 697
9.99999
40
30
7.39314
7.39 315
10.00000
30
30
7.73 090
7.73 090
9.99999
30
40
7.40158
7. 40 158
10.00000
20
40
7.73 479
7.73 480
9.99999
20
50
7.40985
7.40985
10.00000
10
50
7. 73 865
7.73 866
9.99999
10
9
7.41797
7. 41 797
10.00000
51
10
7. 74 248
7.74 248
9.99999
41
10
7.42 594
7.42 594
10.00000
50
10
7.74 627
7.74 628
9.99999
50
20
7.43 376
7. 43 376
10.00000
40
20
7.75 003
7.75 004
9.99999
40
30
7.44145
7.44145
10.00000
30
30
7. 75 376
7. 75 377
9.99999
30
40
7.44900
7.44 900
10.00000
20
40
7. 75 745
7. 75 746
9.99999
20
50
7. 45 ^3
7. 45 643
10.00000
10
60
7. 76 112
7.76113
9.99999
10
10
7.46373
7.46373
10.00000
50
2O0
7.76475
7.76476
9.99999
40
f 9f
log cos
log cot
log Bin
99 f
9 99
log 008
log oot
logsln
ff f
89'
f ft
li^uii
kg tan
leg oot
ff f
f ff
logiin
log taa
log 008
ff f
9O0
7.76475
7.76476
9.99999
04O
8O0
7.94084
7.94086
9.99998
080
10
7.76836
7.76837
9.99999
50
10
7.94325
7.94326
9.99998
60
20
7. 77 193
7. 77 194
9.99999
40
20
7.94564
7.94 566
9.^998
40
SO
7.77 548
7.77 549
9.99999
30
80
7.94802
7.94804
9.99998
30
40
7.77899
7.77900
9.99999
20
40
7.95039
7.95 040
9.99998
20
60
7.78248
7.78249
9.99999
10
60
7.95 274
7.95 276
9.99998
10
2i
7.78594
7.78595
9,99999
089
81
7.95 508
7.95 510
9.99998
29
10
7.78938
7.78938
9.99999
50
10
7.95 741
7.95 743
9.99998
60
20
1.19 zn
7.79279
9.99999
40
20
7.95 973
7.95 974
9.99998
40
30
7.79616
7.79617
9.99999
30
80
7.96203
7.96205
9.99998
30
40
7.79952
7.79952
9.99999
20
40
7.96432
7.96434
9.99998
20
60
7.80284
7.80285
9.99999
10
60
7.96660
7.96662
9.99998
10
280
7.80615
7.80615
9.99999
088
820
7.% 887
7.96889
9.99998
28
10
7.80942
7.80943
9.99999
60
10
7. 97 113
7.97114
9.99998
60
20
7.81268
7. 81 269
9.99999
40
20
7.97337
7.97339
9.99998
40
80
7.81591
7. 81 591
9.99999
30
80
7. 97 560
7.97562
9.99998
30
40
7.81911
7.81912
9.99999
20
40
7.97 782
7.97 784
9.99998
20
60
7.82229
7.82230
9.99999
10
60
7.98003
7.98005
9.99998
10
280
7.82 545
7.82 546
9.99999
087
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n
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r
f
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logoot
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log sin
f
h^'
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41°
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f
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Q
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10—
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f
r logiiB
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f
o
81694
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60
o
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1
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69
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68
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67
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67
4
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66
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66
6
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56
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94 095
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63
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82 649
95 622
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50
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u
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94 222
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12
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13
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14
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46
14
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46
16
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94 299
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16
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46
16
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94 324
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44
16
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44
17
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94 350
05 650
87 590
43
17
82 788
95 875
04 125
86 913
43
18
81955
94 375
05 625
87 579
42
18
82 802
95 901
04 099
86 902
42
19
81 %9
94 401
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87 568
41
19
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95 926
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41
20
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94 426
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40
20
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04 048
86 879
40
21
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94 452
05 548
87 546
39
21
82 844
95 977
04 023
86 867
39
22
82 012
94 477
05 523
87 535
38
22
82 858
%002
03 998
86 855
88
23
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87 524
37
23
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37
24
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36
24
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03 947
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86
26
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85
26
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96 078
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86 821
36
26
82 069
94 579
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87 490
34
26
82 913
%104
03 896
86 809
34
27
82 084
94 604
05 396
87 479
33
27
82 927
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03 871
86 798
33
28
82 098
94 630
05 370
87 468
32
28
82 941
96155
03 845
86 786
32
29
82112
94 655
05 345
87 457
31
29
82 955
96180
03 820
86 775
31
30
82 126
94 681
05 319
87 446
30
30
82 968
96 205
03 795
86 763
30
31
82 141
94 706
05 294
87 434
29
31
82 982
%231
03 769
86 752
29
32
82 155
94 732
05 268
87 423
28
32
82 996
%256
03 744
86 740
28
33
82 169
94 757
05 243
87 412
27
33
83 010
%281
03 719
86 728
27
34
82 184
94 783
05 217
87 401
26
34
83 023
96 307
03 693
86 717
26
35
82 198
94 808
05 192
87 390
25
35
83 037
96 332
03 668
86 705
26
36
82 212
94 834
05 166
87 378
24
36
83 051
96 357
03 643
86 694
24
37
82 226
94 859
05 141
87 367
23
37
83 065
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03 617
86 682
23
38
82 240
94 884
05 116
87 356
22
36
83 078
%408
03 592
86 670
22
39
82 255
94 910
05 090
87 345
21
39
83 092
96 433
03 567
86 659
21
40
82 269
94 935
05 065
87 334
20
40
83 106
96 459
03 541
86 647
20
41
82 283
94 961
05 039
87 322
19
41
83 120
96 484
03 516
86 635
19
42
82 297
94 986
05 014
87 311
18
42
83 133
96 510
03 490
86 624
18
43
82 311
95 012
04 988
87 300
17
43
83 147
96 535
03 465
86 612
17
44
82 326
95 037
04 963
87 288
16
44
83 161
%560
03 440
86 600
16
45
82 340
95 062
04 938
87 277
15
45
83 174
96 586
03 414
86 589
15
46
82 354
95 088
04 912
87 266
14
46
83 188
96 611
03 389
86 577
14
47
82 368
95 113
04 887
87 255
13
47
83 202
96 636
03 364
86 565
13
48
82 382
95 139
04 861
87 243
12
48
83 215
%662
03 338
86 554
12
49
82 396
95 164
04 836
87 232
11
49
83 229
96 687
03 313
86 542
11
60
82 410
95 190
04 810
87 221
10
SO
83 242
96 712
03 288
86 530
10
51
82 424
95 215
04 785
87 209
9
61
83 256
96 738
03 262
86 518
9
62
82 439
95 240
04 760
87 198
8
52
83 270
96 763
03 237
86 507
8
63
82 453
95 266
04 734
87 187
7
53
83 283
96 788
03 212
86 495
7
64
82 467
95 291
04 709
87 175
6
54
83 297
96 814
03 186
86 483
6
55
82 481
95 317
04 683
87164
5
55
83 310
96 839
03 161
86 472
5
66
82 495
95 342
04 658
87 153
4
56
83 324
96 864
03 136
86 460
4
57
82 509
95 368
04 632
87 141
3
57
83 338
96 890
03 110
86 448
3
58
82 523
95 393
04 607
87 130
2
58
83 351
96 915
03 085
86 436
2
59
82 537
95 418
04 582
87119
1
59
83 365
%940
03 060
86 425
1
60
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95 444
9
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10-
87 107
60
83 378
n
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86 413
A
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logoot
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f
f
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— 10
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r
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47c
43°
44°
49
f
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bgcos
f
f
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f
g
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o
83 378
96 966
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59
1
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69
2
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68
2
84 203
98 534
01466
85 669
58
3
83 419
97 042
02 958
86 377
57
3
84 216
98 560
01440
85 657
67
4
83 432
97 067
02 933
86 366
66
4
84 229
98 585
01415
85 645
66
6
83 446
97 092
02 908
86 354
56
5
84 242
98 610
01390
85 632
55
6
83 459
97 118
02 882
86 342
54
6
84 255
98 635
01365
85 620
54
7
83 473
97 143
02 857
86 330
63
7
84 269
98 661
01339
85 608
53
8
83 486
97 168
02 832
86 318
62
8
84 282
98 686
01314
85 596
52
9
83i00
97193
02 807
86 306
61
9
84 295
98 711
01289
85 583
51
10
83 513
97 219
02 781
86 295
50
10
84 308
9&737
01263
85 571
60
11
83 527
97 244
02 756
86 283
49
11
84 321
98 762
01238
85 559
49
12
83 540
97 269-
02 731
86 271
48
12
84 334
98 787
01213
85 547
48
13
83 554
97 295
02 705
86 259
47
13
84 347
98 812
01188
85 534
47
14
83 567
97 320
02 680
86 247
46
14
84 360
98 838
01162
85 522
46
15
83 581
97 345
02 655
86 235
45
15
84 373
98 863
01137
85 510
45
16
83 594
97 371
02 629
86 223
44
16
84 385
98 888
01112
85 497
44
17
83 608
97 396
02 604
86 211
43
17
84 398
98 913
01087
85 485
43
18
83 621
97 421
02 579
86 200
42
18
84 411
98 939
01061
85 473
42
19
83 634
97 447
02 553
86 188
41
19
84 424
98 964
01036
85 460
41
20
83 648
97 472
02 528
86 176
40
20
84 437
98 989
01011
85 448
40
21
83 661
97 497
02 503
86 164
39
21
84 450
99 015
00 985
85 436
39
22
83 674
97 523
02 477
86 152
38
22
84 463
99 040
00 960
85 423
38
23
83 688
97 548
02 452
86 140
37
23
84 476
99 065
00 935
85 411
37
24
83 701
97 573
02 427
86128
36
24
84 489
99 090
00 910
85 399
36
25
83 71i
97 598
02 402
86 116
35
26
84 502
99 116
00 884
85 386
35
26
83 728
97 624
02 376
86104
34
26
84 515
99141
00 859
85 374
34
27
83 741
97 649
02 351
86 092
33
27
84 528
99 166
00 834
85 361
33
28
83 755
97 674
02 326
86 080
32
28
84 540
99191
00 809
85 349
32
29
83 768
97 700
02 300
86 068
31
29
84 553
99 217
00 783
85 337
31
30
83 781
97 725
02 275
86 056
do
30
84 566
99 242
00 758
85 324
30
31
83 79i
97 750
02 250
86 044
29
31
84 579
99 267
00 733
85 312
29
32
83 808
97 776
02 224
86 032
28
32
84 592
99 293
00 707
85 299
28
33
83 821
97 801
02 199
86 020
27
38
84 605
99 318
00 682
85 287
27
34
83 834
97 826
02 174
86 008
26
34
84 618
99 343
00 657
85 274
26
35
83 848
97 851
02 149
85 9%
26
35
84 630
99 368
00 632
85 262
25
36
83 861
97 877
02 123
85 984
24
36
84 643
99 394
00 606
85 250
24
37
83 874
97 902
02 098
85 972
23
37
84 656
99 419
00 581
85 237
23
38
83 887
97 927
02 073
85 960
22
38
84 669
99 444
00 556
85 225
22
39
83 901
97 953
02 047
85 948
21
39
84 682
99 469
00 531
85 212
21
40
83 914
97 978
02 022
85 936
20
40
84 694
99 495
00 505
85 200
20
41
83 927
98003
01997
85 924
19
41
84 707
99 520
00 480
85 187
19
42
83 940
98 029
01971
85 912
18
42
84 720
99 545
00 455
85 175
18
43
83 954
98 054
01946
85 900
17
43
84 733
99 570
00 430
85 162
17
44
83 %7
98 079
01921
85 888
16
44
84 745
99 596
00 404
85 150
16
45
83 980
98 104
01896
85 876
15
45
84 758
99 621
00 379
85 137
15
46
83 993
98 130
01870
85 864
14
46
84 771
99 646
00 354
85 125
14
47
84 006
98 155
01845
85 851
13
47
84 784
99 672
00 328
85 112
13
48
84 020
98 180
01820
85 839
12
48
84 796
99 697
00 303
85 100
12
49
84 033
98 206
01794
85 827
11
49
84 809
99 722
00 278
85 087
11
60
84 046
98 231
01769
85 815
10
60
84 822
99 747
00 253
85 074
10
51
84 059
98 256
01744
85 803
9
51
84 835
99 773
00 227
85 062
9
52
84 072
98 281
01719
85 791
8
52
84 847
99 798
00 202
85 049
8
53
84 085
98 307
01693
85 779
7
53
84 860
99 823
00177
85 037
7
54
84 098
98 332
01668
85 766
6
54
84 873
99 848
00152
85 024
6
55
84 112
98 357
01643
85 754
5
55
84 885
99 874
00 126
85 012
5
56
84 125
98 383
01617
85 742
4
56
84 898
99 899
00101
84 999
4
57
84 138
98 408
01592
85 730
3
57
84 911
99 924
00 076
84 986
3
58
84 151
98 433
01567
85 718
2
58
84 923
99 949
00 051
84 974
2
59
84 164
98 458
01542
85 706
1
59
84 936
99 975
00 025
84 961
1
60
84 177
98 484
01516
85 693
O
eo
84 949
00 000
00 000
84 949.
^
n
in
A
n
lA
n
f
log 008
log cot
— — xu —
log tan
log Bin
/
f
9
log 008
logoot
— lU —
log tan
log sin
f
46'
46'
50
TABLE
IV.
Fob Dstsrmiking with Greater Accuract thah can be done bt
MEANS OF Table UI. :
1. log sin J log tan, and log cot^ when the angle is between O"" and 2^ ;
2. log co«, log tan, and log cot, when the angle is between 88® and 90* ;
S. The value of the angle when the logarithm of the fhnction does not
He between the limits 8. 54 684 and U. 45 816.
♦ ■ '
F0BMULA8 FOR THE USE OF THE NUMBERS 8 AND T.
I. When the angle a is between 0"* and 2** :
log sin a = log o" + S.
log a" = log sin a^ 3j 1
logtano = logo"+ r.
= log tana- r, 1
log cot a = colog tan a. l = colog cot a — r.
n. W hen the angle a is between 88° and 90° :
log cos a = log (90''-a)" + s.
log (90O-a)"= log cos a- 5,
log cot a = log (90O-a)"+ T.
= log cot a— r,
log tan a = colog cot o.
= colog tan a— r,
and o = 90*-(9<y»-a).
VALrES OF S AND T.
tt"
8
loglfaltt
tt"
T
log tan tt
tt
T
logtUItt
__
^_
5146
8.39 713
4.68557
4.68 557
4.68567
2409
4.68556
8.06 740
200
4.68 558
6.98660
5 424
4.68568
8.41999
3417
4.68 555
8.21920
1726
4. 68 559
7.92 263
5689
4.68569
8.44072
3823
4.68555
8.26 795
2432
4.68 560
8.07156
5 941
4.68570
8.45 955
4190
4.68554
8.30776
2976
4.68 561
8. 15 924
6184
4.68 571
8.47697
4840
4.68 553
8.37038
3434
4.68 562
8. 22 142
6417
4.68 572
8.49305
5 414
4.68552
8.41904
3 838
4.68 563
8.26973
6642
4.68573
8.50802
5 932
4.68551
8.45 872
4204
4.68 564
8.30930
6859
4.68 574
8. 52 200
6408
4.68550
8.49 223
4 540
4.68 565
8.34 270
7070
4.68 575
8.53 516
6633
4.68 5i0
8. 50 721
4699
4.68 565
8. 35 766
7173
4.68 575
8. 54 145
6851
4.68549
8.52125
4 853
4.68 566
8.37167
7274
8.54 753
7 267
8.54 684
5146
8.39 713
tt"
8
logiintt
tt"
T
log tan tt
tt
T
logtaatt
61
TABLE V.
THE JSATUKAJL, VAJLUJUS
or
S[M&, COSINES, TANGENTS, AND COTANGENTS,
IN THE UOTT CIRCLB.
0° 8°
o p
■h
tan
oot
008
f o
0.0000
0.0000
infinite
1.0000
09O
10
0.0029
0.0029
343.7737
1.0000
60
20
0.0058
0.0058
171.8854
1.0000
40
30
0.0087
0.0087
114.5887
1.0000
30
40
0.0116
0.0116
85.9398
0.9999
20
60
0.0145
0.0145
68.7501
0.9999
10
1
0.0175
0.0175
57.2900
0.9998
80
o f
CM
oot
tan
tin
r o
o f
tin
tan
oot
008
f
o f
8in
tan
oot
008
f o
1
0.0175
0.0175
57.2900
0.9998
080
5
0.0872
0.0875
11.4301
0.9962
085
10
0.0204
0.0204
49.1039
0.9998
60
10
0.0901
0.0904
11.0594
0.9959
60
20
0.0233
0.0233
42.9641
0.9997
40
20
0.0929
0.0934
10.7119
0.9957
40
80
0.0262
0.0262
38.1885
0.9997
30
30
0.0958
0.0963
10.3854
0.9954
30
40
0.0291
0.0291
34.3678
0.9996
20
40
0.0987
0.0992
10.0780
0.9951
20
60
0.0320
0.0320
31.2416
0.9995
10
60
0.1016
0.1022
9.7882
0.9948
10
2
0.0349
0.0349
28.6363
0.9994
88
6
0.1045
0.1051
9.5144
0.9945
084
10
0.0378
0.0378
26.4316
0.9993
60
10
0.1074
0.1080
9.2553
0.9942
60
20
0.0407
0.0407
24.5418
0.9992
40
20
0.1103
0.1110
9.0098
0.9939
40
30
0.0436
0.0437
22.9038
0.9990
30
30
0.1132
0.1139
8.7769
0.9936
30
40
0.0465
0.0466
21.4704
0.9989
20
40
0.1161
0.1169
8.5555
0.9932
20
60
0.0494
0.0495
20.2056
0.9988
10
60
0.1190
0.1198
8.3450
0.9929
10
3
0.0523
0.0524
19.0811
0.9986
87
7
0.1219
0.1228
8.1443
0.9925
083
10
0.0552
0.0553
18.0750
0.9985
60
10
0.1248
0.1257
7.9530
0.9922
60
20
0.0581
0.0582
17.1693
0.9983
40
20
0.1276
0.1287
7.7704
0.9918
40
30
0.0610
0.0612
16.3499
0.9981
30
30
0.1305
0.1317
7.5958
0.9914
30
40
0.0640
0.0641
15.6048
0.9980
20
40
0.1334
0.1346
7.4287
0.9911
20
60
0.0669
0.0670
14.9244
0.9978
10
60
0.1363
0.1376
7.2687
0.9907
10
4
0.0698
0.0699
14.3007
0.9976
86
8
0.1392
0.1405
7.1154
0.9903
82
10
0.0727
0.0729
13.7267
0.9974
50
10
0.1421
0.1435
6.9682
0.9899
60
20
0.0756
0.0758
13.1969
0.9971
40
20
0.1449
0.1465
6.8269
0.9894
40
30
0.078i
0.0787
12.7062
0.9969
30
30
0.1478
0.1495
6.6912
0.9890
80
40
0.0814
0.0816
12.2505
0.9%7
20
40
0.1507
0.1524
6.5606
0.9886
20
60
0.0843
0.0846
11.8262
0.9964
10
60
0.1536
0.1554
6.4348
0.9881
10
5
0.0872
0.0875
11.4301
0.9962
085
9
0.1564
0.1584
6.3138
0.9877
81
O f
008
oot
tan
Bin
f o
o f
008
cot
tan
sin
f o
81° -89'
52
9°-
26°
O f
lin
ton
oot
008
/ o
o t
lin
tan
oot
008
f o
90
0.1564
0.1584
6.3138
0.9877
81
18
03090
03249
3.0777
0.9511
72
10
0.1593
0.1614
6.1970
0.9872
60
10
0.3118
03281
3.0475
0.9502
60
20
0.1622
0.1644
6.0844
0.9868
40
20
03145
0.3314
3.0178
0.9492
40
30
0.1650
0.1673
5.9758
0.9863
30
30
03173
0.3346
2.9887
0.9483
30
40
0.1679
0.1703
5.8708
0.9858
20
40
03201
03378
2.9600
0.9474
20
60
0.1708
0.1733
5.7694
0.9853
10
50
03228
0.3411
2.9319
0.9465
10
10
0.1736
0.1763
5.6713
0.9848
08O
19
0.3256
0.3443
2.9042
0.9455
71
10
0,1765
0.1793
5.5764
0.9843
50
10
0.3283
0.3476
2.8770
0.9446
60
20
0.1794
0.1823
5.4845
0.9838
40
20
0.3311
03508
2.8502
0.9436
40
30
0.1822
0.1853
5.3955
0.9833
30
30
0.3338
03541
2.8239
0.9426
30
40
0.1851
0.1883
5.3093
0.9827
20
40
0.3365
03574
2.7980
0.9417
20
60
0.1880
0.1914
5.2257
0.9822
10
60
0.3393
0.3607
2.7725
0.9407
10
11
0.1908
0.1944
5.1446
0.9816
Q 79
20
0.3420
0.3640
2.7475
0.9397
7O
10
0.1937
0.1974
5.0658
0.9811
60
10
03448
0.3673
2.7228
0.9387
60
20
0.1965
0.2004
4.9894
0.9805
40
20
0.3475
03706
2.6985
0.9377
40
30
0.1994
0.2035
4.9152
0.9799
30
30
03502
0.3739
2.6746
0.9367
30
40
0.2022
0.2065
4.8430
0.9793
20
40
03529
03772
2.6511
0.9356
20
50
0.2051
0.2095
4.7729
0.9787
10
60
03557
03805
2.6279
0.9346
10
12
0.2079
0.2126
4.7046
0.9781
78
21
03584
03839
2.6051
0.9336
069
10
0.2108
0.2156
4.6382
0.9775
60
10
0.3611
03872
2.5826
0.9325
50
20
0.2136
0.2186
4.5736
0.9769
40
20
03638
0.3906
2.5605
0.9315
40
30
0.2164
0.2217
4.5107
0.9763
30
30
03665
0.3939
2.5386
0.9304
30
40
0.2193
0.2247
4.4494
0.9757
20
40
03692
03973
2.5172
0.9293
20
50
0.2221
0.2278
4.3897
0.9750
10.
60
03719
0.4006
2.4960
0.9283
10
13
0.2250
0.2309
4.3315
0.9744
77
220
03746
0.4040
2.4751
0.9272
068
10
0.2278
0.2339
4.2747
0.9737
50
10
03773
0.4074
2.4545
0.9261
50
20
0.2306
0.2370
4.2193
0.9730
40
20
03800
0.4108
2.4342
0.9250
40
30
0.2334
0.2401
4.1653
0.9724
30
30
0.3827
0.4142
2.4142
0.9239
30
40
0.2363
0.2432
4.1126
0.9717
20
40
0.3854
0.4176
23945
0.9228
20
* 50
0.2391
0.2462
4.0611
0.9710
10
60
0.3881
0.4210
23750
0.9216
10
14
0.2419
0.2493
4.0108
0.9703
76
230
0.3907
0.4245
23559
0.9205
67
10
0.2447
0.2524
3.9617
0.%96
60
10
03934
0.4279
2.3369
0.9194
60
20
0.2476
0.2555
3.9136
0.9689
40
20
0.3961
0.4314
23183
0.9182
40
30
0.2504
0.2586
3.8667
0.9681
30
30
03987
0.4348
2.2998
0.9171
30
40
0.2532
0.2617
3.8208
0.%74
20
40
0.4014
0.4383
2.2817
0.9159
20
60
0.2560
0.2648
3.7760
0.9667
10
60
0.4041
0.4417
2.2637
0.9147
10
15
0.2588
0.2679
3.7321
0.9659
76
240
0.4067
0.4452
2.2460
0.9135
66
10
0.2616
0.2711
3.6891
0.%52
60
10
0.4094
0.4487
2.2286
0.9124
60
20
0.2644
0.2742
3.6470
0.9644
40
20
0.4120
0.4522
2.2113
0.9112
40
30
0.2672
0.2773
3.6059
0.9636
30
30
0.4147
0.4557
2.1943
0.9100
30
40
0.2700
0.2805
3.5656
0.9628
20
40
0.4173
0.4592
2.1775
0.9088
20
50
0.2728
0.2836
3.5261
0.9621
10
60
0.4200
0.4628
2.1609
0.9075
10
16
0.2756
0.2867
3.4874
0.9613
74
250
0.4226
0.4663
2.1445
0.9063
65
10
0.2784
0.2899
3.4495
0.9605
50
10
0.4253
0.4699
2.1283
0.9051
50
20
0.2812
0.2931
3.4124
0.9596
40
20
0.4279
0.4734
2.1123
0.9038
40
30
0.2840
0.2%2
3.3759
0.9588
30
30
0.4305
0.4770
2.0965
0.9026
30
40
0.2868
0.2994
3.3402
0.9580
20
40
0.4331
0.4806
2.0809
0.9013
20
50
0.2896
03026
3.3052
0.9572
10
50
0.4358
0.4841
2.0655
0.9001
10
17
0.2924
0.3057
3.2709
0.9563
73
26
0.4384
0.4877
2.0503
0.8988
64
10
0.2952
0.3089
3.2371
0.9555
60
10
0.4410
0.4913
2.0353
0.8975
50
20
0.2979
0.3121
3.2041
0.9546
40
20
0.4436
0.4950
2.0204
0.8962
40
30
0.3007
0.3153
3.1716
0.9537
30
30
0.4462
0.4986
2.0057
0.8949
30
40
0.3035
0.3185
3.1397
0.9528
20
40
0.4488
0.5022
1.9912
0.8936
20
60
0.3062
0.3217
3.1084
0.9520
10
60
0.4514
0.5059
1.9768
0.8923
10
18
0.3090
0.3249
3.0777
0.9511
72
27
0.4540
0.5095
1.9626
0.8910
63
o t
001
oot
tan
Bin
f o
O f
008
oot
tan
sin
f o
63° -80=
27°-
-44
58
o /
Or
Un
oot
COB
f o
o f
Or
tan
oot
001
f o
27
0.4540
0.5095
l.%26
0.8910
68
86
0.5878
0.7265
13764
0.8090
054
10
0.4566
0.5132
1.9486
0.8897
50
10
0.5901
0.7310
13680
0.8073
50
20
0.4592
0.5]69
1.9347
0.8884
40
20
0.5925
0.7355
13597
0.8056
40
SO
0.4617
0.5206
1.9210
0.8870
SO
SO
0.5948
0.7400
13514
0.8039
30
40
0.4643
0.5243
1.9074
0.8857
20
40
0.5972
0.7445
1.3432
0.8021
20
60
0.4669
0.5280
1.8940
0.8843
10
60
0.5995
0.7490
1.3351
0.8004
10
28
0.4695
0.5317
1.8807
0.8829
062
87
0.6018
0.7536
13270
0.7986
053
10
0.4720
0.5354
1.8676
0.8816
60
10
0.6041
0.7581
13190
0.7969
50
20
0.4746
0.5392
1.8546
0.8802
40
20
0.6065
0.7627
13111
0.7951
40
30
0.4772
0.5430
1.8418
0.8788
SO
30
0.6088
0.7673
1.3032
0.7934
30
40
0.4797
0.5467
1.8291
0.8774
20
40
0.6111
0.7720
1.2954
0.7916
20
50
0.4823
0.5505
1.8165
0.8760
10
50
0.6134
0.7766
1.2876
0.7898
10
29
0.4848
0.5543
1.8040
0.8746
61
380
0.6157
0.7813
1.2799
0.7880
52
10
0.4874
0.5581
1.7917
0.8732
60
10
0.6180
0.7860
1.2723
0.7862
50
20
0.4899
0.5619
1.7796
0.8718
40
20
0.6202
0.7907
1.2647
0.7844
40
30
0.4924
0.5658
1.7675
0.8704
30
30
0.6225
0.7954
1.2572
0.7826
30
40
0.4950
0.5696
1.7556
0.8689
20
40
0.6248
0.8002
1.2497
0.7808
20
50
0.4975
0.6735
1.7437
0.8675
10
50
0.6271
0.8050
1.2423
0.7790
10
aoo
0.5000
0.5774
1.7321
0.8660
60
890
0.6293
0.8098
1.2349
0.7771
51
10
0.5025
0.5812
1.7205
0.8646
50
10
0.6316
0.8146
1.2276
0.7753
50
20
0.5050
O..S851
1.7090
0.8631
40
20
0.6338
0.8195
1.2203
0.7735
40
30
0.5075
0.5890
1.6977
0.8616
SO
30
0.6361
0.8243
1.2131
0.7716
30
40
0.5100
0.5930
1.6864
0.8601
20
40
0.6383
0.8292
1.2059
0.7698
20
60
0.5125
0.5%9
1.6753
0.8587
10
60
0.6406
0.8342
1.1988
0.7679
10
ai
0.5150
0.6000
1.6643
0.8572
058
4O0
0.6428
0.8391
1.1918
0.7660
50
10
0.5175
0.6048
1.6534
0.8557
50
10
0.6450
0.8441
1.1847
0.7642
50
20
0.5200
0.6088
1.6426
0.8542
40
20
0.6472
0.8491
1.1778
0.7623
40
30
0.5225
0.6128
1.6319
0.8526
SO
SO
0.6494
0.8541
1.1708
0.7604
30
40
0.5250
0.6168
1.6212
0.8511
20
40
0.6517
0.8591
1.1640
0.7585
20
60
0.5275
0.6208
1.6107
0.8496
10
50
0.6539
0.8642
1.1571
0.7566
10
82
0.5299
0.6249
1.6003
0.8480
058
41
0.6561
0.8693
1.1504
0.7547
49
10
0.5324
0.6289
1.5900
0.8465
60
10
0.6583
0.8744
1.1436
0.7528
50
20
0.5348
0.6330
1.5798
0.8450
40
20
0.6604
0.87%
1.1369
0.7509
40
30
0.5373
0.6371
1.5697
0.8434
30
30
0.6626
0.8847
1.1303
0.7490
30
40
0.5398
0.6412
1.5597
0.8418
20
40
0.6648
0.8899
1.1237
0.7470
20
50
0.5422
0.6453
1.5497
0.8403
10
60
0.6670
0.8952
1.1171
0.7451
10
33
0.5446
0.6494
1.5399
0.8387
67
42
0.6691
0.9004
1.1106
0.7431
48
10
0.5471
0.6536
1.5301
0.8371
60
10
0.6713
0.9057
1.1041
0.7412
60
20
0.5495
0.6577
1.5204
0.8355
40
20
0.6734
0.9110
1.0977
0.7392
40
30
0.5519
0.6619
1.5108
0.8339
30
30
0.6756
0.9163
1.0913
0.7373
30
40
0.5544
0.6661
1.5013
0.8323
20
40
0.6777
0.9217
1.0850
0.7353
20
60
0.5568
0.6703
1.4919
0.8307
10
60
0.6799
0.9271
1.0786
0.7333
10
340
0.5592
0.6745
1.4826
0.8290
056
430
0.6820
0.9325
1.0724
0.7314
47
10
0.5616
0.6787
1.4733
0.8274
60
10
0.6841
0.9380
1.0661
0.7294
60
20
0.5640
0.6830
1.4641
0.8258
40
20
0.6862
0.9435
1.0.S99
0.7274
40
30
0.5664
0.6873
1.4550
0.8241
30
30
0.6884
0.9490
1.0538
0.7254
30
40
0.5688
0.6916
1.4460
0.8225
20
40
0.6905
0.9545
1.0477
0.7234
20
60
0.5712
0.6959
1.4370
0.8208
10
60
0.6926
0.9601
1.0416
0.7214
10
35
0.5736
0.7002
1.4281
0.8192
055
44
0.6947
0.9657
1.0355
0.7193
46
10
0.5760
0.7046
1.4193
0.8175
60
10
0.6967
0.9713
1.0295
0.7173
50
20
0.5783
0.7089
1.4106
0.8158
40
20
0.6988
0.9770
1.0235
0.7153
40
30
0.5807
0.7133
1.4019
0.8141
30
30
0.7009
0.9827
1.0176
0.7133
30
40
0.5831
0.7177
1.3934
0.8124
20
40
0.7030
0.9884
1.0117
0.7112
20
50
0.5854
0.7221
1.3848
0.8107
10
50
0.7050
0.9942
1.0058
0.7092
10
36
0.5878
0.7265
1.3764
0.8090
054
450
0.7071
1.0000
1.0000
0.7071
45
o r
008
cot
tan
sin
p o
o f
008
cot
tan
am
f o
46° -62°
64
TABLE VL-
-Circumferences and Areas op Circles.
If ^= the radius of the circle, the circamference = «ir^.
If ^= the radius of the circle, the area
= »^'«.
If ^= the circumference of the circle, the radius = ^-ar.
If ^= the circumference of the circle, the area = — N*,
H
3ir^
xN*
h-
.^-
IS
^Tir
tN*
h-
h-*
0.00
0.0
0.000
0.00
60
314. 16
7 854
7.96
198.94
1
6.28
3.1
0.159
0.08
61
320.44
8171
8.12
206.98
2
12.57
12.6
0.318
0.32
62
326. 73
8495
8.28
215. 18
3
18. 8i
28.3
0.477
0.72
53
333.01
8825
8.44
223.53
4
25.13
50.3
0.637
1.27
64
339.29
9161
8.59
232.05
5
31.42
78.5
0.796
1.99
55
345. 58
9 503
8.75
240.72
6
37.70
113.1
0.955
2.86
66
351.86
9852
8.91
249. 55
7
43.98
153.9
1.114
3.90
57
358. 14
10207
9.07
258. 55
8
50.27
201.1
1.273
5.09
68
364.42
10568
9.23
267.70
9
56. 5i
254.5
1.432
6.45
69
370. 71
10936
9.39
277.01
10
62.83
314.2
1.592
7.%
60
376.99
11310
9.55
286.48
11
69.12
380.1
1.751
9.63
61
383.27
11690
9.71
296.11
12
75.40
452.4
1.910
11.46
62
389. 56
12076
9.87
305.90
13
81.68
530.9
2.069
13.45
63
395.84
12469
10.03
315.84
14
87.96
615.8
2.228
15.60
64
402.12
12 868
10.19
325.95
15
94. 2i
706.9
2.387
17.90
65
408.41
13 273
10.35
336. 21
16
100.53
804.2
2.546
20.37
66
414.69
13 685
10.50
346.64
17
106.81
907.9
2.706
23.00
67
420.97
14103
10.66
357. 22
18
113. 10
1017.9
2.865
25.78
68
427.26
14 527
10.82
367. 97.
19
119.38
1 134. 1
3.024
28.73
69
433. 54
14957
10.98
378.87
20
125.66
1 256. 6
3.183
31.83
70
439. 82
15 394
11.14
389. 93
21
131. 9i
1385.4
3.342
35.09
71
446.11
15 837
11.30
401.15
22
138. 23
1 520. 5
3.501
38.52
72
452.39
16286
11.46
412. 53
23
144.51
1661.9
3.661
42.10
73
458. 67
16 742
11.62
424.07
24
150.80
1809.6
3.820
45.84
74
464.96
17203
11.78
435. 77
25
157.08
1963.5
3.979
49.74
75
471. 24
17671
11.94
447.62
26
163.36
2123.7
4.138
53.79
76
477. 52
18146
12.10
459.64
27
169. 6i
2290.2
4.297
58.01
77
483.81
18 627
12.25
471. 81
28
175. 93
2463.0
4.456
62.39
78
490.09
19113
12.41
484.15
29
182. 21
2 642.1
4.615
66.92
79
496.37
19607
12.57
496.64
30
188. iO
2 827.4
4.775
71.62
80
502. 65
20106
12.73
509.30
31
194.78
3 019.1
4.934
76.47
81
508.94
20612
12.89
522. 11
32
201.06
3 217.
5.093
81.49
82
515.22
21124
13.05
535.08
33
207. 35
3 421.2
5.252
86.66
83
521. 50
21642
13.21
548. 21
34
213.63
3 631.7
5.411
91.99
84
527. 79
22167
13.37
561.50
35
219.91
3 848.5
5.570
97.48
85
534.07
22 698
13.53
574. 95
36
226. 19
4071.5
5.730
103. 13
86
540.35
23 235
13.69
588.55
37
232.48
4300.8
5.889
108.94
87
546.64
23 779
13.85
602.32
38
238. 76
4 536.5
6.048
114.91
88
552. 92
24-328
14.01
616. 25
39
245.04
4 778.4
6.207
121.04
89
559. 20
24 885
14.16
630.33
40
251. 33
5 026.5
6.366
127.32
90
565.49
25 447
14.32
644.58
41
257. 61
5 281.0
6.525
133. 77
91
571. 77
26016
14.48
658.98
42
263.89
5 541.8
6.685
140.37
92
578.05
26 590
14.64
673. 54
43
270. 18
5 808.8
6.844
147. 14
93
584.34
27172
14.80
688.27
44
276.46
6082.1
7.003
154.06
94
590.62
27 759
14.96
703. 15
45
282. 74
6361.7
7.162
161. 14
96
596.90
28353
15.12
718. 19
46
289. 03
6647.6
7.321
168.39
96
603.19
28 953
15.28
733. 39
47
295. 31
6939.8
7.480
175. 79
97
609.47
29 559
15.44
748. 74
48
301. 59
7 238.2
7.639
183.35
98
615. 75
30 172
15.60
764.26
49
307. 88
7 543.0
7.799
191.07
99
622.04
30 791
15.76
779.94
50
314. 16
7 854.0
7.958
198.94
100
6^8.32
31416
15.92
795. 77
H
2irJV
irJV»
2ir
^^'
IS
2irJV
irJW«
^-
S^*
WTiich every
A TABLE OF THB ANGIiBS
Point and Quarter Point of the Compass makes with the Meridian.
North..
Pointo.
0-14
1 II
2 48 45
5 37 30
8 26 15
11 15
Points.
1
South. 1
N. by E.
N. by ^W.
S. by E.
S.byW.
N.N.E.
N.N.-W.
i^i
14 3 45
16 52 30
19 41 15
22 30
1-y.
2 • *
S.S.E.
S.S.TAr.
N.E.byN.
N.^W.byN.
2-1^
2-%
3
25 18 45
28 7 30
30 56 15
33 45
^i
S.E.byS.
S.W. by S.
N.E.
N.TAT.
3-%
4
36 33 45
39 22 30
42 11 15
45
4
S.E.
S.TAT.
N.E.byE.
N.'W.by^W.
i^l
47 48 45
50 87 80
53 26 15
56 15
5 *
S.E.byE.
S.'W.byW.
E.N.E.
TAr.N.TAr.
P
59 8 45
61 52 30
64 41 15
67 30
6 *
E.S.E.
^W.S.TAT.
E. by N.
V7. by N.
6-V4
6-iJ
6-%
7
70 18 45
73 7 30
75 56 15
78 45
6-14
6-«2
7
E. by S.
W. by S.
East.
West.
7-%
81 33 45
84 22 30
87 11 15
90
8 *
East.
^West.
56
TABLE VIL
-TRAVERSE TABLE.
BMring.
instance 1.
Distance 2.
Distance 3.
Distance 4.
Distance 5.
Beaxissr.
o f
Lat.
Dep.
Lat.
Dep.
Lat.
Dep.
Lat.
Dep.
Lat.
Dep.
o f
015
1.000
0.004
2.000
0.009
3.000
0.013
4.000
0.017
5.000
0.022
89 45
80
1.000
0.009
2.000
0.017
3.000
0.026
4.000
0.035
5.000
0.044
30
46
1.000
0.013
2.000
0.026
3.000
0.039
4.000
0.052
5.000
0.065
15
1
1.000
0.017
2.000
0.035
3.000
0.052
3.999
0.070
4.999
0.087
89
16
1.000
0.022
2.000
0.044
2.999
0.065
3.999
0.087
4.999
O109
45
30
1.000
0.026
1.999
0.052
2.999
0.079
3.999
O105
4.998
0131
30
46
1.000
0.031
1.999
0.061
2.999
0.092
3.998
0.122
4.998
0153
15
2
0.999
0.035
1.999
0.070
2.998
0.105
3.998
O140
4.997
0.174
88
16
0.999
0.039
1.998
0.079
2.998
0.118
3.997
0157
4.996
0196
46
30
0.999
0.044
1.998
0.087
2.997
0.131
3.996
0174
4.995
0.218
30
46
0.999
0.048
1.998
0.096
2.997
0.144
3.995
0192
4.994
O240
15
8
0.999
0.052
1.997
0.105
2.996
0.157
3.995
0.209
4.993
0.262
87
16
0.998
0.057
1.997
0.113
2.995
O170
3.994
0.227
4.992
0.283
45
30
0.998
0.061
1.996
0.122
2.994
ai83
3.993
0.244
4.991
0.305
30
46
0.998
0.065
1.996
0.131
2.994
0.1%
3.991
0.262
4.989
0327
15
4
0.998
0.070
1.995
0.140
2.993
0.209
3.990
0.279
4.988
0.349
86
15
0.997
0.074
1.995
0.148
2.992
0222
3.989
0.296
4.986
0371
45
30
0.997
0.078
1.994
0.157
2.991
0235
3.988
0314
4.985
0.392
30
46
0.997
0.083
1.993
0.166
2.990
0.248
3.986
0331
4.983
0.414
15
5
0.996
0.087
1.992
0.174
2.989
0261
3.985
0349
4.981
0.436
85
16
0.996
0.092
1.992
0.183
2.987
0275
3.983
0366
4.979
0.458
45
30
0.995
0.096
1.991
0.192
2.986
0288
3.982
0383
4.977
0.479
30
46
0.995
0.100
1.990
0.200
2.985
OJOl
3.980
0.401
4.975
0.501
16
6
0.995
0.105
1.989
0.209
2.984
0.314
3.978
0.418
4.973
0.523
84
16
0.994
0.109
1.988
0.218
2.982
0327
3.976
0.435
4.970
0.544
45
30
0.994
0.113
1.987
0.226
2.981
0340
3.974
0.453
4.968
566
30
46
0.993
0.118
1.986
0.235
2.979
0353
3.972
0.470
4.965
0.588
15
7
0.993
0.122
1.985
0.244
2.978
0.366
3.970
0.487
4.963
O609
83
16
0.992
0.126
1.984
0.252
2.976
0379
3.968
0.505
4.960
0.631
45
30
0.991
0.131
1.983
0.261
2.974
0392
3.966
0522
4.957
0.653
30
46
0.991
0.135
1.982
0.270
2.973
0.405
3.963
0539
4.954
0674
15
8
0.990
0.139
1.981
0.278
2.971
0.418
3.961
0.557
4.951
0.696
82
16
0.990
0.143
1.979
0.287
2.969
0.430
3.959
0.574
4.948
0717
46
30
0.989
0.148
1.978
0.296
2.%7
0.443
3.956
0591
4.945
0.739
30
46
0.988
0.152
1.977
0.304
2.965
0.456
3.953
0.608
4.942
0.761
15
9
0.988
0.156
1.975
0.313
2.963
0.469
3.951
0.626
4.938
0782
81
16
0.987
0.161
1.974
0.321
2.961
0.482
3.948
0.643
4.935
O804
45
30
0.986
0.165
1.973
0.330
2.959
0.495
3.945
0.660
4.931
0825
30
46
0.986
0.169
1.971
0.339
2.957
0.508
3.942
0.677
4.928
0847
15
10
0.985
0.174
1.970
0.347
2.954
0.521
3.939
0695
4.924
0.868
80
16
0.984
0.178
1.968
0.356
2.952
0534
3.936
0712
4.920
O890
45
30
0.983
0.182
l.%7
0.364
2.950
0547
3.933
0729
4.916
0.911
30
46
0.982
0.187
1.965
0.373
2.947
O560
3.930
0.746
4.912
0.933
15
11
0.982
0.191
1.963
0.382
2.945
0.572
3.927
0763
4.908
0.954
79
15
0.981
0.195
1.962
0.390
2.942
0585
3.923
0.780
4.904
0.975
45
30
0.980
0.199
1.960
0.399
2.940
0598
3.920
0.797
4.900
0.997
30
45
0.979
0.204
1.958
0.407
2.937
0.611
3.916
0.815
4.895
1.018
15
12
0.978
0.208
1.956
0.416
2.934
0.624
3.913
0.832
4.891
1.040
78
16
0.977
0.212
1.954
0,424
2.932
0637
3.909
0.849
4.886
1.061
45
30
0.976
0.216
1.953
0.433
2.929
0.649
3.905
0866
4.881
1.082
30
45
0.975
0.221
1.951
0.441
2.926
0.662
3.901
0.883
4.877
1.103
16
13
0.974
0.225
1.949
0.450
2.923
0.675
3.897
0.900
4.872
1.125
77
15
0.973
0.229
1.947
0.458
2.920
0.688
3.894
0.917
4.867
1.146
45
30
0.972
0.233
1.945
0.467
2.917
0.700
3.889
0.934
4.862
1.167
30
46
0.971
0.238
1.943
0.475
2.914
0713
3.885
0.951
4.857
1.188
15
14
0.970
0.242
1.941
0.484
2.911
0.726
3.881
0.968
4.851
1.210
76
16
0.969
0.246
1.938
0.492
2.908
0738
3.877
0.985
4.846
1.231
45
30
0.968
0.250
1.936
0.501
2.904
0.751
3.873
1.002
4.841
1.252
30
46
0.967
0.255
1.934
0.509
2.901
0764
3.868
1.018
4.835
1.273
15
15
0.966
0.259
1.932
0.518
2.898
0.776
3.864
1.035
4.830
1.294
75
f
Dep.
Lat.
Dep.
Lat.
Dep.
Lat.
Dep.
Lat.
Dep.
Lat.
o /
Bearing.
Distance 1.
Distance 2.
Distance 3.
Distance 4.
Distance 5.
Bearing.
75°- 90^
ar-
16^
67
Betting.
Distance 6.
Distance 7.
Distance 8.
Distance 9.
Distance 10.
Bearing.
O f
Lat.
Dep.
Lat.
Dep.
Lat.
Dep.
Lat.
Dep.
Lat.
Dep.
o f
015
6.000
0.026
7.000
0.031
8.000
0.035
9.000
0.039
10.000
0.044
89 46
30
6.000
0.052
7.000
0.061
8.000
0.070
9.000
0.079
10.000
0.087
30
46
5.999
0.079
6.999
0.092
1S99
0.105
8.999
0118
9.999
0131
16
1
5.999
0.105
6.999
0.122
7.999
0.140
8.999
0157
9.999
0175
89
16
5.999
0.131
6.998
0.153
7.998
0.175
8.998
oi%
9.998
0218
46
30
5.998
0.157
6.998
0.183
7.997
0209
8.997
0.236
9.997
0.262
30
46
5.997
0.183
6.997
0.214
7.9%
0.244
8.9%
0275
9.995
0.305
16
2
5.9%
0.209
6.9%
0.244
7.995
0279
8.995
0.314
9.994
0.349
88
16
5.995
0.236
6.995
0.275
7.994
0.314
8.993
0.353
9.992
0.393
46
30
5.994
0.262
6.993
0.305
7.992
0.349
8.991
0.393
9.991
0.436
30
46
5.993
0.288
6.992
0.336
7.991
0.384
8.990
0.432
9.989
0.480
16
3
5.992
0.314
6.990
0.366
7.989
0.419
8.988
0.471
9.986
0523
87
16
5.990
0.340
6.989
0397
7.987
0.454
8.986
0.510
9.984
0567
46
30
5.989
0.366
6.987
0.427
7.985
0.488
8.983
0549
9.981
0.611
30
46
5.987
0.392
6.985
0.458
7.983
0.523
8.981
0.589
9.979
0.654
16
4
5.985
0.419
6.983
0.488
7.981
0558
8.978
0.628
9.976
0.698
86
15
5.984
0.445
6.981
0.519
7.978
0.593
8.975
0.667
9.973
0.741
46
30
5.982
0.471
6.978
0.549
7.975
0.628
8.972
0.706
9.969
0.785
30
46
5.979
0.497
6.976
0.580
7.973
0.662
8.%9
0745
9.966
0.828
16
5
5.977
0.523
6.973
0.610
7.970
0.697
8.966
0.784
9.%2
0.872
85
16
5.975
0.549
6.971
0.641
7.966
0732
8.%2
0.824
9.958
0.915
46
30
5.972
0.575
6.968
0.671
7.963
0767
8.959
0863
9.954
0.959
30
46
5.970
0.601
6.%5
0.701
7.960
0.802
8.955
0.902
9.950
1.002
16
6
5.967
0.627
6.%2
0.732
7.956
0.836
8.951
0.941
9.945
1.045
84
16
5.964
0.653
6.958
0.762
7.952
0.871
8.947
0.980
9.941
1.089
46
30
5.961
0.679
6.955
0.792
7.949
0.906
8.942
1.019
9.936
1.132
30
46
5.958
0.705
6.951
0.823
7.945
0.940
8.938
1.058
9.931
1.175
16
7
5.955
0.731
6.948
0.853
7.940
0.975
8.933
1.097
9.926
1.219
83
16
5.952
0.757
6.944
0.883
7.936
1.010
8.928
1.136
9.920
1.262
46
30
5.949
0.783
6.940
0.914
7.932
1.044
8.923
1.175
9.914
1.305
30
46
5.945
0.809
6.936
0.944
7.927
1.079
8.918
1.214
9.909
1.349
16
8
5.942
0.835
6.932
0.974
7.922
1.113
8.912
1.253
9.903
1.392
82
16
5.938
0.861
6.928
1.004
7.917
1.148
8.907
1.291
9.897
1.435
46
30
5.934
0.887
6.923
1.035
7.912
1.182
8.901
1.330
9.890
1.478
30
46
5.930
0.913
6.919
1.065
7.907
1.217
8.895
1.369
9.884
1.521
16
9
5.926
0.939
6.914
1.095
7.902
1.251
8.889
1.408
9.877
1.564
81
16
5.922
0.964
6.909
1.125
7.8%
1.286
8.883
1.447
9.870
1.607
46
30
5.918
0.990
6.904
1.155
7.890
1.320
8.877
1.485
9.863
1.651
30
46
5.913
1.016
6.899
1.185
7.884
1.355
8.870
1.524
9.856
1.694
15
10
5.909
1.042
6.894
1.216
7.878
1.389
8.863
1.563
9.848
1.737
80
16
5.904
1.068
6.888
1.246
7.872
1.424
8.856
1.601
9.840
1.779
46
30
5.900
1.093
6.883
1.276
7.866
1.458
8.849
1.640
9.833
1.822
30
46
5.895
1.119
6.877
1.306
7.860
1.492
8.842
1.679
9.825
1.865
16
11
5.890
1.145
6.871
1.336
7.853
1.526
8.835
1.717
9.816
1.908
79
16
5.885
1.171
6.866
1.366
7.846
1.561
8.827
1.756
9.808
1.951
46
30
5.880
1.196
6.859
1.3%
7.839
1.595
8.819
1.794
9.799
1.994
30
46
5.874
1.222
6.853
1.425
7.832
1.629
8.811
1.833
9.791
2.036
16
12
5.869
1.247
6.847
1.455
7.825
1.663
8.803
1.871
9.782
2.079
78
16
5.863
1.273
6.841
1.485
7.818
1.697
8.795
1.910
9.772
2.122
46
30
5.858
1.299
6.834
1.515
7.810
1.732
8.787
1.948
9.763
2.164
30
46
5.852
1.324
6.827
1.545
7.803
1.766
8.778
1.986
9.753
2.207
15
13
5.846
U50
6.821
1.575
7.795
1.800
8.769
2.025
9.744
2.250
77
16
5.840
1.375
6.814
1.604
7.787
1.834
8.760
2.063
9.734
2.292
46
30
5.834
1.401
6.807
1.634
7.779
1.868
8.751
2.101
9.724
2.335
30
46
5.828
1.426
6.799
1.664
7.771
1.902
8.742
2.139
9.713
2.377
15
14
5.822
1.452
6.792
1.693
7.762
1.935
8.733
2.177
9.703
2.419
76
16
5.815
1.477
6.785
1.723
7.754
1.969
8.723
2.215
9.692
2.462
46
30
5.809
1.502
6.777
1.753
7.745
2.003
8.713
2.253
9.682
2.504
30
46
5.802
1.528
6.769
1.782
7.736
2.037
8.703
2.291
9.671
2.546
15
15
5.796
1.553
6.761
1.812
7.727
2.071
8.693
2.329
9.659
2.588
75
o f
Dep.
Lat.
Dep.
Lat.
Dep.
Lat.
Dep.
Lat.
Dep.
Lat.
o f
Bearing.
Distance 6.
Distance 8.
Distance 9.
Distance 10.
Bearing.
76°- 90°
58
16^-
30^
)
Beaziag.
Instance 1. 1
Distance 2.
Distances.!
Distance 4.1
Distance 5.
Bearingi
o f
Lat.
Dep.
Lat.
Dep.
Lat.
Dep.
Lat.
Dep.
Lat.
Dep.
o r
1516
0.965
0.263
1.930
0.526
2.894
0.789
3.859
1.052
4.824
1315
74 46
SO
0.964
0.267
1.927
0.534
2.891
0.802
3.855
1.069
4.818
1.336
SO
46
0.962
0.271
1.925
0.543
2.887
0.814
3.850
1.086
4.812
1.357
16
16
0.961
0.276
1.923
0.551
2.884
0827
3.845
1.103
4.806
1378
74
15
0.960
0.280
1.920
0560
2.880
0.839
3.840
1.119
4.800
1.399
45
30
0.959
0.284
1.918
0568
2.876
0852
3.835
1.136
4.794
1.420
SO
46
0.958
0.288
1.915
0576
2.873
0.865
3.830
1.153
4.788
1.441
16
17
0.956
0.292
1.913
0.585
2.869
0877
3.825
1.169
4.782
1.462
73
16
0.955
0.297
1.910
0.593
2.865
O890
3.820
1.186
4.775
1.483
46
SO
0.954
0.301
1.907
0.601
2.861
0.902
3.815
1.203
4.769
1.504
SO
46
0.952
0.305
1.905
0.610
2.857
0.915
3.810
1.220
4.762
1.524
16
18
0.951
0.309
1.902
0.618
2.853
0.927
3.804
1.236
4.755
1.545
72
16
0.950
0.313
1.899
0.626
2.849
0.939
3.799
1.253
4.748
1.566
46
SO
0.948
0.317
1.897
0.635
2.845
0.952
3.793
1.269
4.742
1.587
SO
46
0.947
0J21
1.894
0.643
2.841
0.964
3.788
1.286
4.735
1.607
16
19
0.946
0.326
1.891
0.651
2.837
0.977
3.782
1.302
4.728
1.628
71 X)
16
0.944
0.330
1.888
0659
2.832
0.989
3.776
1319
4.720
1.648
46
SO
0.943
0.334
1.885
0.668
2.828
1.001
3.771
1.335
4.713
1.669
SO
46
0.941
0.338
1.882
0676
2.824
1.014
3.765
1352
4.706
1.690
16
20
0.940
0.342
1.879
0684
2.819
1.026
3.759
1.368
4.698
1.710
70
16
0.938
0.346
1.876
0.692
2.815
1.038
3.753
1384
4.691
1.731
46
SO
0.937
0.350
1.873
0.700
2.810
1.051
3.747
1.401
4.683
1.751
SO
45
0.935
0.354
1.870
0.709
2.805
1.063
3.741
1.417
4.676
1.771
16
21
0.934
0.358
1.867
0717
2.801
1.075
3.734
1.433
4.668
1.792
69
16
0.932
0.362
1.864
0.725
2.796
1.087
3.728
1.450
4.660
1.812
46
30
0.930
0.367
1.861
0.733
2.791
1.100
3.722
1.466
4.652
1.833
SO
46
0.929
0371
1.858
0.741
2.786
1.112
3.715
1.482
4.644
1.853
16
22
0.927
0.375
1.854
0749
2.782
1.124
3.709
1.498
4.636
1.873
68
16
0.926
0.379
1.851
0.757
2.777
1.136
3.702
1.515
4.628
1.893
46
SO
0.924
0.383
1.848
0.765
2.772
1.148
3.696
1.531
4.619
1.913
SO
46
0.922
0.387
1.844
0773
2.767
1.160
3.689
1.547
4.611
1.934
16
23
0.921
0.391
1.841
0781
2.762
1.172
3.682
1.563
4.603
1.954
67
16
0.919
0395
1.838
0789
2.756
1.184
3.675
1.579
4.594
1.974
46
SO
0.917
0399
1.834
0797
2.751
1.196
3.668
1.595
4.585
1.994
SO
46
0.915
0.403
1.831
0.805
2.746
1.208
3.661
1.611
4.577
2.014
16
24
0.914
0.407
1.827
0813
2.741
1.220
3.654
1.627
4.568
2.034
66
16
0.912
0.411
1.824
0.821
2.735
1.232
3.647
1.643
4.559
2.054
46
30
0.910
0.415
1.820
0.829
2.730
1.244
3.640
1.659
4.550
2.073
SO
46
0.908
0.419
1.816
0.837
2.724
1.256
3.633
1.675
4.541
2.093
16
25
0.906
0.423
1.813
0.845
2.719
1.268
3.625
1.690
4.532
2.113
65
16
0.904
0.427
1.809
0853
2.713
1.280
3.618
1.706
4.522
2.133
46
SO
0.903
0.431
1.805
0861
2.708
1.292
3.610
1.722
4.513
2.153
SO
46
0.901
0.434
1.801
0869
2.702
1303
3.603
1.738
4.503
2.172
15
26
0.899
0.438
1.798
0.877
2.696
1315
3.595
1.753
4.494
2.192
64
16
0.897
0.442
1.794
0.885
2.691
1.327
3.587
1.769
4.484
2.211
46
30
0.895
0.446
1.790
0892
2.685
1339
3.580
1.785
4.475
2.231
SO
46
0.893
0.450
1.786
0.900
2.679
1.350
3.572
1.800
4.465
2.250
16
27
0.891
0.454
1.782
0.908
2.673
1.362
3.564
1.816
4.455
2.270
63
16
0.889
0.458
1.778
0.916
2.667
1374
3.556
1.831
4.445
2.289
46
30
0.887
0.462
1.774
0.923
2.661
1.385
3.548
1.847
4.435
2309
SO
46
0.885
0.466
1.770
0.931
2.655
1.397
3.540
1.862
4.425
2.328
15
28
0.883
0.469
1.766
0.939
2.649
1.408
3.532
1.878
4.415
2347
62
15
0.881
0.473
1.762
0.947
2.643
1.420
3.524
1.893
4.404
2367
46
SO
0.879
0.477
1.758
0.954
2.636
1.431
3.515
1.909
4.394
2386
SO
46
0.877
0.481
1.753
0.962
2.630
3.443
3.507
1.924
4384
2.405
16
29
0.875
0.485
1.749
0.970
2.624
1.454
3.498
1.939
4373
2.424
61
16
0.872
0.489
1.745
0.977
2.617
1.466
3.490
1.954
4362
2.443
45
SO
0.870
0.492
1.741
0.985
2.611
1.477
3.481
1.970
4352
2.462
SO
45
0.868
0.496
1.736
0.992
2.605
1.489
3.473
1.985
4.341
2.481
16
30
0.866
0.500
1.732
1.000
2.598
1.500
3.464
2.000
4.330
2.500
60
o t
Dep.
Lat.
Dep.
Lat.
Dep.
Lat.
Dep.
Lat.
Dep.
Lat.
o t
Beamg.
Distance 1.
Distance 2.
Distance 3.
Distance 4.
Distance 5.
Bearing.
60°- 76^
16^-
30^
59
BawlDir'
instance 6.
Distance 7.
Distance 8.1
Distance 9.1
Distance 10.1
Bearing.
o f
Lat
Dep.
Lat.
Dep.
Lat.
Dep.
Lat.
Dep.
Lat.
Dep.
o /
1515
5.789
1.578
6.754
1.841
7.718
2.104
8.683
2.367
9.648
2.630
74 45
30
5.782
1.603
6.745
1.871
7.709
2.138
8.673
2.405
9.636
2.672
30
46
5.775
1.629
6.737
1.900
7.700
2.172
8.662
2.443
9.625
2.714
16
16
5.768
1.654
6.729
1.929
7.690
2.205
8.651
2.481
9.613
2.756
74
16
5.760
1.679
6.720
1.959
7.680
2739
8.640
2.518
9.601
2.798
46
30
5.753
1.704
6.712
1.988
7.671
2.272
8.629
2.556
9.588
2.840
30
45
5.745
1.729
6.703
2.017
7.661
2306
8.618
2.594
9.576
2.882
15
17
5.738
1.754
6.694
2.047
7.650
2339
8.607
2.631
9.563
2.924
78
16
5.730
1.779
6.685
2.076
7.640
2372
8.595.
2.669
9.550
2.%5
46
SO
5.722
1.804
6.676
2.105
7.630
2.406
8.583
2.706
9.537
3.007
30
46
5.714
1.829
6.667
2.134
7.619
2.439
8.572
2.744
9.524
3.049
16
18
5.706
1.854
6.657
2.163
7.608
2.472
8.560
2.781
9.511
3.090
72
16
5.698
1.879
6.648
2.192
7.598
2.505
8.547
2.818
9.497
3.132
46
30
5.690
1.904
6.638
2.221
7.587
2.538
8.535
2.856
9.483
3.173
30
46
5.682
1.929
6.629
2.250
7.575
2.572
8.522
2.893
9.469
3.214
16
19
5.673
1.953
6.619
2.279
7.564
2.605
8.510
2.930
9.455
3.256
71
15
5.665
1.978
6.609
2308
7.553
2.638
8.497
2.967
9.441
3297
46
30
5.656
2.003
6.598
2337
7.541
2.670
8.484
3.004
9.426
3338
30
46
5.647
2.028
6.588
2365
7.529
2.703
8.471
3.041
9.412
3379
16
20
5.638
2.052
6.578
2394
7.518
2.736
8.457
3.078
9.397
3.420
70
15
5.629
2.077
6.567
2.423
7.506
2.769
8.444
3.115
9.382
3.461
46
30
5.620
2.101
6.557
2.451
7.493
2.802
8.430
3.152
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3.502
30
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5.611
2.126
6.546
2.480
7.481
2.834
8.416
3.189
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3.543
16
21
5.601
2.150
6.535
2.509
7.469
2.867
8.402
3.225
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3.584
69
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5.592
2.175
6.524
2.537
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5.582
2.199
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6.479
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2.296
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5.533
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6.455
2.707
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3.094
8.300
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2.344
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2.735
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8.285
3.517
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3.907
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6.432
2.763
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3.158
8.269
3.553
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3.947
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5.502
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2.791
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8.254
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9.171
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5.492
2.416
6.407
2.819
7.322
3.222
8.238
3.625
9.153
4.028
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5.481
2.440
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8.222
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9.136
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2.464
6.382
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6.292
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6.278
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2.863
6.152
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Distance 6.
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3.383
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0.537
1.687
1.075
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3.374
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6.327
45
30
4.630
3.816
5.401
4.453
6.173
5.089
6.945
5.725
7.716
6.361
30
45
4.613
3.837
5.382
4.476
6.151
5.116
6.920
5.755
7.688
6.394
15
40
4.596
3.857
5.362
4.500
6.128
5.142
6.894
5.785
7.660
6.428
50
15
4.579
3.877
5.343
4.523
6.106
5.169
6.869
5.815
7.632
6.461
45
30
4.562
3.897
5.323
4.546
6.083
5.1%
6.844
5.845
7.604
6.495
30
46
4.545
3.917
5.303
4.569
6.061
5.222
6.818
5.875
7.576
6.528
15
41
4.528
3.936
5.283
4.592
6.038
5.248
6.792
5.905
7.547
6.561
49
15
4.511
3.956
5.263
4.615
6.015
5.275
6.767
5.934
7.518
6.594
45
30
4.494
3.976
5.243
4.638
5.992
5.301
6.741
5.964
7.490
6.626
30
45
4.476
3.995
5.222
4.661
5.968
5.327
6.715
5.993
7.461
6.659
15
42
4.459
4.015
5.202
4.684
5.945
5.353
6.688
6.022
7.431
6.691
48
15
4.441
4.034
5.182
4.707
5.922
5.379
6.662
6.051
7.402
6.724
45
30
4.424
4.054
5.161
4.729
5.898
5.405
6.635
6.080
7.373
6.756
30
46
4.406
4.073
5.140
4.752
5.875
5.430
6.609
6.109
7.343
6.788
15
43
4.388
4.092
5.119
4.774
5.851
5.456
6.582
6.138
7.314
6.820
47
15
4.370
4.111
5.099
4.796
5.827
5.481
6.555
6.167
7.284
6.852
46
30
4.352
4.130
5.078
4.818
5.803
5.507
6.528
6.195
7.254
6.884
30
45
4.334
4.149
5.057
4.841
5.779
5.532
6.501
6.224
7.224
6.915
15
44
4.316
4.168
5.035
4.863
5.755
5.557
6.474
6.252
7.193
6.947
46
15
4.298
4.187
5.014
4.885
5.730
5.582
6.447
6.280
7.163
6.978
46
30
4.280
4.206
4.993
4.906
5.706
5.607
6.419
6.308
7.133
7.009
30
46
4.261
4.224
4.971
4.928
5.681
5.632
6.392
6.336
7.102
7.040
15
45
4.243
4.243
4.950
4.950
5.657
5.657
6.364
6.364
7.071
7.071
45
o f
Dep.
Lat.
Dep.
Lat.
Dep.
Lat.
Dep.
Lat.
Dep.
Lat.
o f
Bearing.
Distance 6.
Distance 7.
Distance 8.
Distance 9.
Distance 10.
Bearing.
45°- 60^
Live Music Archive
Librivox Free Audio
Metropolitan Museum
Cleveland Museum of Art
Internet Arcade
Console Living Room
Books to Borrow
Open Library
TV News
Understanding 9/11