Horatio N. (Horatio Nelson) Robinson.

# Elements of plane and spherical trigonometry : with numerous practical problems online

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KOBIKSON'S MATHEMATICAL SERIES.

ELEMENTS

PLANE AND SPHERICAL

TEiaO^OMETEY,

WITH

NUMEROUS PRACTICAL PROBLEMS.

BY ;

HORATIO N. ROBii^^'O.K, h%;D

AUTHOR OF A FULL Cdtr^te'OF ll^^^^C^.', . ,

IVISON, BLAKEMAN, TAYLOR & CO.,

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NEW YOEK AND CHICAGO.

1880.

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NOTICE.

Upon the suggestion of many- Teachers, the Publishers have thought
6 est to bind in a separate volume this Treatise on Plane and Spherical
Trigonometry ; continuing, however, as heretofore, to bind it up with
Robinson's New Geometry, in one volume.

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Teachers and Students who do' not wish to take up the Geometry in con-
nection with it, or who desire to use this Treatise on Trigonometry and
some other author on Geometry.

924227

TRIGONOM-ETEY

PART I.

PLANE TRIGONOMETRY.

SECTION I.

ELEMENTARY PRINCIPLES.

Trigonometry, in its literal and restricted sense, has
for its object the measurement of triangles. When it
treats of plane triangles it is called Plane Trigonometry .
In a more enlarged sense, trigonometry is the science
which investigates the relations of all possible arcs of the
circumference of a circle to certain straight lines, termed
trigonometrieal lines or circular functions^ connected with
and dependent on such arcs, and the relations of these
trigonometrical lines to each other.

The measure of an angle is the arc of a circle inter-
cepted between the two lines which form the angle — the
center of the arc always being at the point where the
two lines meet.

-..The arc is measured by degrees j minutes, and seconds;
IJi^re being 360 degrees to the whole circle, 60 minutes
hl'one degree, and 60 seconds in one minute. Degrees,
nliHutes, and seconds, are designated by °, ', " ; thus,
270 -^^f 2\fr^ is read 27 degrees 14 minutes 21 seconds.
.'The circumferences of all circles contain the same
n«mber of degrees, but the greater the radius the greater

SECTION I. - 245

is the absolute length of a degree. The circumference of
a carriage wheel, the circumference of the earth, or the
still greater and indefinite circumference of the heavens,
has the same number of degrees ; yet the same number
of degrees in each and every circumference is the meas-
ure of precisely the same angle.

DEFINITIONS.

1. The Complement of an arc is 90° minus the arc.

2. The Supplement of an arc is 180° minus the arc.

3. The Sine of an angle, or of an arc, is a line drawn
from one end of an arc, perpendicular to a diameter
drawn through the other end. Thus, BF is the sine of
the arc AB, and also of the arc BDE, BK is the sine
of the arc BD.

H

4. The Cosine of an arc is the per- j) ^^

pendicular distance from the center of
the circle to the sine of the arc ; or, it is
the same in magnitude as the sine of
the complement of the arc. Thus, OF
is the cosine of the arc AB; but CF=
KB, which is the sine of BD. ^

5. The Tangent of an arc is a line touching the circle
in one extremity of the arc, and continued from thence, to
meet a line drawn through the center and the other ex-
tremity. Thus, AH is the tangent to the arc AB, and
DL is the tangent of the arc DB»

6. The Cotangent of an arc is the tangent of the com-
plement of the arc. Thus, DL, which is the tangent of
the arc DB, is the cotangent of the arc AB.

Remark. — The co is but a contraction of the word complement.

7. The Secant of an arc is a line drawn from the center
of the circle to the extremity of the tangent. Thus, CR
is the secant of the arc AB, or of its supplement BDE,

8. The Cosecant of an arc is the secant of the comple-
ment. Thus, CLy the secant of J5i>, is the cosecant oiAB.

21*

J

246 PLANE TRIGONOMETRY.

9. The Versed Sine of an arc is the distance from tiio
extremity of the arc to the foot of the sine. Thus, Al
s the versed sine of the arc AB^ and I)K is the versed
line of the arc BB,

For the sake of brevity, these technical terms are con-
tracted thus : for sine AB^ we write nn. AB ; for cosine
A S, T7e write co«. AB\ for tangent AB^ we write tan,
AB, etc.

From the preceding definitions we deduce the follow
mg obvious consequences :

1st. That when the arc AB becomes insensibly small,
or zero, its sine, tangent, and versed sine are also
nothing, and its secant and cosine are each equal to

2d. The sine and versed sine of a quadrant are each
equal to the radius ; its cosine is zero, and its secant and
tangent are infinite.

3d. The chord of an arc is twice the sine of one half
the arc. Thus, the chord, J5(r, is double the sine, BF,

4th. The versed sine is equal to the difierence between

5th. The sine and cosine of any arc form the two sides
of a right-angled triangle, which has a radius for its
hypotenuse. Thus, (TFand FB are the two sides of the
right-angled triangle, CFB.

Also, the radius and tangent always form the two
sides of a right-angled triangle, which has the secant of
the arc for its hypotenuse. This we observe from the
right-angled triangle, OAH,

To express these relations analytically, we write

sin.' -h COS.' = i? ( 1 )

R' -I- tan.' = sec' (2)

From the two equiangilar triangles CFB^ CAH^ we

Have

OF : FB ^ OA : AH,

SECTION I. 247

That id,

COS. : sin. = R : tan. ; whence, tan. «= — _ I (3^

cos

Also, OF , CB ^ CA : CH,

That is,

COS. : 72 = jR : sec. ; whence, cos. sec. = -B*. (4)
The tivo equiangular triangles, CAH and ODL, give

CA I AH = DL I DC.

That is,

R : tan. = cot. : i2; whence, tan. cot. =■ iT. (5)
Also, OF I FB =. DL I DQ.

That is,
COS. : sin. = cot. : i2; whence, cos. R = sin. cot. (6)
From equations (4) and (5), we have

COS. sec. = tan. cot. (7)

Or, COS. : tan. = cot. : sec.

We also have ver. sin. = R — cos. (8)

The ratios between the various trigonoraetrical lines
are always the same for arcs of the same number of
degrees, whatever be the length of the radius; and we
may, therefore, assume radius of any length to suit our
convenience. The preceding equations will be more con-
equal unity. This supposition being made, we have, for
equations 1 to 6, inclusive.

tan. =
tan. =

sin." + COS.*

= 1.

(1)

1 + tan.'

= sec'

(2)

«Jfh (3)

COS.

COS. = —

sec

(•»)

\ (^)

COS. = sin. cot.

(6)

cot.

Let the circumference, AEDH, be divided into fouj
equal parts by the diameters, AD and EH, the one hori

248

PLANE TRIGONOMETRY.

zontal and the other vert-
ical. These equal parts
they may be distinguished
as the first, seccnd, third,
The center of the circle
taken as the orig:in of

18

w

^^^- —

^^

B

71

\

m'

m,

\

\

n^

/

B^/

^ 1

B///

distances, or the zero point,
and the different directions
in which distances are esti-
mated from this point are indicated by the signs + and
— . 1^ those from Q to the right be marked +, those
from C to the left must be marked — ; and if distances
from Q upwards be considered plus, those from down-
wards must be considered minus.

If one extremity of a varying arc be constantly at A,
and the other extremity fall successively in each of the
above rule, the algebraic signs of the sines and cosines
of all arcs from 0° to 360°. Now, since all other trigo-
nometrical lines can be- expressed in terms of the sine
and cosine, it follows that the algebraic signs of all the
circular functions result from those of the sine and
cosine.

We shall thus find for arcs terminating in the

sin. COS. tan. cot.

1st quadrant, + -f -f +

2d " + — __ —

3d " — __ + -f

Itli " — + — — 4-

sec.

easec.

rers.

+

■f

-f

•1-

4

•f

+

-f-

PROPOSITION 1.

The chord of 60° and the tangent of 45° are each equal to
radius ; the sine of 30°, the versed sine of 60°, and the oo-
fine of 60° are each equal to one half the radius*

SECTION 1

24S

With (7 as a center, and CA as a
ladius, describe the arc ABF, and
from A lay off the arcs AD = 45°,
AB = 60°, and AE = 90° ; then
is EB = 30°.

1st. The side of a regular in-
scribed hexagon is the radius of
the circle, (Prob. 28, B. IV), and as the arc subtended
by each side of the hexagon contains 60°, we have the
chord of 60° equal to the radius.

2d. The triangle CAII is right-angled at J., and the
angle (7 is equal to 45°, being measured by the arc AD\
bence the angle at H is also equal to 45°, and the trian-
gle is isosceles. Therefore AH — QA = radius of the
circle.

3d. The triangle ABQ is isosceles, and Bn is a per-
pendicular from the vertex upon the base ; hence An =
nC — Bm, But Bm is the sine of the arc BE, On is the
cosine of the arc AB, and An is the versed sine of the
same arc, and each is equal to one half the radius.

Hence the proposition ; the cJwrd of 60°, etc.

PROPOSITION II.

Given, the sine and the cosine of ttvo arcs, to find the sine
and the cosine of the sum and of the difference of the same
arcs expressed by the sines and cosines of the separate arcs.

Let G be the center of the
circle, CD the greater arc,
and DF the less, and denote
these arcs by a and h re-
spectively.

the arc DE equal to the arc
DF, and draw the chord EF,
From ^ and ^, the extremi-
ties, and I, the middle point

Q M

NO

250 • PLANE TRIGONOMETRY.

of the chord, let fall the perpeiuliciilars FM, EP, and
IN, on the radius GO, Also draw DO, the sine of the
arc 01), and let fall the perpendiculars III on FM, and
EK on IK

Now, by the definition of sines and cosines, BO =
&\n.a\ CfO « cos.a; FI = sin. 6; GI = co&.b. We are
to find

FM = sin. {a -\- h); GM = cos. [a -{■ h); -
EP = sin. {a~b); GP = cos. (a — h).

Because IN is parallel to DO, the two a's, GBO^
GIN, are equiangular and similar. Also, the A FHI is
Biinilar to the A GIN; for the angles, FIG and HIN,
are right angles; from these two equals, taking away tKe
common angle IIIL, we have the angle FIR— the angle
GIN The angles at H and iV^are right angles; there-
Ibre, the A's FHI, GIN, and GDO, are equiangular
and similar; and the side RI is homologous to IN
and DO.

Again, as FI = IE, and IK is parallel to FM^

FH = IK, and HI = KE.

^j similar triangles we have

GD : DO = GI : IN
That is, R : sin.a = cos.b : IN; or, IN== ^^"-y.— . ( 1 )
Also, GD : GO = FT : FH,

That is, 72 : cos.a = « sin.6 : HF; or, FH-
Also, GD : GO = GI : GN

That is, '^ : cos.a = cos,b ; (7iV^; or, (7iV^= co_s^cos^6^ ^ 3 ^

Also, GD : DO = FI : IH,

That is, R : sin.a = sin.6 : IH; or, Zff = !liifJl!.^:^. (4)

R

By adding the first and second of these equations, we
bave

IN 4 FH = F3I « sin. (a + ft).

That is, R : cos.a-* sin.6 : iTi^; or, FH= g ^^-^j^ lllj, (2)

SEOTION I. 251

«,, , . • / . 7\ sin.a cos.b -f- cos.a sin. 6
That IS, sin. (a + o) == ^ .

By subtracting the second from the first, since

Z^— FJI^ IN^ IK^ EP, we have

, ,. sin.a C0S.6 — cos.a sin.ft
8111. (a ^ J) = _

By subtracting the fourth from the third, we have

GN— IE = GM = COS. (a + h) for the first member.

TT / , T\ COS. a C0S.5 — sin.a sin. J ,-»
Ilence, cos. (a + o) = ^ . (5)

15y adding the third and fourth, we have

aN-\- in= GN+NF= GF =^ co8.{a—h),

rr / IS COS. a cos. 6 + sin.a sin. 5 /^»

Hence, cos.(a — 5) = ^ (o)

Collecting these four expressions, and considering the
ra lius unity, we have ^;^. :.

fsin.(a + h)= sin.a cos.5 + cos.a sin. 6 ( 7 )
sin. (a — b)= sin.a cos.6 — cos.a sin. 6 (8)
cos.(a -hb) — cos.a coa.b — sin.a sin.i ( 9 )
cos.(a — 6) = cos.a cos.6 + sin.a sin.5 (10)

FormuljB (A) accomplish the objects of the proposi
tion, and from these equations many useful and import
ant deductions can be made. The following are the
most essential :

By a Iding ( " ) to ( 8 ), we have ( n ) ; subtracting ( 8 )
fr^>m ( 7 ) gives ( 12 ). Also, ( 9 ) added to ( 10 ) gives ( 13 ) ,
( 9 ) taken from ( 10 ) gives ( H ).

f sin. (a + b)-{- sin. (a — b) = 2sin.a cos.6 ( H )

6in.(a + J) — sin. (a — 5) = 2cos.a sin. 5 (12)

cos.(a + b) + cos.(a — b) = 2cos.a cos.6 ( 13 )

cos.(a — b) — cos.(a + 6) = 28in.a sin.6 ( 14 )

If we put a-\-b = A, and a — b = B, then ( H ) become^
( 15 ), ( 12 ) becomes ( 16 ), ( 13 ) becomes ( 17 ), and ( H ) be
comes (18).

(B)

262 PLANF TRIGONOMETRY.

\.A H-sin.5=2sin. t

(^)

sm.

2 ^ ^ 2
sin.^ - sin.^ == 2cos. (-± -) sin. (^V^) ( 1^ )

oos.^ + cos.^ = 2cos. (^ -) COS. (^-~) ( 17 )

C08.J5 — COS J.

..i^^).i.(^) .

18)

sin.

If we divide ( 15 ) by (16 )j (observing that — '- = tan.,

cos.

cos 1

ttnd -V— ^ = cot. = as we learn by equations (6) and

sin. tan.

( 5 ), we shall have

A^^B

A—B^

rA+B>

sin. J. + sin

.B ''•^•(-y-) "'°^-(-2-) t^-^-C^-)

sin.^ — sin.i>

Whence,

A—B^

A—B^

cos.(— 2-j ^^m.(— -j tan.(-^)

(19)

A+Bs

sin.JL+sin.^ : sin.J. — sin.5 = tan.f ) : tan. ( )

A^B^

2 / ^ 2'

That is : The sum of the sines of any two arcs is to the dif-
ference of the same sines, as the tangent of one half the sum
of the same arcs is to the tangent of one half their difference.
By operating in the same way with the different equa-
tions in formulae {0), we find,

^sin.J. + sin.^ /A-{-B\

(^)

sin.^ + sin.^ /A — - ^\

cos.^ — cosTZ ^ ^^^' V 2~~)
mn.A — sin.^ /^ _ ^.

cos.tI + co~sr^ ^ *^^- \ 2 )
sin.J.— sin.^ _ /^ + ^A

cos.^— cos.^ ~ ^^^- { 2~)

cos.tI + cos..g _ ^^^' \~^ /
eos.i — COS. J. ~ ^ yA—B\

(20)
(21)
(22)
(23)

(24)

tan.

V-T

SECTiOxX I. 253

These equations are all true, whatever be the value
of the arcs designated by A and B ; we may, therefore,
assign any possible value to either of them, and if in
equations (20), (21), and (24), we make J9= 0, we shall
have,

sin.^ .A 1 /OCX

= tan. — = — -; ( 25 )

i -f cos.^ 2 cot.JJ.

Bin. J. .A 1 f^^.

= cot. -pr- = — T ( 26 )

{§ 1 — cos.^ 2 tan.J^

1 + cos.^ cot.^^ 1

27)

1— cos.^ tan. J J. tan'. J J.
K we now turn back to formulse (A), and divide equa-
tion ( 7 ) by ( 9 ), and ( 8 ) by ( 10 ), observing at the same

tame that — - = tan., we shall have,
cos. '

. / , ,x sin. a cos. J 4- cos.a sin. 6
tan. (a -f 6) =

tan. (a — 5) =

cos.a C0S.6 — sin. a sin.i
sin. a cos. J — cos.a sin. 6

cos.a cos.^ + sin. a sin. 6
By dividing the numerators and denominators of the
second members of these equations by (cos.a cos.5), we
find,

sin. a COS. 6 co8.a sin. 5

, ,v cos.a COS. 6 cos.a cos.i tan.a-f-tan.J

tan.(a+M= — , ; -. — z=:- — (28)

cos.a COS. sm.a sin.6 1 — tan.a tan.6

cos.a COS. 6 cos.a COS. 5
sin. a COS. 5 cos.a sin.6

, ,, cos.a cos.6 cos.a C0S.6 tan. a — tan. 5

tau.(a— 6)= -, — -. . =-— - -^^ (29)

^ ' cos.a COS. sin. a sin. 6 l+tan.atan.5

cos.a COS. 6 cos.a COS. 6
If in equation (H), formulae (B), we make a == J, we
ihall have,

sin. 2a = 2sin.a cos.a (30)

Making the same hypothesis in equation (13), gives,
C08.2a + 1= 2co8'.a (31)
22

254 PLANE TRIGONOMETRY.

The same hypothesis reduces equation (14) to

l_cos.2a = 2siIl^a (32)

The same hypothesis reduces equation ( 28 ) to

, o 2tan.a ,qq\

tan.2a = ~ — - —- (33)

1 — tan\a

If we substitute a for 2a in (31) and (32)^ we shall have

1 + cos.a = 2cos.'Ja. (34)

and 1 — cos.a = 2sin.'Ja. (35) ^

PROPOSITION III.

In any right-angled plane triangle, we may have the fol-
lowing proportions :

1st. The hypotenuse is to either side, as the radius is to the
sine of the angle opposite to that side,^

2d. One side is to the other side, as the radius is to the tan-
gent of the angle adjacent to the first side,

3d. One side is to the hypotenuse, as the radius is to the
secant of the angle adjacent to that side.

Let CAB represent any right-
angled triangle, right-angled at
A.

(Here, and in all cases hereafter, we shall represent the angles of a
triangle by the large letters A, B, C, and the sides opposite to them,
by the small letters a, b, c.)

From either acute angle, as C, take any distance, as
CD, greater or less than OB, and describe the arc BU.
This arc measures tlie angle 0. From B, draw BF par-
allel to BA ; and from B, draw BG, also parallel to BA
or BF.

By the definitions of sines, tangents, secants, etc, BF
is the sine of the angle 0; EG is the tangent, CQ the
secant, and CF the cosine.

SECTION I.

255

Now, Ly proportional triangles we have,

GB: BA=CD: DF or, a : c = R : 8in.(7
CA: AB= CU: EG or, b : c =- E : tan.O
CAiCB ^ CE : Ca or, b : a =^ R : sec.C

Hence the proposition.

ScnoLiuM. — If the hypotenuse of a triangle is made radius, one side
is the sine of the angle opposite to it, and the other side is the cosine
of the same angle. This is obvious from the triangle CDF.

PROPOSITION IV.

In any triangle, the sines of the angles are to one another
as the sides opposite to them.

Let ABC be any tri-
angle. From the points
A and B, as centers,
Hcribe the arcs meas-
urinf): these anorjes, and
draw pa, CD, and mn,
perpendicular to AB,

Then, pa = sin. J., and mn = sin.^.

By the similar A*s, Apa and ACD, we have,
R : sin.^ ^biCD\ or, R{CD) = b sin.A

By the similar a*8, Bmn and BCD, we have,

R : sm.B = a: CD; or, R(CD) = a sm.B

By equating the second members of equations ( 1 )
and (2)

h sin. -4 = a sin. 5.

Hence, sin.^ : sin. 5 = a : b

Or, a : b = sin.A : sin.^.

ScnoLiUM 1. — When either angle is 90°, its sine is radius.

ScnoLiUM 2. — W^hen CB is less than AC, and the angle B, acute,
the triangle is represented by ACB. When the angle B becomes B^,
it is obtuse, and the triangle is A CB^ ; but the proportion is equally

(1)

(2)

256 PLANE TRIGONOMETRY.

true with either triangle ; for the angle CB^D = CBA, and the siu^
of CB^D is the same as the sine of AB^C. In practice we can deter-
mine which of these triangles is proposed, by the side AB being
greater or less than AC; or, by the angle at the vertex C being large,
as A CB, or small, as A CB^.

In the solitary case in which AC, CB, and the angle A, are given,
and CB less than AC, we can determine both of the A's JCB and
A CB'' ; and then we surely have the right one.

PROPOSITION V.

If from any angle of a triangle, a perpendicular he let fall
on the opposite aide, or base, the tangents of the segments of
the angle are to each other as the segments of the base.

Let ABO be the triangle. Let fall
the perpendicular CD, on the side
AB.

Take any radius, as Cn, and de-
scribe the arc which measures the A av.^^^/B
angle 0. From n, draw qnp parallel to AB. Then it is
obvious that np is the tangent of the angle DCB, and nq
is the tangent of the angle ACD.

Now, by reason of the parallels AB and qp, we have,
qn : 7ip = AD : DB

That is, tan.J.aZ> : tan.Z>(7^ = AD z DB.

PROPOSITION VI.

If a perpendicular he let fall from any angle of a triangh
to its opposite side or base, this base is to the sum of the other
two sides, as the difference of the sides is to the difference of
the segments of the base.

(See figure to Proposition 5.)

Let AB be the base, and from 0, as a center, with the
shorter side as radius, describe the circle, cuttiiig AB in
Q^, and AC in F; produce AC to U.

SECTION I. 257

It is obvious that AE is the sum of the sides AC and
CB, and ^i^is their difference.

Also, AD \q one segment of the base made by the per-
pendicular, and BB — BCr is the other; therefore, the
difference of the segments is AG,

As J. is a point without a circle, by Cor. Th. 18, B.
m, we have

AE X AF = AB X AG

Hence, AB : AE ^ AF : AG,

PROPOSITION VII.

17ie sum of any two sides of a triangle is to their difference^
as the tangent of one half the sum of the angles opposite to
these sides, is to the tangent of one half their difference.

Let ABQ be any plane triangle. ^

Then, by Proposition 4, we have,

BQ: AQ= sin.J. : sin.j5.

Hence, A B

^C + ^C:^C— ^(7=sin.^+sin.i?:sin.^— sin.^(Th.9,B.II).
But,

tan. ( — i — ) : tan. ( — - — ^ = sin.^ -f sin.^ : sin.-4

— sin.jS, (eq. (10), Trig.)

Comparing the two latter proportions, (Th. 6, B. H),
we have,

Hence the proposition.

PROPOSITION VIII.

Given, the three sides of any plane triangle, to find some
relation which they must hear to the sines and cosines of thi
respective angles,

22* R

258

PLANE r 1 1 I G N M E T R Y.

Let ABO be
the triangle, and
let the perpen-
dic alar fall either
upon, or without
the base, as shown
m the

figures.

C « B
By recurring to Th. 41, B. I, we shall find

2a
Kow, by Proposition 3, we have

R : COS. = b : OJ),

Therefore, OB = —^, (2)

Equating these two values of OB, and reducing, we
have

E (a' + b" — c'^

COS. =

2ab

( m

In this expression we observe, that the part c, whose
square is found in the numerator with the minus sign, is
tlie side opposite to the angle ; and that the denominator
is twice the rectangle of the sides adjacent to the angle..
From these observations we at once draw the following
expressions for the cosine A, and cosine B :

COS. A —

R{b' + c' — a'
2fo

eos.^ = ^i^-^llZ.^-!).

in)

{P)

:aG

As these expressions are not convenient for loganth-
mic compu'-ation, we modify them as follows:
If we put 2a — A, in equation ( 31 )j we have

COS. Jl + 1 = 2cos.'JJ..
In the preceding expression, {n)^ if we consider radius
Uxuty, and add 1 to both members, we shall have

COS. J. + 1 = 1 -f
Therefore, 2cos.'JJ. =

SECTION I. 259

5' 4. (7> — a"

2hc
^hc-^h' + c^ — a*

2bc

^ {h + cY — g'
2bc
Considering 6 + <? as one quantity, and observing that
(5 4- cy — a' is the difference of two squares, we have
(t-|-c)> - a«=(5+c-|-a) {b-]-c—a) ; but (6+r— a)=i4-c+a— 2a.

Hence, 2cos.'i^ = [ b ^ c + a)( l> +^c + a-2a )^

/h -{- c + a\ /h -}- c -h a \

n . 1 J ^ 2 ) V 2— "";.

Or, cos.*J^ = J

By putting = «, and extracting square root,

the final result for radius unity is

COS. lA = v/l5p).
For any other radius we must write

«^« 1 A 4 /i^'s (« — a)
cos. 1^^ = \/ i- i..

^ be

By inference, cos. ^B = a/^M£Z1^

Also, COS. iC = \/^MiEI).

a6

In every triangle, the sura of the three angles is equal
to 180° ; and if one of the angles is small, the other
two must be comparatively large ; if two of them are
email, the third one must be large. The greater angle
is always opposite the greater side; hence, by merely
inspecting the given sides, any person can decide at
once which is the greater angle; and of the three pre-

260 PLANE TRIGONOMETRY.

ceding equations, that one should be taken whicli applies