Edward Olney.

Elements of trigonometry, plane and spherical online

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Font size log cot 88 17' 44" = 8.473588.

4. Having given the logarithmic sine 7.246481 to find the angle.

SOLUTION. From Table II. we find 6' 5" = 365" as the angle. But this is
subject to the inaccuracy exhibited in (73). To obtain the correct result from
Table III., we have (8O),

log x = 7.246481 - 4.685575 + = 2.500906.
/. x = 363.8; or the angle is 6' 3" .8.

5. Given the logarithmic tangent 7.805487, to find the correspond-
ing angle.

SOLUTION. Table II. gives 21' 58".3 = 1318".3 as the angle. From Table III.
the Diff. corresponding to this is 6 (millionths). Hence,

log x log tan x" 4.685575 Dili'. (82)
becomes, log x = 7.805487 4.685575 .000006 = 3.119906.
/. x = 1318; and the angle is 1318" = 21' 58".

6. Given the logarithmic cotangent 12.197148, to find the corre-
sponding arc.

Table IL gives the angle 21' A9".8, but the true angle as given by
Table III. is 21' 50".

PLANE TRIGONOMETRY.

SECTION V.

TRIGONOMETRICAL SOLUTION OF PLANE TRIANGLES.

83. There are six parts in every plane triangle: three sides ami
three angles; one side and any other two of which being given, the
remaining parts can be found by means of the relations which exist
between the sides and tabulated trigonometrical functions. To
exhibit these highly important practical operations is the object of
this section. We shall treat first of right angled plane triangles, and
then of oblique angled plane triangles.

OF RIGHT ANGLED TRIANGLES.

Prop. 84. The relations between the sides and the trigonomet
rical functions of the oblique angles of a right angled triangle are .
follows :

_ side opposite
hypotenuse '

~~ hypotenuse '

/ox . side opposite

(3) tangent = -.-y- - :

(1) sine

(2) cosine

cosecant =

secant =

. oiuc eiu ttueiit

cotangent = -^

01 Ho r\ TM"\ /-* o TT *

hypotenuse
side opposite '
hypotenuse
side opposite*

OEM. Let CAB, Fig. 12, be a triangle, right
angled at A. Let aM be the measuring arc of the
angle B, PD = sin B,and BD cos B. From the

PD
similar triangles PDB and CAB, we have ^= =

CA B side opposite . DQ

; i. e., sm B - , since BP = 1.
BC hypotenuse

From the same triangles |P =^~\ i. e., cos B
BP BC

side adjacent^ T nt bemg j to sine divicled
hypotenase

Fra.

bycoslne, we nave tan B = side PP site + lide^^cent ^j^
other fuiuuious being the reciprocals of these three, are as given in the proposi-
tion. Fiually, as a similar construction could be made about the other oblique
angle, 0, this demonstration may be considered general. Q. B. D.

SOLUTION OF RIGHT ANGLED PLANE TRIANGLES. 45

Sen. 1. These formula are so important that it is well to have them fixed
In the memory, not only as written above, but also as follows :

(1). Side-opp = hy x sin, or , or tan x side-adj, or ;

cosec' cot

(2). Side-adj hy x cos, or , or cot x side-opp, or ;

of which the relations side-opp = hy x sin, and side-adj = hy x cos are of thf
most frequent use.

Sen. 2. The six ratios given in this proposition are frequently made the def-
initions of the trigonometrical functions. Thus, referring to a right angled tri-
angle, a sine of an angle may be defined to be the ratio of the side opposite to
the hypotenuse ; the cosine as the ratio of the side adjacent to the hypotenuse,
etc.

Sen. 3. The student will be aided in remembering these important relations
by observing that the side opposite the angle is analogous to the sine, and the

its analogue

side adjacent to the cosine. Now, the sine = , and so also the co-
sine. Tangent equals sine divided by cosine, and in this case it is the part
analogous to the sine, divided by the part analogous to the cosine. One should

hy

not make the blunder of s&ying that sin =: r-: , since that would make

side-opp

the sine always more than 1 ; but we have sec a that it never can exceed 1.
Similar checks against error may be made in the case of the other relations.

EXERCISES.

. The first five of these exercises are mainly designed to illustrate the
proposition, and familiarize the mind with the relations.]

1. In a right angled triangle whose sides are 3, 4, and 5, what are
the trigonometrical functions of the angles ? What are the functions
when the sides are 6, 8, and 10 ?

2. In a right angled triangle the hypotenuse is 12, and the angl *
at the base sm~~ a . What are the sides ? What is the sine of th^
other angle ?

SUG. Represent the angles by B, A, and C, A being the right angle; anil

b b

the sides opposite by &, a, and c. Then sin B = -^or^ = - whence, b 6

a LA

Sin C = cos B = %^ c = G/v/37

3. In a right angled triangle whose hypotenuse is 12, aud the
angle at the base tan -1 2, what are thu other parts ?

AIM.

46 PLANE TEIGONOMETEY.

4. The sides of a right angled triangle are 20 and 32. What are
the angles and hypotenuse? Obtain the hypotenuse by means of
the secant.

Ans. Tan- 1 1, tan" 1 !, and 4 A/89-

5. The hypotenuse of a right angled triangle is 120, and one
side 100. Show that the angle opposite the latter is tan-V^VTi, the
adjacent angle cosec" 1 -|^-V / ll> ^ u( l the remaining side 20\XH- Obtain
tnese results in the order given. Use a trigonometrical function to
obtain the last.

EXAMPLES.

(a) BY MEANS OF THE TABLE OF tfATUKAL FUNCTIONS.

1. In a right angled triangle ABC, the hypotenuse BC is 235, and
the angle B is 43 25'. Find the angle C, and the sides AB and AC.

C = 46 35' ; AB = 170.7 ; AC = 161.52.

SOLUTION To find C, we have but to remember tliat the angles of a right
angled triangle are complements of each other ; whence, C = 90 B = 46 35'.

To find AB, we have cos B = ^' or AB = 235 x cos 43 25'. Now, from the table
of natural functions we find cos 43 25' - .72637 ; whence AB = 235 x .7263?

- 170.7. To find AC, we have sin B - || ; whence AC = 235 x .6873 = 161.52.

2. In a right angled triangle ABC, the hypotenuse AC is 94.6, and
the angle C is 56 30'. Find the angle A, and the sides AB and BC.

A =33 30'; AB =78.88; BC = 52.21.

3. In a right angled triangle BDF, the hypotenuse BF is 127.9, and
the angle B is 40 10' 30". Find the angle F, and the sides BD and
DF.

F = 49 49' 30"; BD = 97.72; DF = 82.51.

4. In the triangle CDE, right angled at E, given the side DE 75, the
side CE 50.59, to find the other parts.

Hypotenuse = 90.47. *

5. In the right angled triangle CDE, given the hypotenuse CD 264,
the side CE 135.97, to find the other parts.

DE = 226.28.

6. Given the hypotenuse 435, and one of the acute angles 44, to
find the other parts.

7. Given the hypotenuse 64, and the base 51.778, to find the other
parts.

i

SOLUTION OF U1G1IT ANGLED PLANE TRIANGLES. 47

8. Given the hypotenuse 749 feet, and the base 548.255 feet, to
find the other parts.

9. Given the hypotenuse 125.7 yards, and one of the acute angles
75 12 23", to find the other parts/

10. Given one side 388.875, and the adjacent angle 27 38' 50", to
find the other parts of a right angled triangle.

(b) BY MEANS OF A TABLE OF LOGARITHMIC FUNCTION'S.

11. In a right angled triangle, given an oblique angle 54 27' 39",
rind the side opposite 56.293, to find the other parts.

SOLUTION. The other oblique angle is 90 -54 27' 39" = 35 32' 21".

To find the hypotenuse, we have sin 54 27' 39" = :^r, O r hy = -.

hy : sin 54 27' 39"

Applying logarithms to facilitate computation, log hy = log 56.293 log
sin 54 27' 39" + 10. The 10 is added since the log sin 54 27' 39" found in the
table is 10 too great. Now, looking in Table L, we find log 56.293 = 1.750454
and in Table II., log sin 54 27' 39" = 9.910473. Hence,

log hy = 1.750454 - 9.910474 + 10 = 1.839980, and hy = 69.18.

To find the other side we have, tan angle = ^^, or tan 54 27' 39" r-

Applying logarithms, log^o^

= log 56.293 - log tan 54 27' 39" + 10 = 1.750454 - 10.146104 + 10 = 1.604350.

12. In a right angled triangle ABC, the hypotenuse AC is 340, and
the side AB is 200. Find the acute angles A and c, and the perpen-
dicular BC.

A = 53 58' 6" ; C = 36 1' 54' ; BC = 274.95.

13. In a right angled triangle ABC, the perpendicular AB is 736.3
and BC 500. Find the acute angles A and C, and the hypotenuse AC.

A = 34 10' 45" ; c = 55 49' 15" ; AC = 890.02.

14. In a right angled triangle BDF, the perpendicular BD is 246.32,
and DF 380.07. Find the acute angles E and F, and the hypote-
nuse BF.

B = 57 3' 11"; F = 32 56' 49" ; BF = 452.91.

15. In a right angled triangle ABC, the side AB is 249, and the
angle A is 29 14'. Find the perpendicular BC, and the hypotenuse

*C.

BC = 139.35 ; AC = 285.341,

48 PLANE TRIGONOMETRY.

16. Iii a right angled triangle ABC, the hypotenuse AC is 95.75, and
t'ne side BC 60. Find the acute angles A and c, and the perpen-
dicular AB.

A = 38 48' 7" ; C = 51 11' 53" ; AB = 74.62.

17. In a right angled triangle ABC, the side BC is 364.3, and the
angle A is 50 45'. Find the perpendicular AB, and the hypotenuse

AC.

AB = 297.645 ; . AC = 470.433

[NOTE. The first ten examples may be solved by logarithms if additional
exercises are needed, or these by means of the natural functions. Also any one
of the examples will afford several others by giving and requiring different
parts. Thus, from Ex. 17, we can give AB = 297.645, A = 50 45', and require
the other parts, etc.].

GENERAL APPLICATIONS.

1. Find the area of a parallelogram whose adjacent sides fire 28
and 30 feet, and the included angle 75.

SUG. First find the altitude.

2. A railroad track 463 feet 3 inches in length has a uniform grade
of 3. Show that the vertical rise is 24 feet 3 inches, nearly.

3. A railroad track makes a vertical rise of 150 feet, by uniform

4. Find the apothem and radius of the circumscribed circle of a
regular heptagon one of whose sides is 12 feet.

5. Find the area of a regular dodecagon inscribed in a circle whoae

6. The angle of elevation to the top of a steeplf is 47 30', a&
measured from a point in the same horizontal plane as its base, and
at a distance of 200 feet from it. What is the height of the steeple?

Ans. 218.26. ft.

7. A tower 103 feet high throws a shadow 51.5 feet long, upon the
horizontal plane of its base : what is the angle of elevation of the
sun?

8. The angle at the vertex of a right cone is 52 23', and the
slant height 126 feet : what is the diameter of the base, and what
the altitude?

SOLUTION OF OBLIQUE ANGLED PLANE ilUANGLES.

4<J

FIG. 13.

9. In Fig. 13, letting EO rep-
resent the earth and M the
moon, the radius of the earth
EO = 3956.2, and the angle
EMO * 57', required to fiud
the distance OM, E being a
right angle.

The distance of the moon from the earth, as given by this compu-
tation, is 238,613 miles.

10. In Fig. 13, letting ON represent a tangent to the moon's disc
at N, the angle MOM is readily measured, being half the moon's
apparent diameter. The apparent diameter of the moon being
31' 20", and its distance from the earth as found in the last example,
what is the diameter ?

Ans. 2176 miles.

OF OBLIQUE ANGLED PLANE TRIANGLES.

IMPORTANT RELATIONS EXISTING BETWEEN THE SIDES AND TRIGO-
NOMETRICAL FUNCTIONS OF THE ANGLES OF OBLIQUE ANGLED
PLANE TRIANGLES.

85. Prop. TJie sides of any plane
triangle are proportional to the sines of
the angles opposite.

DEM. Let ABC be any plane triangle. Let
fall from either angle, as C, a perpendicular
upon the opposite side, or upon that side
produced. Designate the angles by A, B, and
C, the sides opposite by a, 6, and c, and the per-
pendicular by P.

Now, from the right angled triangle ADC
we have P = b sin A ; also from CDB, P = A
a sin B ; sin ABC in the second figure being =
suiCBD. Hence, equating the values of P, Pie. 14.

b sin A = a sin B, or a : b : : sin A : sin B. Q. E. D.

* This angle is called the moon's horizontal parallax, and is readily measured. Some ruOe
notion of the manner in which parallax becomes apparent, may be got from considering the
difference in direction from two observers to the moon, one observer standing directly under
the moon, as at F, and the other at E, seeing the moon in his horizon. The angular displace
raent of the moon due to these different points of observation ie horizontal parallax.

4

50 PLANE TRIGONOMETRY.

86. Prop. T/ie sum of any two sides of a plane triangle is to
their difference, as the tangent of half the sum of the angles opposite
if to the tangent of half their difference.

DEM. Letting a and b represent any two sides of a plane triangle, and A
and 6 the angles opposite, we have a : b : : sin A : sin B. Taking this both by
composition and division, we have a + b : a b : : sin A + sin B : sin A sin B.
But from (6*0), sin A + sin B : sin A sin B : : tan|(A + B) : tan (A B).
/. a + b : a b : : tan |(A + B) : tan i(A B). Q. B. D.

87. Prop. TJie tangent of half of any angle of a plane triangle
equals .fc, divided by half the perimeter of the triangle minus the side
opposite the angle ; in which k is the radius of the inscribed circle,
and equals the square root of the continued product of half the peri-
meter minus each side separately, divided by half the perimeter.

DEM. Let ABC be any plane triangle.
Represent the angles by A, B, and C, the
sides opposite by a, 6, and c, the perimeter
fcy p, and the segments of the sides made
ny the radii of the inscribed circle, by x, y^
and 2, as in the figure.

Then a + b + c = 2x + 2y + 2z = p, or

>

Whence x = \$p a, y =\$p 5, and z Fl0t 15p

= iP c ; since y + z = a,x + z = b, and x + y c.

Now from the triangles AOD, DOB, and COE, tan iA=^- k _ , tan

if if if if

B = - = and tan C = - = -p-^ .
y \P-V . z kp-c

To find k. |A + 46 + C = 90, or iA = 90 - QB + iC) ; whence, tan A
= = tan [90 - B + |C)] = cot (*B 4 |C) = * B (^ DKM.

-

Substituting for tan IB, and tan iC, -, and -, we have - = - ^

y z x k k'

_ y ~*>

^ = ^l ^ y~ J ( x + y + 3& = *yz ; and k = y^ - . In this value of A;, sub

stituung for aj, y, and z their values, we have k = j^ ~ a) (" (^ - c \

iP

SOLUTION OF OBLIQUE ANGLED PLANE TRIANGLES. 51

88. Sen. 1. These three propositions (85, 8V, 87), furnish the most
elegant and expeditious means for finding the unknown parts of an oblique
angled plane triangle, when a sufficient number of parts are given or known (&.'{)

89. Sen. 2. Important Practical Suggestions.

1st Two angles of a triangle being given, the third is known by implication,
It being the supplement of the sum of the other two.

2nd. When two of the known, or given, parts are opposite each other, the
frrst proposition (85) effects the solution.

3rd. When two sides and the included angle are given, the solution is effected
by means of the second proposition (86).

4th. When the three sides are given, the angles are found by the third
proposition (7).

EXERCISES.

1. In the plane triangle CDE, given the angle D = 15 19' 51",
C = 72 44' 05", and the side c, opposite c,
250.4, to find the other parts. E

SOLUTION. First, E = 180- (D + C) = 91 56' 04" _
(89, 1st).

Second, To find side d opposite angle D (89, 2nd).
sin C : sin D : : c : d, or
sin 72 44' 05" : sin 15 19' 51" :: 250.4 : d.

This proposition may be solved for d, by taking the natural sine of 15 19' 51"
multiplying it by 250.4, and dividing the product by the natural sine of
72 44' 05"; or, more expeditiously, by logarithms, as follows :
log 250.4 = ............. 2.398634

log sin 15 19' 51" = .......... 9.422249

log sin 72 44' 05" (ar. comp.)* = ..... 0.020024

logd= ............. f 1.840907 .\ d= 69.328.

Third, To find side e opposite angle E (89, 2nd).

sin C : sin E : : c : e, or
sin 72 44' 05" : sin 91 56' 04" : : 250.4 : e.
Making the computation by logarithms,
log 250.04 = ......... ... 2.398634

log sin 91 56' 04" t .......... 9.999752

log sin 72 44' 05" (ar. comp.) = ..... 0.020024

log e = .............. 2.418410. .'. e = 262.066.

* See Introduction to Table I. (17).

t The student must bear in mind the fact that all the log. trig, funcs. are 10 too large, and
aiaet pee exactly what corrections to make in his results, on this account. In this case the
ar. comp. is 10 too small, since the logarithm we took from the table for log sin 72 44' 05" was
10 too large. But our log sin 15 1!)' 51" is 10 too large, and just corrects the latter. Hence, we
have to reject only 10 from the entire sr in 11.840907, and this on account of the use of ar. comp.

f Tnke the sine of the supplement, or the cosine of the given angle minus 90

52 PLANE TRIGONOMETRY.

2. In the plane triangle ABC, A = 35 42', B = 76 27', anl AB =
142. What are the other parts ?

Ans. C = 67 51'; AC = 149.05; BC = 89.47.

3. Given two angles of a plane triangle, 23 40' 32" and G9 39' 51",
and the included side 100, to find the other parts.

4. In a plane triangle ABC, the side AB is 254.3, the side AC 39G.8,
and the angle B 94 29'. Find the angles A and C, and the side BC.

A= 45 48' 21"; c = 39 42' 39"; BC = 285.37.

Sue's. To find the angle C, we have

396.8 : 254.3 : : sin 94 29' : sin C.

From this proportion we get log sin C = 9.805443. Now, as we have seen
before, there are an infinite number of angles corresponding to any given sine,
how shall we know what one to take in this case ? First, no angle of a triangle
can exceed 180 ; hence, there are but two angles, one an acute angle, and the
other its supplement, which can come into consideration in the solution of plane
triangles. But which of tJiese two are we to take ? Thus, in this case, both the
angles 39 42' 39" and 140 17' 21" correspond to log sin 9.805443. In this ex-
ample the ambiguity is resolved by observing that the given angle B is obtuse,
and a plane triangle can have but one obtuse angle. /. C = 39 42' 39".

5. In a plane triangle BDF, the angle B is 40, the side BD is 400,
and the side DF 350. Find the angles D and F, and the side BF.

BUG'S. To find F, we have

350 : 400 : : sin 40 : sin F,

from which log sin F = 9.866059, and F = 47 16' 28", and its supplement
132 43' 32". How are we to determine which of these to take? The given
angle is 40 ; hence, as far as that is concerned, either of the two will meet the
conditions. There are, therefore, two angles, F = 47 16' 28" and F = 132 43' 32",
which fulfill the conditions. We therefore solve two triangles, one having two
D of its sides 400 and 350, and the angles 40",

47 16' 28", and 92 43' 32" ; and another triangle
with the same given parts and the two required
angles 132 43' 32", and 7 16' 28". This is readily
illustrated geometrically. Thus, lay off DBF' = 40.

' Take BD 400. Then from D as a centre, with
FlG - 16 - radius 350 describe an arc cutting BF'. It is evident

that if B is an acute angle the following cases may arise depending upon the
value of DF:

1st. If DF is less than the perpendicular p, the problem is impossible.

2nd. If DF = p, the triangle is right angled at F.

3rd. If DF > p and <BD there are two triangles, one with an acute angle at
F', as DF'B, and the other with an obtuse angle at F, as DFB, both of which
fulfil the conditions of the problem.

4th. If DF > DB there is but one triangle which fulfills the conditions, viz.,
the one with an acute angle at F'.

SOLUTION OF OBLIQUE ANGLED PLANE TRIANGLES. 53

The results in the above examples are, for triangle DF'B, angle DF'B =
47 16' 28", B DP = 92 43' 32", and side BP = 543.89; for triangle DFB,
angle DFB = 132 43' 32", angle BDF = 7 16' 28", and side BF = 68.94.

yO. COR. In applying trigonometrical formula to the solution of
triangles, if the part sought is found in terms of its SINE, the result
is ambiguous, and we are to determine whether there really are two
solutions to the problem in a geometrical sense, by certain geometrical
considerations, or else by trying both values for the angle determined
by its sine. This ambiguity arises only when an angle is determined
by its sine, as will appear hereafter.

6. Given two sides of a plane triangle 201 and 140, and the angle
opposite the latter 36 44'. Find the other parts.

Results. There are two triangles.

Parts of the first, 120 49' 49", 22 26' 11", and 89.34;
Parts of the second, 59 10' 11", 84 5' 49", and 232.84.

7. Given two sides of a plane triangle 180, 100, and the angle op-
posite the former 127 33', to find the other parts.

There is but one triangle, and the parts are 26 7' 59", 26 19' 1",
and 100.65.

8. Given two sides of a plane triangle 30.8 and 54.12, and the
angle opposite the latter 36 42' 11", to find the other parts. Why
hut one triangle ?

9. Given two sides of a plane triangle 600 and 250, and the
angle opposite the latter 42 12'. Find the other parts.

SUG. Attempting to get the angle opposite 600, we find log sin = 10.207400,
which is impossible. It is in some such way that a trigonometrical solution
shows a geometrical absurdity.

10. Given two sides of a plane triangle 1337.5 and 493, and the
angle opposite the former 69 46'. Find the other parts.

11. In a plane triangle, given two sides 1686 and 960, and the in-
cluded angle 128 04', to find the other parts.

c

SOLUTION. Let a = 1686, b = 960, and
C = 128 04'. (See 89, 3rd.) The sum of

the angles A and B is 180 - 128 04' =

51 56', and \$(A + B) = 25 58'. From (86) A c

w have, Fia. 17.

a + b : a b : : tan UA + B) : tan^(A B), or
2646 : 726 : : tan 25 58' : tan |(A - B).

54 PLANE TRIGONOMETRY.

Making the computations by logarithms, we find log tan \$(A B) = 9.125887
Hence, i(A B) = 7 36' 40", the angle found in the table, or its supplement
But i the difference of two angles of a triangle is less than 90 ; consequently
KA - B) = 7 36' 40".

Now having i(A + B) = 25 58', and i(A - B) = T 30' 40", we find A =
33 34' 40", and B = 18 21' 20".

The side c can be found by (85) as two opposite parts are now known,
-5 = 2400.

12. In a plane triangle ABC, the side AB is 304, BC 280.3, and the
included angle B is 100. Find the angles A and C, and the side AC.

A = 38 3' 3" ; c = 41 56' 57" ; AC = 447.856

13. In a plane triangle ABC, the side AB is 103, AC 126, and the
included angle A is 56 30'. Find the angles B and C, and the side BC

B = 72 20' 15" ; C = 51 9' 45" ; BC = 110.267.

14. In a plane triangle ABC, given the three sides, a = 3459, I
4209, and c = 6030.4, to find the angles.

SOLUTION. Applying (87), we have,

t or

log A: = | {log (\$p -a) + log (\$p - b) + log (fr c)
Also, tan |A = k _ ^ tan \B = k _ ^ and tan \$C =jr- , r log tan

= log k - log (ip - a), log tan 4B = log k - log (\$p - b), and log tan |C
log k - log (4p - c).

COMPUTATION.
a= 3459
b= 4209
c= 6030.4
p = 13698.4

\$p = 6849.2 (ar. comp.) log = 6.164360

ft - a = 3390.2 ..... log = 3.530226

i \p - b = 2640.2 ..... log = 3.421637

lp _ c - 818.8 ..... log = 2.913178

2)6.029401
log k = 3.014700

log tan |A = log k - log (& - a) = 9.484474 .-. A = 33 56' 10".5
log tan|B = log k - log (\$p - b) = 9.593063 /. B = 42 47' 25".3
log tan iC = log k - log (ip - c) = 10.101522 /. ^=103 16' 24".2

Proof, A + B + C = 180* 00' 00"

SOLUTION OF OBLIQUE ANGLED PLANE TRIANGLES,

55

15. In a plane triangle ABC, the side A3 is 95.G, BC is 27o, and AC
300. Find the angles A, B, and C.

A= 65 47' 55"; B = 95 42' 52"; C = 18 29' i:j".

16. In a plane triangle BDF, the side BD is 500, DF is 403.7, and
8F 395.75. Find the angles B, D, and F.

B =52 0' 3"; D = 50 34' 45"; F = 77 25' 12",

OBLIQUE ANGLED TRIANGLES SOLVED BY MEANS OF RIGHT
ANGLED TRIANGLES.

[NOTE. Articles 91-94: inclusive, may be omitted in an elementary course,
if thought desirable. Or 91 and its applications may be taken instead of #(*-
88. 85 should be included in any course. It is too elementary and important
to be omitted.]

91. Prop. All cases of oblique angled plane triangles may be
solved by the solution of right angled triangles.

DEM. Of the three given parts we may affirm that they are, 1st, All

1st. When the given parts are all adjacent ; i. e. t when they are two sides and
the included angle, or two angles and the in-
cluded side. To solve the first let fall a perpen-
dicular from the extremity of one of the given
sides upon the other given side, or upon that
side produced. There will thus be formed two
right angled triangles which can be computed,
and from the parts of which the parts of the
required triangle can be found. Thus, let A be
the given angle, and b and c the given sides.

Online LibraryEdward OlneyElements of trigonometry, plane and spherical → online text (page 5 of 28)