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° +

.

Now every sine is less

(60° + A) for the three values of x

~ 1 / 3 p must be greater than gol, or

q

than radius, hence 2 » p

p

27 1 must be greater than-q*

From which it follows that any equation of the third degree falling under the irreducible case, may be solved by this method.

Rx3 973 36. Call R the tabular radius, and we shall have

for

2 PNP

Rx 3 av 3. But A being the tabular sine of the arc A, or sin A=

2 PnP 1

-) 3

3 given; and consequently the three roots of the proposed equation (reducing these sines to those which correspond to the radius

known, sin A, sin (60°—___ A), and—sin (60°+q), are also

3

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p) are

27 $ P sin } A

X

R
2w $ Psin (60°-4A)

R and

27 $ Psin (60° +1 A)

R EXAMPLE 1. Let the proposed equation be x_3 x+1=0 Here P=3,q=1; whence sin A={ R, and A=30°. Therefore the three values of x are

2 sin 100

= -0.3472964
R
2 sin 50°

=1.5320888
R
-2 sin 70°

-1.8793952

R
Ex. II. Let xi-x+
We shall have p=1, ,95

'sin 3 R А

=

X =

X =

=0.

et x-x+
-x

=> AV 3. Therefore A=60°; and the three values of x are

2 Sin 20°
X =

0,394931
RN 3
2 sin 40°

0.742227
RJ 3
-2 sin 800

-1.137158
RN 3

and

+ 5' ")

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Ex. III. Let the equation be x5 x+3=0. This gives p=5,
R 35

5 938, sin A=

and log sin A=log R+ log 3_log 2 53

2 3

log 5=9.843318=log sin (44° 11' 52"); therefore A= 2 449 Il' 52" and the three values of the unknown quantity are

2 5 sin (14° 43' 57")

R 3
2 5 sin (45° 16' 3")

RM 3
-2 5 sin (74° 43' 57")

R3
Performing the calculations indicated, we obtain

x=0. 656625, X-1.834238, x=-2.490863 Scholium. Though there are an infinity of other arcs whose sines are the same as that of the arc A, yet they are such that the sines of their third parts may be reduced to one of the three forms

1: 3

3 Therefore any cubic equation solved by this method will never have more than three roots, as ought to be the case.

Examples for Practice. 1. Given x-12 X=15; required x Answer, x=3.971963, or-1.577032, or-2.394930 2. Given x-27 X=36; required x Answer, x=5.765722, or-4.320685, or-1.445038 SOLUTION OF THE CASES OF PLANE

TRIANGLES.

sin į A, sin (604– A), –sin (600 + A)

3

a

37. Any right-lined triangle may be inscribed in a circle; and this being done, each side of the triangle will be the chord of an arc double of that which measures the opposite angle; that is, double the sine of that angle measured in the circle ; therefore the sides of the triangle are to each other as the sines of the opposite angles measured in the same circle, and consequently as the sines of the same angles measured in a circle whose radius is that of the tables. Hence the following proposition of such frequent use in the practice of trigonometry.

In any triangle the sines of the angles are proportional to the sides opposite those same angles.

38. Therefore 1° in any right-angled triangle BAC, the sine of the right-angle A, or radius, is to the hypothenuse BC :: sin C. AB :: sin B: AC.

II. Since in all right-angled triangles the sine of one acute angle C, is the cosine of the other acute angle B, we have sin C=cos B and reciprocally.

A B

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:

2

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Therefore instead of the proportion sin B : sin C::AC: AB, we may always substitute sin B : cos B :: AC: AB. But since (10)...sin B : cos B :: lang B : R, we have

AC: AB : tang B:R:: cot C:R These principles suffice to solve all the cases of the right-angled triangle ABC, when besides the right-angle A, we know any two of the five parts B, C, AB, AC, BC; provided they are not the two angles. In this latter case we can only determine the ratios of the three sides. See the following

TABLE
For the solution of right-angled Triangles.
Given. Required. |

Formulas.
AB, AC

BC BC=NAB? + AC
B AB : BC::R:tang B

с AC: AB ::R: tang C
AB, BC AC AC=NBC2 — AB?

B BC: AB:: R: cos B

С BC: AB::R: sin C
AC, BC AB

AB=N BCP-AC?
B BC : AC::R: sin B

C BC : AC::R: cos C
AC
R: tang B:: AB:

AC
BC cos B: R:: AB: BC
AC, B

AB R: cot B:: AC: AB

BC sin B: R:: AC: BC
AB, C

AC R: cot C;: AB : AC

BC sin C:R:: AB : BC
AC, C AB

R ; tang C:: AC: AB

BC cos C:R:: AC : BC
BC, B

AB R: cos B:: BC : AB

AC R: sin B :: BC: AC
BC, C

AB R sin C:: BC: AB
AC R: cos C::BC : AC

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AB, B

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Examples of the solution of right-angled Triangles. Ex. 1. In the right-angled plane triangle ABC, "given the base AB=140, and the hypothenuse BC=335, required the perpendicular AC, and the angles B and C.

By Euclid (47. 1) and (38) of the present article, or by the preceding table we have

AC=V (BC-AB)=v (335-140")=304. 34 and BC: AB :: R: cos B, whence 000 B = RX AB 140

=.41791 BC

335 and by inspecting the table of natural sines and cosines at the end of this work, we shall find that the angle whose cosine=.417913 to radius 1 is 65° 18'; consequently the other acute angle=90° — 63° 18'-24° 42'.

But in working trigonometrical examples, it is better to avail ourselves of the logarithmic tables, as in most cases they enable us to dispense with the tedious multiplications and divisions which commonly result from employing the natnral sines and tangents of angles and arcs. Thus in the above example, since AC=V BC-AB”, log AC=

1

2

log VBC*—AB®= {log (BC+AB)+log (BC—AB)}= {log 475 +log 195}

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= 10.

=

But log 475=2.676694 log 195=2.290035

2 14.966729 log AC=2.483364 whence AC=304. 34

RX AB Again cos B=

and therefore log cos B=log R +log AB

BC - log BC. But log R..... log AB (140) = 2.146128

12.146128 log BC (335) 2.525045 log cos B

9.621083 Whence the angle B=65° 18', and therefore ? C=24° 42', as before.

Example 2. In the right-angled plane triangle ABC (eame figure) given BC=33.249, and the angle B=17° 12' 51"; required the base AB, the perpendicular AC, and the angles B and C. By (38) or by table

BX
R: cos B :: BC : AB (=cos By BC)
B

R
R: sin B :: BC: AC (=

sin B x BC

BC)

R But log cos B (17° 12' 51")= 9.980097 log BC (33. 249)

= 1.521784

11.501881 log radius

10. log AB

= 1.501881 and AB=31. 76

Again to find AC

= 10.

log AC

= Io.

og sln B (17° 12' 51") = 9.471209
log BC (33. 249)

1.521783

10.992992 log radius

0.992990 whence AC=9.84

EXAMPLE 3. In the right-angled triangle ABC (same figure) given the perpendicular AC=43, and the base AB=55, required the hypothenuse BC, and the angles B and C

By Euclid (47.1), or by table BC=VAB+ AC=552 + 43 - 69.81

RXAC and by (38) or by table, tang B

AB But log radius log AC (43) = 1.633469

:

11.633469 log AB (55) = 1.740363

log tang B 9.893106 whence Z B=38° 1' 8", and LC=519 58' 52"

Instead of subtracting a logarithm it is usual to employ the arithmetical complement of that logarithm, which enables us to add at once the three terms which enter the proportion, without interrupting the operation on account of the subtraction.

The arithmetical complement of a logarithm is found from the tabular logarithm, by subtracting the right-hand figure of this latter from ten, and every one of the remaining figures from 9. Thus the arithmetical complement of log (55) or 1.740363 is 8.259637; and this number added to the logarithms of radius and of the side AC (43) will produce the same result as before, provided we reject ten from the integer part of the aggregate. Thus log raclius

log AC (43) 1.633469 arith. com. log AB (55) 8.259637

19.893106 The same result as before, if we reject 10 from the integer part of the sum.

Instead of rejecting ten from the result on account of the arithmetical complement, I recommend the pupil to incorporate the rejectaneous quantity by prefixing an unit to the arithmetical complement, and writing it in the form 18.259637. Here the figure 1 standing in the place of tens, is equivalent to ten, and the negative sign by which it is surmounted intimates that in the aggregation of the terms this single figure must be considered as negative, and consequently must be subtracted. The operation will stand thus

= 10.

=

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