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(60° + —— A) for the three values of x. Now every sine is less than radius, hence 2 √p must be greater than 34, or

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1

27

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From which it follows that any equation of the third degree falling under the irreducible case, may be solved by this method. 36. Call R the tabular radius, and we shall haveRx3q√3

the tabular sine of the arc A, or sin A=

for

2p/P Rx393. But A being

2P P

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

3

3

given; and consequently the three roots of the proposed equation (reducing these sines to those which correspond to the radius 2 √ — p) are

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R

EXAMPLE 1. Let the proposed equation be x3-3x+1=0 Here p 3, q=1; whence sin AR, and A=30°. Therefore the three values of x are

2 sin 10°

=-0.3472964

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=1.5320888

R

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

= 0,394931

and

X=

2 Sin 20°

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R 3
-2 sin 80°

-1.137158

Ex. III. Let the equation be x3-5 x+3=0. This gives p=5,

R

q=8, sin A=

2

35 53

5

and log sin A=log R+

log 3-log 2

2

3

2

log 59.843318=log sin (44° 11′ 52"); therefore A= 449 11′ 52" and the three values of the unknown quantity are 25 sin (14° 43′ 57′′)

X=

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5 sin (74° 43′ 57′′)
R 3

Performing the calculations indicated, we obtain

x=o. 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

sin A, sin (60a—— A), —sin (60° +/- A)

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 x3-12 x=15; required x

Answer, x=3.971963, or-1.577032, or-2.394930 2. Given x3-27 x=36; required

Answer, x=5.765722, or-4.320685, or-1.445038

SOLUTION OF THE CASES OF PLANE
TRIANGLES.

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 conséquently 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 I in any right-angled triangle C 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

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: tang 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

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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 (BC-AB*)=√ (335-1402) 304. 34 and BC: AB :: R: cos B, whence

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RX AB 140

BC 335

=.41791

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°— 65° 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-√ BC-AB2, log AC=

log ✅BC*—-AB3 — — { log (BC+AB)+log (BC—AB) } =

2

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

EXAMPLE 2. In the right-angled plane triangle ABC (came 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.

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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=√AB2+AC2=√552 + 432 -69.81

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

RX AC
AB

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log AB (55)

= 1.740363

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log tang B = 9.893106

whence B 38° 1′ 8′′, and ≤ C=51° 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

= 10.

log radius 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

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