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because CD=c-b: and, by the lemma, this becomes bc: (s —b) (s — c) : : R2 : sin2 A.

Cor. Hence, taking R = 1, dividing the product of the means by the first extreme, and extracting the square root, we find ; and it is plain, that we should find

sin A = ✓ (s—b) (s

bc

c)

in a similar manner, sin B =

| | (s — a) (s—c), and sin‡C =

ac

=

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THE rectangle under two sides of a triangle is to the rectangle under half the perimeter and its excess above the third side, as the square of the radius, to the square of the cosine of half the angle contained by the two sides.

Let ABC be a triangle, and let s = } (a + b + c): then bc: s (sa): R: cos A.

E

B

G

Produce the less side CA, through A, making AD equal to AB; join BD; draw AE, CF perpendicular, and AG parallel to BD. Then BD (I. 47. cor. 5.) is bisected in E; and the angle BAC being (1. 32.) equal to the two equal angles D and ABD, each of them is equal to half the angle BAC, that is, half the angle A in the triangle ABC: ́and (I. 29.) F GAC is equal to D. Now, it would be shown, as in the preceding proposition, that the rectangle under half the sum and half the difference of DC, CB is equal to the rectangle BE.EF. But (APP. III. 1.) AB: BE::R: cos ABE, or cos&A; and AC: AG, or EF:: R: cosGAC, or cos A; whence (V. Sup. 15.) AC.AB: BE.EF, or (DC +CB). (DC — CB) :: R: cos2A; or be: (a+b+c). § (b + c − a) :: R2 : cos2)A, because DC=b+c. Hence, by the lemma,

bcs (sa) :: R2: cos2 A.

Cor. 1. Hence we find, as in the corollary to the preceding ; and it would be proved

proposition, that cos}A = s (s—a)

in a similar manner, that

s

/ ཙ (8

bc

cos ¡B = √3 (s—b), and cos‡C = √ 8 (s — c) ̧

ac

ab

Cor. 2. From the sixth corollary to the definitions of this book, it is plain, that when the radius is unity, if the sine of an

arc be divided by its cosine, the quotient is its tangent. Hence, by dividing the expression for the sine of A in the corollary to the preceding proposition, by the value of its cosine in the last corol

lary, we obtain, tan § A = √(s—b) (s—c);

s (s—a)

and we should obviously find in a similar manner, that tanB =

and tan C = · √ (s—a) (s—b).

s (s—c)

(s—a) (s—c), s(s—b)

Cor. 3. By dividing the values of tan! B, tan C, in the preceding corollary, each by that of tan A, we obtain tan B s-a tan C s-a

tan A S

=

b

and

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PROP. VIII. PROB.

GIVEN the radius of a circle, and the cosine of an arc, less than a quadrant; to compute the cosine of half the arc.

Let CD, the cosine of the arc AEB, less than a quadrant, be given; it is required to compute the cosine of its half.

B

E

Draw the chord AB, and perpendicular to it draw CFE; draw also FG parallel to BD: then (III. 3. and App. III. def. 5. and 7.) AF or FB, is the sine, and CF the cosine of AE, the half of AEB. Also (VI. 2.) DG is equal to GA, since BF is equal to FA; and DA = 2DG: to each of these add 2CD; then CA + CD = 2CG, and consequently CG (CA + CD);

1

DGA

or, if the radius be taken as unity, and the arc AEB be denoted by A, CG = (1+cos A). Again, in the similar triangles ACF, CFG, AC: CF :: CF: CG; whence (VI. 17.) CF2 AC.CG; that is, cos A = (1 + cos A). Hence, to compute cos A, add 1 to cos A, take half the sum, and extract the square root.

PROP. IX. THEOR.

IF A and B be any two arcs, R: cos B: sin A : sin (AB) + sin (A + B).

Make AC equal to A, and BC, CD each equal to B; draw

D

BE, CF, DH, the sines of AB, AC, AD; join BD, and through the centre K draw KNC: then KN is evidently the cosine of BC. Draw also BML, NMG parallel to AK, DH. Now, in the similar triangles DLB, NMB, since DB is double of NB, DL (VI. 4.) is double of NM; to DL add LH, BE, and to 2NM add what is equivalent, 2MG; then DH+ BE = 2NG; wherefore NG is equal to half the sum of BE and DH, that is, tosin (AB) + sin (A+B). Again, in the similar triangles CFK, NGK, we have (VI. 4. and alternately,) CK: NK:: CF NG; that is, R: cosB: sinA : sin(A-B) + sin(A + B).

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=

L

B

M

IC H GF E

A

Cor. 1. Hence, if R 1, we have, by doubling the second and fourth terms, and by taking the products of the extremes and means, sin (A — B) + sin (A + B) = 2 cos B sin A; whence, sin (A+B) 2 cos Bsin A-sin(AB).

=

Cor. 2. If B A, the last expression becomes simply sin 2A = 2 sin A cos A.

PROP. X. PROB.

To investigate a method of computing tables of sines, tangents, and secants.

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The chord of 60o (IV. 15. cor.) is equal to the radius 1, and therefore (APP. III.defs. cor. 1.) the sine of 30° is 0·5; and (by the fourth of those corollaries,) cos 30o = √ (1 sin 300); whence cos 300 may be computed. Then, (APP. III. 8.) from the cosine of 30°, we can compute that of 150; from the cosine of 15°, that of 7o 30'; and thus we may proceed as far as we please, by continual bisections of the arc. By eleven such operations, we should find the cosine of 52"-734375: and its sine would then be computed by means of the fourth corollary to the definitions of this book. Now (APP. I. 40. schol.) a polygon may be inscribed in a circle with so many sides, that its perimeter will differ from the circumference by a quantity as small as we please; and hence it follows, that a very small arc and its sine will have to one another very nearly the ratio of equality, and consequently that the sines of very small arcs are very nearly as the arcs themselves. Hence, therefore, as 52".734375 to 60", so is the sine of 52"-734375 to the sine of 60", or one minute.

The sine of 1' being thus obtained, take B, in the last formula in the first corollary to the preceding proposition, constantly equal to l', and A successively equal to 1', 2', 3', &c., and there will be

obtained the following expressions, by means of which the sines of the succeeding minutes may be computed :—

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By this method the sines of all the successive minutes of the quadrant, may be computed: and they comprehend also the cosines; sin l', sin 2', &c., being cos 89° 59′, cos 89° 58', &c. Then, (APP. III. defs. cor. 5. and 6.) the tangents of 1', 2', 3', &c., will be found by dividing the sine of each by its cosine, and their secants by dividing 1 by their cosines.

The numbers which are computed in the manner now explained, are called natural sines, tangents, &c., to distinguish them from their logarithms, which are called logarithmic sines, tangents, &c.* By the nature of logarithms, the sum of the logarithms of two numbers is equal to the logarithm of their product; and, conversely, if the logarithm of one number be taken from the logarithm of another, the remainder is the logarithm of the quotient obtained by dividing the latter number by the former. Hence, the logarithmic sines, &c., are generally employed in practice; as by their means multiplication and division are exchanged for the easier operations of addition and subtraction. In these tables, the radius is taken for convenience, as 10,000,000,000; and therefore all the sines, tangents, &c., to the radius 1, are multiplied by this number, and the logarithm of the radius is 10.

Schol. 1. The principles that have been thus established, enable us to solve all the elementary cases of plane trigonometry. Now, of the three sides and three angles of a triangle, some three, and those not the three angles, must be given to determine the triangle;† and the resolution of plane triangles may therefore be reduced to the three following cases :

I. When a side and the opposite angle, and either another side, or another angle, are given:

* In trigonometrical tables, the sines are generally placed, for convenience, in one column, and the cosines in the adjoining one, or at least in the same page. Now it is plain that the sine is one leg, and the cosine another, of a right-angled triangle, of which the radius is the hypotenuse; and hence the radius, sine, and cosine, may be regarded merely as numbers proportional to the three sides of a right-angled triangle, having one of its angles equal to that to which the sine and cosine belong: and the same is the case respecting the radius, tangent, and secant. Hence trigonometrical tables may be regarded as a register, exhibiting the ratios of the sides of a great number of right-angled triangles, in connexion with the magnitudes of their oblique angles.

It may be remarked, that while the method of computation explained above will show the student how trigonometrical tables may be calculated, there are other methods much easier, particularly by means of series. The investigation

of such methods, however, would be unsuitable here.

† See the note to the 26th proposition of the first book.

II. When two sides and the contained angle are given:
III. When the three sides are given.

The FIRST CASE is solved on the principle (APP. III. 2.) that the sides are proportional to the sines of the opposite angles. Thus, if A, B, a be given, add A and B together, and take the sum from 1800: the remainder (I. 32.) is C. Then b and c will be found by the following analogies: sinA : sinB::a: b; and sin A: sinC:: a: c.

If, again, a, b, A be given, we compute B by the analogy, ab sinA sin B. Then C is found by subtracting the sum of A and B from 180°, and c by the analogy, sinA : sinC :: a : c.

:

When in this case two unequal sides, and the angle opposite to the less, are given, the angle opposite to the greater (APP. III. defs. cor. 8.) may be either that which is found in the table of sines, or its supplement; and thus the problem admits of two solutions. To illustrate this, suppose the angle B, the side BA, and the side AC, less than BA, to be given: then, AB and the angle B being made of the given magnitude, if an arc described from A as centre, with the other given side as radius, cut the side opposite to the angle A, in one point C, it will cut it also in another C'; therefore each of the triangles ABC, ABC' answers equally to the data, and (I. 13.) the angle ACB is the supplement of ACC', or (I. 5.) its equal AC'C.

B

If in this case one of the angles be a right angle, the solution is rather easier; as, by the second corollary to the definitions, the sine of that angle is equal to the radius. The same conclusion may also be obtained by means of the first proposition of this book.*

* To exemplify the solution of this case, let a = 13 yards, b = 15 yards, and 53° 8',-to resolve the triangle; and the operation by means of logarithms will be as follows:

A

-:

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In these operations, to find the fourth term, the second and third terms are added together, and the first is taken from the sum. This may be done very easily, in a single operation, by adding the figures of the second and third terms successively to what remains after taking the right hand figure of the first term from 10, and each of the rest from 9, and rejecting 10 from the final result. Thus, in the first operation, we have 8 and 1 are 9, and 7 are 16; then 1 and 9 are 10 and 5 are 15, &c. It is still easier, however, when the quantity to be subtracted is a sine, to use the cosecant, and when it is a cosine, to use the secant, each diminished by 10, and then to add all the terms together. The reason of this is evident from the nature of logarithms, and from the fifth corollary to the definitions of this book. In like manner, when the number to be subtracted is a tangent, or cotangent, we may use in the former case a cotangent—in the latter a tangent, subtracting in each case 10, either at first or afterwards.

This example evidently belongs to the doubtful case; and hence we have two values for each of the quantities B, C, and c; and therefore two analo, "gies are requisite for finding the values of c.

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