the arcs, of their sum, and their difference, be what they may. This might be rendered evident by a suitable modification of the diagram; and will still farther appear in the fourth chapter of this work.

PROP. X.

10. As the difference or sum of the square of the radius and the rectangle under the tangents of two arcs, is to the square of the radius; so is the sum or difference of their tangents, to the tangent of the sum or difference of the arcs.

Let AB, AD, be the two arcs; AT, AR, their tangents; also let вs in the first figure be the tangent of the sum of those arcs, and BS in the second figure the tangent of their difference: and from Tthe point R let Rн, in both figures, be drawn parallel to BS, or perpendicular to the radius CB.

B

H

R A

R T

Then, because of the similar triangles TAC, THR, we shall have,

TC: CA:: TR: RH, whence TC. RH = CA. TR,

TC: TA :: TR: TH, whence TC. TH=TA. TR. Each of the last equal rectangles being taken from the square of rc, there will remain, TC2 TC. TH, or TC. (TC TH), or TC.CH = TC2 -TA. TR. Now,

TC. CH (or TC2

TA.

TR): TC. RH (or CA. TR) :: CHI : RH:: CB (or CA): BS :: CA2: CA. BS; whence, alternando, TC2 - TA .TR: CA2:: CA.TR: CA. BS: TR: BS. Of this proportion, the first term TC2. TA. TR, because TC2 CA2 + AT2, may also be expressed by CA 4 AT2 -TA. TR, or by CA2 + AT2 TA (TAAR), or lastly, by CAAT. AC; whence the proposition is manifest.

PROP. XI.

11. As the sum of the sines of two unequal arcs, is to their difference, so is the tangent of half the sum of those two arcs, to the tangent of half their difference.

Let AE and AB be two unequal arcs, of which EK and BG, are the sines; and let Eк be produced to cut the circle in D, and BI be drawn parallel to the diameter A AA'. Draw ID, IE, from the centre 1 with the radius CA of the assumed circle, describe an arc dbe, and through b draw Loм parallel

D

B N

G K

L

M

BG

BCD

to DE. Then, it is evident that NE KE + BG, is the sum of the sines of AE and AB, and ND = KE KD KN, is their difference. Also, since BIE (at the circumference) BCE (at the centre); and BID= (Euc. iii. 20); bм is equal to the tangent of (AE+AB), and b to the tangent of (AE-AB), that is, of (AD - AB.) But, by reason of the parallels DE, LM, we have EN: DN :: bм: bL; which is evidently the theorem enunciated above.

Cor. The sum of the cosines of two arcs, is to their difference, as the cotangent of half the sum of those two arcs, is to the cotangent of half their difference.

For, the cosines being the sines of the complements, it follows from the proposition that the sum of the cosines, is to their difference, as the tangent of half the sum of the complements, is to the tangent of half their difference. But half the sum of the complements of two arcs is the complement of half the sum of those two arcs, and half the difference of the complements is the same as the complement of half the difference; whence the truth of the corollary..

PROP. XII.

12. Of any three equidifferent arcs, it will be, as

radius, to the cosine of their common difference, so is the sine of the mean arc, to half the sum of the sines of the extremes; and, as radius, to the sine of the common difference, so is the cosine of the mean arc to half the difference of the sines of the two extremes.

Let AD', AB, AD, (in the figure to prop. 8, of this chapter), be the three equidifferent arcs. Then DF

DF', is the sine of their common difference, and cr its cosine. Also FG, being an arithmetical mean between the sines DK, D'K', of the two extreme arcs, is equal to half their sum, and DE equal to half their difference. By reason of the similar triangles CBH, cfe,

DFE,

we have, CB: CF :: BH:FG,

and CB: DF:: CH: DE;

which are the analogies in the proposition.

Cor. 1. From the preceding proportions, we have

CB

Cor. 2. Hence, if the mean arc AB be one of 60°, its cosine CH, will (prop. 5) be equal to CB, and DK — d'K'′ DF: consequently DK will in that case equal DF + D'K'. From this conjointly with the preceding corollary result these two theorems:

(A). If the sine of the mean of three equidifferent arcs (radius being unity) be multiplied into twice the cosine of the common difference, and the sine of either extreme be deducted from the product, the remainder will be the sine of the other extreme.

(B). The sine of any arc above 60°, is equal to the sine of another arc as much below 60°, together with the sine of its excess above 60°.

Remark. From this latter corollary, the sines below 60° being known, those of arcs above 60° are determináble by addition only.

Thus, sin 60° 1′ = sin 59° 59′+ sin l',

13. In any right angled triangle, the hypothenuse is to one of the legs, as the radius to the sine of the angle opposite to that leg; and one of the legs is to the other, as the radius to the tangent of the angle opposite to the latter.

Let ABC be a triangle, right-angled at B, and let AR on the leg AB, be the radius of the tables. With centre A and radius AR, describe an arc to cut the hypothenuse in D, and draw DH, TR, рerpendicular to AB. Then DH is the sine, TR the tangent, and AT the secant, of the AHR arc DR, or angle A: and the similar triangles ABC, ART, AHD, give

AC: CB:: AD: DH:: rad: sin a;

and AB: BC :: AR: RT :: rad: tan A.

Cor. To the hypothenuse as a radius, each leg is the sine of its opposite angle; and to one of the legs as a radius, the other leg is the tangent of its opposite angle, and the hypothenuse is the secant of the same angle.

PROP. XIV.

14. In any plane triangle, as one of the sides, is to another, so is the sine of the angle opposite to the former, to the sine of the angle opposite to the latter. Let ABC be a plane triangle,

in which it is required to determine the relation which subsists between the sides AC and BC.

'c

With the angular point A as a Gƒ Ad H

B

centre, and the distance ac, equal to the radius of the tables, describe the semicircle GecH. From c draw cb parallel to CB, and cd perpendicular to AB: draw also CA parallel to CB, and from the point of intersection e (of that line with the circle) demit the perpendicular ef on BA produced. Then cd is the sine of the angle CAB, to the radius AC, and ef is the sine of the angle eAG = angle CBA, to the same radius.

But AC: BC: Ac: bc, because of the parallels BC and bc, :: Ae: bc, because Ae = AC,

Aef, bcd.

:: ef: cd, because of the similar triangles

That is, AC: BC :: sin B: sin a.

In a similar manner it may be shewn that

AC: AB:: sin B: sin c, and AB: BC: sin c: sin A. And by drawing a figure for each case, it will be seen that the circumstance of any one of the angles being obtuse will make no difference in the demonstration.

Otherwise.

From c the vertical angle of the triangle let fall the perpendicular CD upon AB, or AB produced, according as the angles A and B are both acute, or one obtuse.

A

Then (prop. 13) AC: CD:: rad: sin a; also CD: CB:: sin B: rad;

.. ex æquo pertur. AC: CB :: sin B : sin A.

PROP. XV.

15. In any plane triangle it will be, as the sum of the sides about the vertical angle, is to their difference, so is the tangent of half the sum of the angles at the base, to the tangent of half their difference.

By the preceding prop. AC: BC :: sin B: sin A, .. by comp, and div. AC + BC: AC - BC :: sin B + sin a

:sin B

sin A.