CHI, FOH, FOM, are equiangular; in the first place, CD: DK :: CO: OP, which, consequently, is known. Also we have CD: CK:: FO: FM; and so, likewise, this will be known. But because FO=EO, then will FM=MG= ON; and so OP+FM=FI=sine of the sum of the arcs; And OP-FM: that is, OP-ON-EL-sine of the difference of the arcs: which were to be found.

COR. Because the differences of the arcs BE, BD, BF, are equal, the arc BD is an arithmetical mean between the ares BE, BF.

PROP. VI. THEOR.

THE same things being supposed, the radius is to double the cosine of the mean arc, as the sine of the difference is to the difference of the sines of the extremes.

For we have CD: CK::FO: FM; whence by doubling Fig. 30. the consequents, CD: 2 CK:: FO: (2 FM, or) to FG, which is the difference of the sines EL, FI. Q.E.D.

COR. If the arc BD be 60 degrees, the difference of the sines FI, EL, will be equal to the sine FO of the difference. For, in this case, CK is the sine of 30 degrees; the double whereof is equal to the radius (by 15. 4.); and so, since CD 2 CK, we shall have FO=FG. And, consequently, if the two arcs BE, BF, are equidistant from the are of 60 degrees, the difference of the sines will be equal to the sine of the difference FD.

COR. 2. Hence, if the sines of all arcs distant from one another by a given interval, be given, from the beginning of a quadrant to 60 degrees, the other sines may be found by one addition only. For the sine of 61 degrees the sine of 59 degrees+the sine of 1 degree; and the sine of 62 degrees the sine of 58 degrees+the sine of 2 degrees. Also, the sine of 63 degrees the sine of 57 degrees + the sine of 3 degrees, and so on.

COR. 3. If the sines of all arcs, from the beginning of a quadrant, to any part of a quadrant, distant from each other by a given interval, be given, thence we may find the sines of all arcs to the double of that part. For example: let all the sines to 15 degrees be given; then, by the preceding analogy, all the sines to 30 degrees may be found. For the

radius is to double the cosine of 15 degrees, as the sine of 1 degree is to the difference of the sines of 14 degrees, and 16 degrees: So, also, is the sine of 3 degrees to the difference between the sines of 12 and 18 degrees; and so on continually, until you come to the sine of 30 degrees.

. After the same manner, as the radius is to double the cosine of 30 degrees, or to double the sine of 60 degrees, so is the sine of 1 degree to the difference of the sines of 29 and 31 degrees :: sine 2 degrees to the difference of the sines of 28 and 32 degrees :: sine 3 degrees to the difference of the sine of 27 and 33 degrees. But, in this case, the radius is to double the cosine of 30 degrees, as 1 to 3.

For (see the figure for Prop. 15, Book IV. of the Elements) the angle BGC=60 degrees, as the arc BC, its measure, is a sixth part of the whole circumference; and the straight line BC=R. Hence it is evident that the sine of 30 degrees is equal to half the radius; and therefore, by Prop. 2.

=√/3R2=R×√√/3. Consequently, radius is to double the cosine of 30°::R: R× √3::1 : √/3.

And, accordingly, if the sines of the distances from the arc of 30 degrees be multiplied by 3, the differences of the sines will be had.

So, likewise, may the sines of the minutes in the beginning of the quadrant be found, by having the sine and cosines of one and two minutes, given. For, as the radius is to double the cosine of 2' :: sine l': difference of the sines of 1' and 3 sine 2: difference of the sines of O' and 4'; that is, to the sine of 4'. And so, the sines of the four first minutes being given, we may thereby find the sines of the others to 8', and from thence to 16', and so on.

Fig. 31.

PROP. VII. THEOR.

IN small arcs, the sines and tangents of the same arcs are nearly to one another, in a ratio of equality.

For, because the triangles CED, CBG, are equiangular, CE: CB:: ED: BG. But as the point E approaches B, EB will vanish in respect of the arc BD; whence CE will

become nearly equal to CB, and so ED will be also nearly equal to BG. If EB be less than the

part of

10000000 the radius, then the difference between the sine and the 1 tangent will be also less than the 10000000

tangent.

of the

COR. Since any arc is less than the tangent, and greater than its sine, and the sine and tangent of a very small arc are nearly equal; it follows, that the arc will be nearly equal to its sine: And so, in very small arcs, it will be, as arc is to arc, so is sine to sine.

PROP. VIII. PROB.

To find the sine of the arc of one minute.

The side of a hexagon inscribed in a circle, that is, the subtense of 60 degrees, is equal to the radius (by Coroll. 15th of the 4th); and so the half of the radius will be the sine of the arc of 30 degrees. Wherefore the sine of the arc of 30 degrees being given, the sine of the arc of 15 degrees may be found (by Prop. 3). Also the sine of the arc of 15 degrees being given (by the same Prop.) we may have the sine of 7 degrees 30 minutes. So, likewise, can we find the sine of the half of this, viz. 3 degrees 45 minutes; and so on, until 12 bisections being made, we come to an arc of 522, 443, 034, 45, whose cosine is nearly equal to the radius; in which case (as is manifest from Prop. 7.) arcs are proportional to their sines: and so, as the arc of 52, 443, 03*, 455, is to an arc of one minute, so will the sine before found be to the sine of an arc of one minute, which therefore will be given. And when the sine of one minute is found, then (by Prop. 2. and 4.) the sine and cosine of two minutes will be had.

PROP. IX. THEOR.

If the angle BAC, being in the periphery of a cir- Fig. 32. cle, be bisected by the right line AD, and if AC be produced until DE AD meets it in E; then will CE AB.

In the quadrilateral figure ABDC (by 22. 3.) the angles

B and DCA are equal to two right angles DCE + DCA (by 13. 1.) whence the angle B-DCE. But, likewise, the angle EDAC (by 5. 1.)=DAB, and DC=DB: Wherefore the triangles BAD and CED are congruous, and so CE is equal to AB. Q. E. D.

PROP. X. THEOR.

Fig. 35. LET the arcs AB, BC, CD, DE, EF, &c. be equal; and let the subtenses of the arcs, AB, AC, AD, AE, &c. be drawn; then will AB: AC:: AC: AB +AD::AD: AC+AE::AE: AD+AF::AF: AE+AG.

Let AD be produced to H, AE to I, AF to K, and AG to L, so that the triangles ACH, ADI, AEK, AFL, be isosceles ones: Then because the angle BAD is bisected, we shall have DHAB (by the last Prop.): so likewise EI AC, FK AD, also GLAE.

But the isosceles triangles ABC, ACH, ADI, AEK, AFL, because of the equal angles at the bases, are equiangular: Wherefore it will be, as AB: AC :: AC: (AH=) ÅB+AD :: AD : (AI=) AC+AE;:AE:(AK=) AD+ AF:: AF: (AL=) AE+AG. Q. E. D.

COR. 1. Because AB is to AC, as radius is to double the cosine of the arc AB, (by Coroll. Prop. 4.) it will also be, as radius is to double the cosine of the arc AB, so is AB: AC: AC: AB+ AD:: AD: AC+ AE :: AE : 1⁄2 AD + AF, &c. Now let each of the arcs AB, BC, CD, &c. be 2'; then will AB be the sine of one minute, AC the sine of 2 minutes, AD the sine of 3 minutes, AE the sine of 4 minutes, &c. Whence, if the sines of one and two minutes be given, we may easily find all the other sines in the following manner.

Let the cosine of the arc of one minute, that is, the sine of the arc of 89 deg. 59', be called Q; and make the following analogies; R 2 Q:: Sin. 2: S. 1' + S. 3'. Wherefore the sine of 3 minutes will be given. Also, R.: 2 Q :: S. 3': S. 2' + S. 4'. Wherefore the S. 4' is given. And R.: 2Q:: S. 4': S. 3'+ S. 5'; and so the sine of 5' will be had.

Likewise, R.: 2 Q : : S. 5′ : S. 4' + S. 6'; and so we shall have the sine of 6'. And in like manner, the sines of every minute of the quadrant will be given. And because the radius, or the first term of the analogy, is unity, the ope

rations will be with great ease and expedition calculated by multiplication, and contracted by addition. When the sines are found to 60 degrees, all the other sines may be had by addition only, by Cor. 1. Prop. 6.

The sines being given, the tangents and secants may be found from the following analogies (see Figure 3, for the definitions); because the triangles BDC, BAE, BHK, are equiangular, we have

BD: DC :: BA: AE; that is, Cos.: S.:: R.: T.

AE: BA:: BH: HK; that is, T.: R.:: R.: Cot. BD: BC:: BA: BE; that is, Cos.: R.:: R.: Secant. CD: BC:: BH; BK; that is, S.: R.:: R.: Cosec.