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The Sine DE of any Arc D B being giveu, to find
DM or B M, ibe Sine of Half the Arc.
DE being given, CE (by the last Prop.) will be
given, and accordingly EB, which is the Difference between the Coline and Radius. Therefore DE, EB, being given, in the Right-angled Triangle DBE, there will be given D B, whose Half D M is the Sine of the Arc DL= the Arc B D.
Ibe Sine B M of the Arc B L being given, to find
the Sine of double ihai Arc.
HE Sine B M being given, there will be given by
Prop. 2.) the Cofine C M. But the Triangles CBM, DBE, are equiangular, because the Angles at E and M are Right Angles, and the Angle at B common: Wherefore (by 4. 6.) we have ČB:CM:: (BD, or) 2 B M:DE. Whence, since the three first Terms of this Analogy are given, the fourth also, which is the Sine of the Arc DB, will be known.
Coroll. Hence CB:2 CM::BD: 2 DE ; that is,
the Radius is to double the Coline of one Half of the Arc D B, as the Subtense of the Arc D B is to the Subt«nse of double that Arc. Allo CB: 2CM::(2 B M:2 DE::) BM:DE:: CIB : CM. Wherefore the Sine of an Arc, and the Sine of its Double, being given, the Cofine of the Arc itself is given.
to find F I, the Sine of the Sum, as likewise
ET the Radius CD be drawn, and then CO is
the Cofine of the Arc FD, which accordingly is given, and draw O P thro' O parallel to DK; alfo let OM, G E, be drawn parallel to CB: Then, because the Triangles C DK, COP,CHI, FOH, FOM, are equiangular; in the first Place CD:DK: CO:0 P, which, confequently, is known. Also, we have CD:CK::F0: FM; and lo, likewise, this shall be known. But because FO-EO, then will FM=MG=ON; and fo O P + F M=FI=Sine of the Sum of the Arcs : And OP-FM; that is, OP-ONEEL=Sine of the Difference of the Arcs ; which were to be found. Coroll. Because the Differences of the Arcs BE, BD,
BF, are equal, the Arc B D shall be an Arithmerical Mean between the Arcs B E, B F.
double the Cofine of the mean Arc, as the Sine
of the Extremes. FOR we have CD;CK::FO:FM; whence,
by doubling the Confequents, CD:2 CK:: FO:(2 FM, or to FG, which is the Difference of the Sines EL, FI. W. W. D. Coroll. If the Arc B D be 60 Degrees, the Difference
of the Sines FI, EL, shall be equal to the Sine FO, of the Difference. For, in this Cafe, CK is the Sine of 30 Degrees; the double whereof is equal to the Radius; and fo, fince CD=2C K, we shall have
FO=F G. And, consequently if the two Arcs BE,
Coroll. 2. Hence, if the Sines of all Arcs diftant from
one another by a given Interval, be given, from
of 57 Degrees + the Sine of 3 Degrees, and so on.
of a Quadrant to any Part of a Quadrant, distant
Coine of 30 Degrees, or to double the Sine of 60 De-
33 Degrees. But, in this case, the Radius is
See Fig. of the DEFINITIONS. # Let B D be an Arc of 30 Degrees :
Rad. Tan. Cofine Sine
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 Be
ginning 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 Cofine of 2': : Sine I': Difference of the Sines of 1' and 34:: Sime 2: Difference of the Sines of o' and 4'; that is, to the Sine of 4!. And so, the Sines of the four firft Minutes being given, we may thereby find the Sines of the others to 8', and from therce to 16, and so on.
Arcs are nearly to one another, in a Ratio of
FOR, because the Triangles CED, CB G, are
equiangular, CE:CB::ED:CG. But as the Point E approaches B, E B will vanish in respect of the Arc BD; whence C E will become nearly equal to CB, and so E D will be also nearly equal to BG. If
I E B be less than them - Part of the Radius,
10,000,000 then the Difference between the Sine and the Tangent
I will be also less than the
-Part of the Tan.
Coroll. 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 shall be nearly equal to its Sine: And so, in very small Arcs, it fhall be, as Arc is to Arc, fo is Sine to Sine.
PROBLEM. To find the Sine of the Arc of one Minute. T! HE Side of a Hexagon inscribed in a Circle, that
is, the Subtense of 60 Degrees, is equal to the Ra. dius (by Coroll. 15th of the 4tb): and fo the Half of the Radius shall 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.) Allo the Sine or 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'; and so on, until twelve Bisections being made, we come to an Arc of 522, 44", 034, 45', whose Cofine is nearly equal to the Radius ; in which Cale (as is manifest from Prop. 7.) Arcs are proportional to their Sines: And so, as the Arcos 52", 44', 0,4, 45', is to an Arc of one Minute, lo hall 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 Cofine of two Minutes will be had.
THEOREM. if the Angle B A C, being in the Periphery of a
Circle, be biseEted by the Right Line AD, and if AC be produced until D E=AD meets it in E; then all C E=A B.
IN N 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 E=DAC by 5. 1.)=DA B, and D CODB: Wherefore the Triangles BAD and C E D are congruous, and fo C E is equal to A B. W. W. D.